Hình thang này có đáy nhỏ bằng \(2\), đáy lớn bằng \(4\) và chiều cao bằng \(6-1=5\).

Do đó diện tích là: \(S=\dfrac{(2+4)\times 5}{2}=15\).

Ta có \[ \begin{aligned} I=\ & \int_0^1(x^2-2x+2)\mathrm{\,d}x = \left(\frac{1}{3}x^3-x^2+2x\right)\Big|_0^1\\ =\ & \left(\dfrac{1}{3}\cdot 1^3-1^2+2\cdot 1\right)- \left(\dfrac{1}{3}\cdot 0^3- 0^2+2\cdot 0\right)=\dfrac{4}{3}. \end{aligned} \]

Ta có \[ \begin{aligned} I=& \int_0^{\frac{\pi}{2}}\left(2\sin x-x\right)\mathrm{\,d}x = \left(-2\cos x-\dfrac{1}{2}x^2\right)\Big|_0^{\frac{\pi}{2}}\\ =& \left(-2\cos\dfrac{\pi}{2}-\dfrac{1}{2}\cdot \left(\frac{\pi}{2}\right)^2\right)- \left(-2\cos 0-\dfrac{1}{2}\cdot 0^2\right)\\ =& -\dfrac{{\pi}^2}{8}-(-2\cdot 1)=-\dfrac{{\pi}^2}{8}+2. \end{aligned} \]

Ta có \[ \begin{aligned} I=& \displaystyle\int_2^14x^2\mathrm{\,d}x=\dfrac{4}{3}x^3\Big|_2^1= \dfrac{4}{3}\cdot 1^3-\dfrac{4}{3}\cdot 2^3 = -\dfrac{28}{3}. \end{aligned} \]

Ta có \[ \begin{aligned} I=\ & \displaystyle\int_1^\mathrm{e}\left(\dfrac{2}{x}+\dfrac{1}{x^2}\right)\mathrm{\,d}x=\left(2\ln x-\dfrac{1}{x}\right)\Big|_1^e\\ =\ & \left(2\ln e-\dfrac{1}{e}\right)-\left(2\ln1-\dfrac{1}{1}\right)\\ =\ & 2-\dfrac{1}{e}+1=3-\dfrac{1}{e}. \end{aligned} \]

\(I=\displaystyle\int_0^1x\left(x^2+1\right)^5\mathrm{\,d}x=\displaystyle\int_0^1\left(x^2+1\right)^5x\mathrm{\,d}x\)
Đặt \(t=x^2+1\Rightarrow dt=2xdx\Rightarrow xdx=\dfrac{1}{2}dt.\)
Đổi cận: \(x=0\Rightarrow t=1\), \(x=1\Rightarrow t=2\).
\[I=\int_1^2t^5\dfrac{1}{2}dt=\dfrac{1}{2}\cdot \dfrac{1}{6}t^6\Big|_1^2=\dfrac{21}{4}.\]

\(I=\displaystyle\int_1^\mathrm{e}\dfrac{2\ln x+1}{x}\mathrm{\,d}x=\displaystyle\int_1^\mathrm{e}\left(2\ln x+1\right)\dfrac{1}{x}\mathrm{\,d}x\)
Đặt \(t=2\ln x+1\Rightarrow dt=2\cdot\dfrac{1}{x}dx\Rightarrow \dfrac{1}{x}dx=\dfrac{1}{2}dt.\)
Khi \(x=1\Rightarrow t=1\), khi \(x=e\Rightarrow t=3\). \[I=\int_1^3 t\cdot \dfrac{1}{2}dt=\dfrac{1}{2}\cdot\dfrac{1}{2}t^2\Big|_1^3=2.\]

\(I=\displaystyle\int_0^{\frac{\pi}{2}}\sin ^3x\cdot \cos x\mathrm{\,d}x\).
Đặt \(t=\sin x\Rightarrow dt=\cos xdx\).
Khi \(x=0\Rightarrow t=0\), khi \(x=\dfrac{\pi}{2}\Rightarrow t=1\). \[I=\int_0^1t^3dt=\dfrac{t^4}{4}\Big|_0^1=\dfrac{1}{4}.\]

Ta có \(I=\displaystyle\int_0^{\frac{\pi}{4}}\dfrac{\sqrt{3\tan x+1}}{\cos^2x}\mathrm{\,d}x=\displaystyle\int_0^{\frac{\pi}{4}}\sqrt{3\tan x+1}\cdot\dfrac{1}{\cos^2x}\mathrm{\,d}x\).
Đặt \(t=\sqrt{3\tan x+1}\Rightarrow t^2=3\tan x+1\Rightarrow 2tdt=\dfrac{3}{\cos^2x}dx\Rightarrow \dfrac{1}{\cos^2x}dx=\dfrac{2}{3}tdt\).
Khi \(x=0\Rightarrow t=1\), khi \(x=\dfrac{\pi}{4}\Rightarrow t=2\). \[I=\int_1^2t\cdot\dfrac{2}{3}tdt=\dfrac{2}{3}\int_1^2t^2dt=\dfrac{2}{3}\cdot \dfrac{t^3}{3}\Big|_1^2=\dfrac{14}{9}.\]

Đặt \(t=2x-1\Rightarrow dt=2dx\Rightarrow dx=\dfrac{1}{2}dt\).
Khi \(x=1\Rightarrow t=1\), khi \(x=3\Rightarrow t=5\). \[I=\int_1^5f(t)\dfrac{1}{2}dt=\dfrac{1}{2}\int_1^5f(t)dt=\dfrac{1}{2}\cdot 8=4.\]

Đặt \(\left\{\begin{aligned}&u=2x\\ &dv=\cos xdx\end{aligned}\right.\Rightarrow \left\{\begin{aligned}&du=2dx\\ &v=\sin x.\end{aligned}\right.\)
\(\begin{aligned} I=&\ 2x\sin x\Big|_0^1-\int_0^1\sin x\cdot 2dx=2\sin 1-0+2\cos x\Big|_0^1\\ =&\ 2\sin1+2\cos1-2\cos0=2\sin1+2\cos1-2. \end{aligned}\)

Đặt \(\left\{\begin{aligned}&u=2-x\\ &dv=e^xdx\end{aligned}\right.\Rightarrow \left\{\begin{aligned}&du=-dx\\ &v=e^x.\end{aligned}\right.\)
\[ \begin{aligned} I=&\ (2-x)e^x\Big|_0^{\ln2}+\displaystyle\int_0^{\ln2}e^x\mathrm{\,d}x\\ =&\ (2-\ln2)e^{\ln2}-(2-0)e^0+e^x\Big|_0^{\ln2}\\ =&\ (2-\ln2)2-2+e^{\ln2}-e^0=4-2\ln2-2+2-1=3-2\ln2. \end{aligned} \]

Đặt \(\left\{\begin{aligned}&u=\ln x\\ &dv=dx\end{aligned}\right.\Rightarrow \left\{\begin{aligned}&du=\dfrac{1}{x}dx\\ &v=x.\end{aligned}\right.\)
\[ \begin{aligned} I=&\ x\ln x\Big|_1^e-\displaystyle\int_1^ex\cdot\dfrac{1}{x}\mathrm{\,d}x\\ =&\ e\ln e-\ln1-\int_0^e\mathrm{\,d}x=e-x\Big|_1^e\\ =&\ e-(e-1)=1. \end{aligned} \]

Ta có \[ \begin{aligned} I=& \int_{-1}^15x^4\mathrm{\,d}x = x^5\Big|_{-1}^1=2. \end{aligned} \]

Ta có \[ \begin{aligned} I=& \displaystyle\int_1^4\sqrt{x}\mathrm{\,d}x=\displaystyle\int_1^4x^{\frac{1}{2}}\mathrm{\,d}x=\dfrac{2}{3}\cdot x^{\frac{3}{2}}\Big|_1^4=\dfrac{14}{3}. \end{aligned} \]

\[ I=\displaystyle\int_0^{\frac{\pi}{6}}\sin2x\mathrm{\,d}x=-\dfrac{1}{2}\cos2x\Big|_0^{\frac{\pi}{6}}=\dfrac{1}{4}. \]

\[ I=\displaystyle\int_0^{\frac{\pi}{4}}\dfrac{2}{\cos^2x}\mathrm{\,d}x=2\tan x\Big|_0^{\frac{\pi}{4}}=2. \]

\[ I=\displaystyle\int_0^1e^{2x}\mathrm{\,d}x=\dfrac{1}{2}e^{2x}\Big|_0^1=e^2-1. \]

\[ I=\displaystyle\int_0^1\dfrac{2}{3x+1}\mathrm{\,d}x=\dfrac{2}{3}\ln |3x+1|\Big|_0^1=\dfrac{2}{3}\ln4-\dfrac{2}{3}\ln1=\dfrac{2}{3}\ln4. \]

Ta có: \[ \begin{aligned} \displaystyle\int\limits_{-\pi}^1 f\left(x\right)\mathrm{\,d}x &=& \displaystyle\int\limits_{-\pi}^0 \left(x\sin x\right)\mathrm{\,d}x+\int\limits_0^1 \left(2x^2+x\right)\mathrm{\,d}x\\ &=& -\displaystyle\int\limits_{-\pi}^0 x\mathrm{\,d}\left(\cos x\right)+\left(\displaystyle\frac{2}{3}x^3+\displaystyle\frac{1}{2}x^2\right)\Big|_0^1\\ &=& \displaystyle \left(-x\cos x\right)\Big|_{-\pi}^0+\int\limits_{-\pi}^0 \cos x\mathrm{\,d}x+\displaystyle\frac{7}{6}\\ &=& \pi+\displaystyle\frac{7}{6}+\left(\sin x\right)\Big|_{-\pi}^0=\pi+\displaystyle\frac{7}{6} \end{aligned} \]

\[ \begin{aligned} \displaystyle \int\limits_{a}^{b}f'(x) \mathrm{\, d}x= & \ 3\sqrt{5} \Rightarrow f(b)-f(a)=3\sqrt{5} \Rightarrow f(a)\\ = & \ f(b) - 3\sqrt{5}= \sqrt{5}(\sqrt{5}-3). \end{aligned} \]

\[ \begin{aligned} I=& \ \displaystyle \int \limits_{-5}^{3}\left[7f(x)-x\right] \mathrm{d} x = 7F(x)\bigg|_{-5}^3 - \displaystyle\frac{x^2}{2}\bigg|_{-5}^3 = 2. \end{aligned} \]

\[ \begin{aligned} I=& \ \displaystyle \int_{0}^{1} f(x) \mathrm{\, d}x=F(1)-F(0)=3. \end{aligned} \]

\[ \begin{aligned} I=& \ \displaystyle\int\limits_1^3 f'(x)\mathrm{\,d}x=f(x) \big|_1^3=f(3)-f(1)=9-2=7. \end{aligned} \]

\[ \begin{aligned} I=& \ \displaystyle \int \limits_a^b f(x) \mathrm{\,d}x=\displaystyle \int \limits_a^c f(x) \mathrm{\,d}x+\displaystyle \int \limits_c^b f(x) \mathrm{\,d}x\\ =&\ \displaystyle \int \limits_a^c f(x) \mathrm{\,d}x-\displaystyle \int \limits_b^c f(x) \mathrm{\,d}x=17-(-11)=28. \end{aligned} \]

Ta có \(\displaystyle\int\limits_9^0 g(x)\mathrm{\,d}x=16\Rightarrow \displaystyle\int\limits_0^9 g(x)\mathrm{\,d}x=-16\). \[ \begin{aligned} I=& \ \displaystyle\int\limits_0^9 2f(x)\mathrm{\,d}x+\displaystyle\int\limits_0^9 3g(x)\mathrm{\,d}x=2\displaystyle\int\limits_0^9 f(x)\mathrm{\,d}x+3\displaystyle\int\limits_0^9 g(x)\mathrm{\,d}x\\ =& \ 2\cdot 37+3\cdot(-16)=26. \end{aligned} \]

\[ \begin{aligned} I=& \ \int\limits_{1}^{5}f(x)\mathrm{\, d}x=\int\limits_{1}^{2}f(x)\mathrm{\,d}x+ \int\limits_{2}^{5}f(x)\mathrm{\, d}x=3+(-1)=2. \end{aligned} \]

\[ \begin{aligned} I=& \ \displaystyle \int \limits_2^7f(x)\mathrm{d}x=\displaystyle \int \limits_2^5f(x)\mathrm{d}x+\displaystyle \int \limits_5^7f(x)\mathrm{d}x=3+9=12. \end{aligned} \]

\[ \begin{aligned} K=& \ 2\displaystyle\int\limits_1^3 f(x)\mathrm{\,d}x-3\displaystyle\int\limits_1^3 g(x)\mathrm{\,d}x=2\cdot9-3\cdot(-5)=33. \end{aligned} \]

\[ \begin{aligned} I=& \ \displaystyle\int\limits_1^2 f'(x)\mathrm{\,d}x =f(x)\bigg|_1^2=f(2)-f(1)=2-1=1. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{5}^{2}\left[2-4f(x)\right]\mathrm{\,d} x=\int\limits_{5}^{2}2\mathrm{\,d} x-4\int\limits_{5}^{2}f(x)\mathrm{\,d} x\\ =&\ 2(2-5)-4\cdot(-10)=34. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_a^b [2f(x)-3g(x)] \mathrm{\,d}x=2\int\limits_a^b f(x) \mathrm{\,d}x-3\int\limits_a^b g(x) \mathrm{\,d}x\\ =&\ 2\cdot (-2)-3\cdot 3=-13. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle \int \limits_{-5}^{3}\left[7f(x)-x\right] \mathrm{d} x = 7F(x)\bigg|_{-5}^3 - \displaystyle\frac{x^2}{2}\bigg|_{-5}^3 = 2. \end{aligned} \]

\[ \begin{aligned} &\ -1=\displaystyle\int\limits_{1}^{2}[kx-f(x)]\mathrm{\,d}x=\displaystyle\frac{kx^{2}}{2}\Big |_{1}^{2}-4\Rightarrow\displaystyle\frac{3k}{2}=3\Rightarrow k=2. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{-1}^26x^2\mathrm{\,d}x =\left.\left(2x^3\right)\right|_{-1}^2\\ =&\ 2\left[2^3-(-1)^3\right]=18 \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_0^2 \left(x^2-3x \right)\mathrm{\,d}x=\left(\displaystyle\frac{x^3}{3}-\displaystyle\frac{3x^2}{2}\right)\bigg|_0^2=\displaystyle\frac{-10}{3} \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_1^6\displaystyle\frac{1}{2x+1}\mathrm{\, d}x =\displaystyle\frac{1}{2}\ln \left|2x+1\right|\big|_1^6\\ =&\ \displaystyle\frac{1}{2}\ln 13-\displaystyle\frac{1}{2}\ln 3=\ln \sqrt{\displaystyle\frac{13}{3}} \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_0^1 \mathrm{e}^{2x}\mathrm{\,d}x=\left. \displaystyle\frac{\mathrm{e}^{2x}}{2}\right| _0^1=\displaystyle\frac{\mathrm{e}^{2}-1}{2}. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_0^1 \mathrm{e}^{-x}\mathrm{\,d}x=\left. -\mathrm{e}^{-x}\right|_0^1=\displaystyle\frac{\mathrm{e}-1}{\mathrm{e}}. \end{aligned} \]

Áp dụng công thức \(\displaystyle\int \displaystyle\frac{1}{ax+b}\mathrm{\,d}x=\displaystyle\frac{1}{a}\ln |ax+b|+C.\)
Ta có \(\displaystyle\int\limits_1^2 \displaystyle\frac{\mathrm{\,d}x}{3x+1}=\left.\displaystyle\frac{1}{3}\ln |3x+1|\right|_1^2=\displaystyle\frac{1}{3}\ln 7-\displaystyle\frac{2}{3}\ln 2\).
Do đó \[ \begin{aligned} &\ a=\displaystyle\frac{1}{3},\,b=-\displaystyle\frac{2}{3}\Rightarrow a+b=-\displaystyle\frac{1}{3}. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle \int\limits_{0}^{1} \displaystyle\frac{x+4}{x+3} \mathrm{d}x = \int\limits_{0}^{1} \left ( 1+\displaystyle\frac{1}{x+3} \right )\mathrm{d}x = (x+\ln |x+3| )\Big|_0^1=1+\ln \displaystyle\frac{4}{3}. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_0^2 \sqrt{4x+1} \mathrm{\,d}x = \left. \displaystyle\frac{\sqrt{(4x+1)^3}}{6} \right|^2_0=\displaystyle\frac{13}{3}. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{0}^{\tfrac{\pi}{4}}\sin 3x\mathrm{\,d}x=-\displaystyle\frac{\cos 3x}{3}\Big |_{0}^{\tfrac{\pi}{4}}\\ =&\ \displaystyle\frac{1+\displaystyle\frac{\sqrt{2}}{2}}{3}=\displaystyle\frac{2+\sqrt{2}}{6} \end{aligned} \]

\[ \begin{aligned} &\ f(2)-f(1)=\displaystyle\int\limits_{1}^{2}f'(x)\mathrm{\,d}x=\left(x^{2}+\displaystyle\frac{2}{x}\right)\Big |_{1}^{2}=2. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_0 ^4 \displaystyle\frac{1}{\sqrt{2x+1}}\mathrm{\,d}x=\displaystyle\frac{1}{2}\displaystyle\int\limits_0 ^4 \left(2x+1 \right)^{-\tfrac{1}{2}}\mathrm{\,d}\left(2x+1 \right)\\ =&\ \displaystyle\frac{1}{2}\cdot 2\left(2x+1\right)^{\tfrac{1}{2}}\big|_0^4=2 \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle \int \limits_{0}^{\tfrac{\pi}{3}}\cos 2x \mathrm{\,d}x=\displaystyle\frac{1}{2}\sin 2x\big|_0^{\tfrac{\pi}{3}}=\displaystyle\frac{\sqrt{3}}{4}. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle \int\limits_0^2 \displaystyle\frac{x^2}{x+1} \mathrm{d}x=\displaystyle \int\limits_0^2 \left(x-1+\displaystyle\frac{1}{x+1}\right) \mathrm{d}x=\left(\displaystyle\frac{x^2}{2}-x+\ln \left|x+1\right|\right)\bigg|_0^2=\ln 3. \end{aligned} \]

Suy ra, \(a=0\) và \(b=3\). Do đó, \(S=3 \in \left(2;4\right)\).

\[ \begin{aligned} I=&\ \displaystyle \int\limits_{0}^{1}\left(\displaystyle\frac{1}{2x+1}-\displaystyle\frac{1}{3x+1}\right)\mathrm{\,d}x =\displaystyle\frac{1}{2}\ln|2x+1|\big|_0^1-\displaystyle\frac{1}{3}\ln|3x+1|\big|_0^1\\ =&\ \displaystyle\frac{1}{2}\ln3-\displaystyle\frac{1}{3}\ln4=\displaystyle\frac{1}{6}(3\ln3-2\ln4) =\displaystyle\frac{1}{6}\ln\displaystyle\frac{27}{16}. \end{aligned} \]

Suy ra \(a=27, b=16\Rightarrow a+b=43>22\).

\[ \begin{aligned} I=&\ \displaystyle \int\limits_{0}^{1}\displaystyle\frac{2x+3}{2-x}\mathrm{\,d}x =\displaystyle \int\limits_{0}^{1}\displaystyle\frac{-2(-x+2)+7}{2-x}\mathrm{\,d}x\\ =&\ \displaystyle \int\limits_{0}^{1}\left(-2+\displaystyle\frac{7}{-x+2}\right)\mathrm{\,d}x =2x\big|_0^1-7\ln|-x+2|\big|_0^1=-2+7\ln2. \end{aligned} \]

Suy ra \(a=7,b=-2\Rightarrow a+2b=7-4=3\).

\[ \begin{aligned} I=&\ \displaystyle \int\limits_0^1 f(x)\, \mathrm{d}x = \displaystyle\frac{-a \cos \pi x}{\pi} + bx \bigg|_0^1 = \displaystyle\frac{2a}{\pi} + 2 = 4. \Rightarrow a = \pi. \end{aligned} \]

Vậy \(a = \pi, b = 2\).

\[ \begin{aligned} &\ \displaystyle\int\limits_{0}^{\frac{1}{2}}{\cos \left(\pi x\right)\mathrm{\,d}x=2a + 1}\\ \Leftrightarrow\ & \left. \displaystyle\frac{1}{\pi}\sin \left(\pi x\right)\right|_{0}^{\frac{1}{2}}=2a + 1\Leftrightarrow \displaystyle\frac{1}{\pi}=2a + 1\Leftrightarrow a=\displaystyle\frac{1 - \pi}{2\pi}. \end{aligned} \]

\[ \begin{aligned} &\ \displaystyle\int\limits_0^a{\sqrt{\mathrm{e}^x}\mathrm{\,d}x=}\displaystyle\int\limits_0^a{\mathrm{e}^{\frac{1}{2}x}\mathrm{\,d}x=}\left. 2\mathrm{e}^{\frac{1}{2}x}\right|_0^a=2\mathrm{e}^{\frac{a}{2}} - 2 =2(\mathrm{e} - 1)\\ \Leftrightarrow\ & \mathrm{e}^{\frac{a}{2}}=\mathrm{e}\Leftrightarrow a=2. \end{aligned} \]

\[ \begin{aligned} H=&\ \displaystyle\int\limits_{0}^{\frac{\pi}{4}} \tan^2x\mathrm{\,d}x=\displaystyle\int\limits_{0}^{\frac{\pi}{4}} \left(\displaystyle\frac{1}{\cos^2x}-1\right)\mathrm{\,d}x\\ =\ & (\tan x-x)\Big|_{0}^{\frac{\pi}{4}}=1-\displaystyle\frac{\pi}{4}. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{0}^{\frac{1}{2}}{\cos \left(\pi x\right)\mathrm{\,d}x=2a + 1}\Leftrightarrow \left. \displaystyle\frac{1}{\pi}\sin \left(\pi x\right)\right|_{0}^{\frac{1}{2}}=2a + 1\Leftrightarrow \displaystyle\frac{1}{\pi}=2a + 1\Leftrightarrow a=\displaystyle\frac{1 - \pi}{2\pi} \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_0^a{\sqrt{\mathrm{e}^x}\mathrm{\,d}x=}\displaystyle\int\limits_0^a{\mathrm{e}^{\frac{1}{2}x}\mathrm{\,d}x=}\left. 2\mathrm{e}^{\frac{1}{2}x}\right|_0^a=2\mathrm{e}^{\frac{a}{2}} - 2=2(\mathrm{e} - 1)\\ \Leftrightarrow\ & \mathrm{e}^{\frac{a}{2}}=\mathrm{e}\Leftrightarrow a=2. \end{aligned} \]

\[ \begin{aligned} H=&\ \displaystyle\int\limits_{0}^{\frac{\pi}{4}} \tan^2x\mathrm{\,d}x=\displaystyle\int\limits_{0}^{\frac{\pi}{4}} \left(\displaystyle\frac{1}{\cos^2x}-1\right)\mathrm{\,d}x =(\tan x-x)\Big|_{0}^{\frac{\pi}{4}}=1-\displaystyle\frac{\pi}{4}. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_1^4\left( x^2+3\sqrt{x}\right)\mathrm{\,d}x=\left.\left( \displaystyle\frac{x^3}{3}+2x\sqrt{x}\right)\right|_1^4= \left( \displaystyle\frac{64}{3}+16\right) -\left( \displaystyle\frac{1}{3}+2\right)=35. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle \int \limits_0^{\pi}(x-\sin 2x)\mathrm{\,d}x=\left( \displaystyle\frac{x^2}{2}+\displaystyle\frac{1}{2}\cos 2x\right) \Bigg|_0^\pi=\displaystyle\frac{\pi^2}{2}+\displaystyle\frac{1}{2}-\displaystyle\frac{1}{2}=\displaystyle\frac{\pi^2}{2}. \end{aligned} \]

Suy ra \(a=1\), \(b=2\) khi đó \(a+b=3\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits _{\frac{\pi}{6}}^{\frac{\pi}{4}}\sin 2x \mathrm{\,d}x=\displaystyle\frac{1}{4}=F\left(\displaystyle\frac{\pi}{4}\right)-F\left(\displaystyle\frac{\pi}{6}\right)\\ \Rightarrow\ & F\left(\displaystyle\frac{\pi}{6}\right)=F\left(\displaystyle\frac{\pi}{4}\right)-\displaystyle\frac{1}{4}=1-\displaystyle\frac{1}{4}=\displaystyle\frac{3}{4}. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{0}^{2}f(x)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}3x^2\mathrm{\,d}x+\displaystyle\int\limits_{1}^{2}(4-x)\mathrm{\,d}x=\displaystyle\frac{7}{2}. \end{aligned} \]

Đặt \(t=2x\Rightarrow dt=2\mathrm{\,d}x\).

Với \(x=0\) thì \(t=0\) và với \(x=2\) thì \(t=4\).

\[ \begin{aligned} I=&\ \int\limits_0^4 f(t)\displaystyle\frac{\mathrm{\,d}t}{2}=\displaystyle\frac{1}{2}\int\limits_0^4 f(t)\mathrm{\,d}t=\displaystyle\frac{1}{2}\int\limits_0^4 f(x)\mathrm{\,d}x=8. \end{aligned} \]

Đặt \(t=\displaystyle\frac{x}{4}\Rightarrow x=4t \Rightarrow \mathrm{d}x=4\mathrm{d}t\).

Khi \(x=4\Rightarrow t=1; x=12\Rightarrow t=3\).

\[ \begin{aligned} I=&\ \displaystyle \int_{4}^{12} f\left(\displaystyle\frac{x}{4}\right) \mathrm{\, d}x=4\displaystyle \int_{1}^{3} f\left(t\right) \mathrm{\, d}t=4\cdot 8=32. \end{aligned} \]

Đổi biến \(t=3x\) suy ra \(\mathrm{d}t = 3\mathrm{d}x\Rightarrow dx=\dfrac{1}{3}dt\).

Khi \(x = 0\) thì \(t = 0\); khi \(x = 2\) suy ra \(t = 6\). Do đó

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{0}^{6}f\left(t\right)\cdot\dfrac{1}{3} \mathrm{\,d}x = \dfrac{1}{3}\cdot 1 = \displaystyle\frac{1}{3}. \end{aligned} \]

Đặt \(x=2t\), suy ra \(\mathrm{d}x=2\mathrm{d}t\).

Với \(x=0\), ta có \(t=0\), với \(x=4\), ta có \(t=2\).

\[ \begin{aligned} I=&\ \displaystyle \int\limits_0^4 f(x) \mathrm{d}x=2\displaystyle \int\limits_0^2 f(2t) \mathrm{d}t=2\displaystyle \int\limits_0^2 f\left(2x\right) \mathrm{d}x=16. \end{aligned} \]

Đặt \(t=2x\Rightarrow \mathrm{\,d} t=2\mathrm{\,d} x\Rightarrow \mathrm{\,d} x=\displaystyle\frac{1}{2}\mathrm{\,d} t\).

Khi \(x=0\Rightarrow t=0\), khi \(x=2\Rightarrow t=4\).

\[ \begin{aligned} I=&\ \int\limits_{0}^{2}f(2x)\mathrm{\,d} x=\int\limits_{0}^{4}f(t)\cdot \displaystyle\frac{1}{2}\mathrm{\,d} t=\displaystyle\frac{1}{2}\int\limits_{0}^{4}f(t)\mathrm{\,d} t=\displaystyle\frac{1}{2}\cdot 12=6. \end{aligned} \]

Đặt \(t=3x \Rightarrow \textrm{d}t=3 \textrm{d}x\).

Đổi cận \(x=0\Rightarrow t=0\), \(x=2\Rightarrow t=6\).

\[ \begin{aligned} I=&\ \displaystyle \int\limits_0^2 f(3x) \textrm{\,d}x =\displaystyle\frac{1}{3}\int\limits_0^6 f(t) \textrm{\,d}t=\displaystyle\frac{1}{3} \cdot 12=4. \end{aligned} \]

Đặt \(t=3x \Rightarrow dx=\displaystyle\frac{\mathrm{\,d}t}{3}\).

Đổi cận \(x=0\Rightarrow t=0\) và \(x=1\Rightarrow t=3\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{0}^{3}f(t)\displaystyle\frac{\mathrm{\,d}t}{3}=\displaystyle\frac{1}{3}\displaystyle\int\limits_{0}^{3}f(t)\mathrm{\,d}t=3. \end{aligned} \]

Đặt \(t=2-3x\Rightarrow \mathrm{\,d}t=-3\mathrm{\,d}x\).

Khi \(x=-1\) thì \(t=5\), khi \(x=1\) thì \(t=-1\).

\[ \begin{aligned} \ & \displaystyle\int\limits_{-1}^{1}f(2-3x)\mathrm{\,d}x=a \Leftrightarrow -\displaystyle\frac{1}{3}\displaystyle\int\limits_{5}^{-1}f(t)\mathrm{\,d}=a\\ \Leftrightarrow\ & \displaystyle\int\limits_{-1}^{5}f(t)\mathrm{\,d}t =-\displaystyle\frac{a}{3} \Leftrightarrow \displaystyle\int\limits_{-1}^{5}f(x)\mathrm{\,d}x=-\displaystyle\frac{a}{3}\\ \Leftrightarrow\ &a=-3. \end{aligned} \]

Đặt \(x=2u\Rightarrow \mathrm{\,d}x=2\mathrm{\,d}u\).

Đổi cận \(u=2\Rightarrow x=4\), \(u=5\Rightarrow x=10\).

Khi đó \(\displaystyle \int \limits_2^5 f’(2u)du=\displaystyle \int \limits_2^5 f’(x)\displaystyle\frac{\mathrm{\,d}x}{2}=\displaystyle\frac{1}{2} \displaystyle \int \limits_2^5 f’(x)\mathrm{\,d}x\).

Mà \(\displaystyle \int \limits_2^5 f’(2u)du=6\Rightarrow \displaystyle \displaystyle\frac{1}{2} \int \limits_2^5 f’(x) \mathrm{\,d}x=6\Rightarrow \displaystyle \int \limits_2^5 f’(x)\mathrm{\,d}x=12\).

\[ \begin{aligned} I=&\ \displaystyle \int \limits_0^{10} f'(x)\mathrm{\,d}x=\displaystyle \int \limits_0^4 f'(x)\mathrm{\,d}x+\displaystyle \int \limits_4^{10} f'(x)\mathrm{\,d}x=5+12=17. \end{aligned} \]

Mà \(\displaystyle \int \limits_0^{10} f'(x)\mathrm{\,d}x=f(10)-f(0) \Rightarrow f(10)=20\).

Đặt \(t = 3x-3 \Rightarrow \mathrm{d}t = 3\mathrm{d}x\).

Đổi cận: \(x = 1 \Rightarrow t = 0\); \(x = 4 \Rightarrow t = 9\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits_1^4 f(3x-3)\mathrm{\,d}x = \displaystyle\frac{1}{3}\displaystyle\int\limits_0^9 f(t)\mathrm{\,d}t = \displaystyle\frac{1}{3} \cdot 9 = 3. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{2}^{6} f\left(\displaystyle\frac{x}{2}\right)\mathrm{\,d}x = 2\displaystyle\int\limits_{2}^{6} f\left(\displaystyle\frac{x}{2}\right)\mathrm{\,d}\left(\displaystyle\frac{x}{2}\right)\\ =\ & 2\displaystyle\int\limits_{1}^{3} f(t)\mathrm{\,d}t=2\cdot 12=24. \end{aligned} \]

Đặt \(t=2x+1 \Rightarrow \mathrm{\,d}t=2 \mathrm{\,d}x\).

Đổi cận: \(x=1 \Rightarrow t=3\); \(x=2 \Rightarrow t=5\).

\[ \begin{aligned} I=&\ \Rightarrow I=\displaystyle\int\limits_3^5 \displaystyle\frac{1}{2}f(t) \mathrm{\,d}t=\displaystyle\frac{1}{2}\displaystyle\int\limits_3^5 f(x) \mathrm{\,d}x=\displaystyle\frac{1}{2}a. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{-1}^1f(|2x-1|)\mathrm{\, d}x = \displaystyle\int\limits_{-1}^{\tfrac{1}{2}}f[(1-2x)]\mathrm{\, d}x+\displaystyle\int\limits_{\tfrac{1}{2}}^1f[(2x-1)]\mathrm{\, d}x=I_1+I_2. \end{aligned} \]

\(\bullet\quad\) Tính \(I_1= \displaystyle\int\limits_{-1}^{\tfrac{1}{2}}f[(1-2x)]\mathrm{\, d}x\):

Đặt \(u=1-2x\Rightarrow \mathrm{\, d}u=-2\mathrm{\, d}x\).

Với \(x=-1\Rightarrow u=3\), với \(x=\displaystyle\frac{1}{2}\Rightarrow u=0\).

Khi đó \(I_1= -\displaystyle\frac{1}{2}\displaystyle\int\limits_3^0f(u)\mathrm{\, d}u =\displaystyle\frac{1}{2}\displaystyle\int\limits_0^3f(x)\mathrm{\, d}x=4\).

\(\bullet\quad\) Tính \(I_2= \displaystyle\int\limits_{\tfrac{1}{2}}^1f[(2x-1)]\mathrm{\, d}x\):

Đặt \(u=2x-1\Rightarrow \mathrm{\, d}u=2\mathrm{\, d}x\).

Với \(x=1\Rightarrow u=1\), với \(x=\displaystyle\frac{1}{2}\Rightarrow u=0\).

Khi đó \(I_2= \displaystyle\frac{1}{2}\displaystyle\int\limits_0^1f(u)\mathrm{\, d}u =\displaystyle\frac{1}{2}\displaystyle\int\limits_0^1f(x)\mathrm{\, d}x=1\).

Vậy \(\displaystyle\int\limits_{-1}^1f(|2x-1|)\mathrm{\, d}x=I_1+I_2=5\).

Đặt \(3x-1=t \Rightarrow \mathrm{\,d}x= \displaystyle\frac{\mathrm{\,d}t}{3}.\)

Khi \(x=-1\) thì \(t=-4\); khi \(x=1\) thì \(t=2\).

\[ \begin{aligned} I=&\ \displaystyle\frac{1}{3} \displaystyle\int\limits_{-4}^2 f\left(|t|\right)\mathrm{\,d}t \Rightarrow 3I= \displaystyle\int\limits_{-4}^0 f\left(|t|\right)\mathrm{\,d}t +\displaystyle\int\limits_{0}^2 f\left(|t|\right)\mathrm{\,d}t\\ =\ & \displaystyle\int\limits_{-4}^0 f\left(-t\right)\mathrm{\,d}t+\displaystyle\int\limits_{0}^2 f\left(t\right)\mathrm{\,d}t=J+1. \end{aligned} \]

Tính \(J= \displaystyle\int\limits_{-4}^0 f\left(-t\right)\mathrm{\,d}t\).

Đặt \(-t=x \Rightarrow \mathrm{\,d}t =-\mathrm{\,d}x\).

Khi \(t=-4\) thì \(x=4\); khi \(t=0\) thì \(x=0\).

Suy ra \(J=-\displaystyle\int\limits_{4}^0 f\left(x\right)\mathrm{\,d}x = \displaystyle\int\limits_{0}^4 f\left(x\right)\mathrm{\,d}x=3.\)

Vậy \(3I=4 \Leftrightarrow I=\displaystyle\frac{4}{3}.\)

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{-1}^1f(|2x-1|)\mathrm{\, d}x = \displaystyle\int\limits_{-1}^{\tfrac{1}{2}}f[(1-2x)]\mathrm{\, d}x+\displaystyle\int\limits_{\tfrac{1}{2}}^1f[(2x-1)]\mathrm{\, d}x=I_1+I_2. \end{aligned} \]

\(\bullet\quad\) Tính \(I_1= \displaystyle\int\limits_{-1}^{\tfrac{1}{2}}f[(1-2x)]\mathrm{\, d}x\):

Đặt \(u=1-2x\Rightarrow \mathrm{\, d}u=-2\mathrm{\, d}x\). Với \(x=-1\Rightarrow u=3\), với \(x=\displaystyle\frac{1}{2}\Rightarrow u=0\).

Khi đó \(I_1= -\displaystyle\frac{1}{2}\displaystyle\int\limits_3^0f(u)\mathrm{\, d}u =\displaystyle\frac{1}{2}\displaystyle\int\limits_0^3f(x)\mathrm{\, d}x=4\).

\(\bullet\quad\) Tính \(I_2= \displaystyle\int\limits_{\tfrac{1}{2}}^1f[(2x-1)]\mathrm{\, d}x\):

Đặt \(u=2x-1\Rightarrow \mathrm{\, d}u=2\mathrm{\, d}x\). Với \(x=1\Rightarrow u=1\), với \(x=\displaystyle\frac{1}{2}\Rightarrow u=0\).

Khi đó \(I_2= \displaystyle\frac{1}{2}\displaystyle\int\limits_0^1f(u)\mathrm{\, d}u =\displaystyle\frac{1}{2}\displaystyle\int\limits_0^1f(x)\mathrm{\, d}x=1\).

Vậy \(\displaystyle\int\limits_{-1}^1f(|2x-1|)\mathrm{\, d}x=I_1+I_2=5\).

Đặt \(u=3x+2\Rightarrow \mathrm{\,d}u = 3 \mathrm{\,d}x\).

Khi \(x=0\Rightarrow u=2\), khi \(x=2\Rightarrow u=5\).

\[ \begin{aligned} I=&\ \displaystyle\frac{1}{3}\displaystyle \int \limits_2^5 f(u)\mathrm{\,d}u =\displaystyle\frac{1}{3} \displaystyle \int \limits_2^5 f(x)\mathrm{\,d}x= \displaystyle\frac{2018}{3}. \end{aligned} \]

Đặt \(t=x^2+1\Rightarrow\mathrm{\,d}t=2x\mathrm{\,d}x\Rightarrow x\mathrm{\,d}x=\displaystyle\frac{1}{2}\mathrm{\,d}t\).

Khi đó \(x=2\Rightarrow t=5\) và \(x=1\Rightarrow t=2\).

\[ \begin{aligned} &\ \displaystyle\int\limits_1^2 f(x^2+1)x\mathrm{\,d}x=\displaystyle\frac{1}{2}\displaystyle\int\limits_2^5 f(t)\mathrm{\,d}t=2 \Leftrightarrow I=4. \end{aligned} \]

Ta có \(8 =\displaystyle\int\limits_{0}^{1}f(2x) \textrm{d}x =\displaystyle\frac{1}{2} \displaystyle\int\limits_{0}^{1} f(2x) \textrm{d}(2x)=\displaystyle\frac{1}{2} \displaystyle\int\limits_{0}^{2} f(t) \textrm{d}t \Rightarrow \displaystyle\int\limits_{0}^{2} f(t) \textrm{d}t =16\).

Đặt \(t=x^2 \Rightarrow \textrm{d}t=2x\textrm{d}x\). Suy ra

\[ \begin{aligned} I=&\ \displaystyle\frac{1}{2}\displaystyle\int\limits_{0}^{2} f(t)\textrm{d}t=8. \end{aligned} \]

Đặt \(t=\ln x\). Ta có \(\mathrm{\,d}t= \displaystyle\frac{\mathrm{\,d}x}{x}\).

Khi \(x=1\) thì \(t(1)=0\), khi \(x=\mathrm{e}\) thì \(t(\mathrm{e})=1.\)

\[ \begin{aligned} I=&\ \displaystyle\int\limits_1^\mathrm{e} \displaystyle\frac{f(\ln x)}{x}\mathrm{\,d}x = \displaystyle\int\limits_0^1 f(t)\mathrm{\,d}t =\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x = \mathrm{e}. \end{aligned} \]

Đặt \(t=\sqrt{x}\Rightarrow \mathrm{,d}t=\displaystyle\frac{\mathrm{\,d}x}{2\sqrt{x}}\).

Đổi cận: \(x=1\Rightarrow t=1,x=4\Rightarrow t=2\).

\[ \begin{aligned} I=&\ 2\displaystyle\int\limits_1^2 f(t)\mathrm{\,d}t=2. \end{aligned} \]

Đặt \(t=\sin x\Rightarrow \mathrm{\,d}t=\cos x \mathrm{\,d}x \).

Đổi cận: \(x=0\Rightarrow t=0\), \(x=\displaystyle\frac{\pi}{2}\Rightarrow t=1\).

\[ \begin{aligned} I=&\ \Rightarrow I=\displaystyle\int\limits_0^{\frac{\pi }{2}}\cos x\cdot f\left( \sin x \right)\mathrm{\,d}x =\displaystyle\int\limits_0^1 f(t) \mathrm{\,d}t=\displaystyle\int\limits_0^1 f(x) \mathrm{\,d}x=7. \end{aligned} \]

Ta có \(\displaystyle \int\limits_1^2 f\left(x^2 + 1\right) x \, \mathrm{d}x = \displaystyle\frac{1}{2} \int\limits_1^2 f\left(x^2 + 1\right)\, \mathrm{d} \left(x^2 + 1\right) = 2\).

Do đó \(\displaystyle\frac{1}{2}F\left( x^2 + 1\right)\bigg|_1^2 = 2\), hay \(F(5) - F(2) = 4\).

\[ \begin{aligned} I= &\ \displaystyle \int\limits_2^5 f(t)\, \mathrm{d}t = 2 \int\limits_1^2 f \left(x^2 + 1\right) x \, \mathrm{d}x = 4. \end{aligned} \]

\[ \begin{aligned} &\ 8 =\displaystyle\int\limits_{0}^{1}f(2x) \textrm{d}x =\displaystyle\frac{1}{2} \displaystyle\int\limits_{0}^{1} f(2x) \textrm{d}(2x)=\displaystyle\frac{1}{2} \displaystyle\int\limits_{0}^{2} f(t) \textrm{d}t \Rightarrow \displaystyle\int\limits_{0}^{2} f(t) \textrm{d}t =16. \end{aligned} \]

Đặt \(t=x^2 \Rightarrow \textrm{d}t=2x\textrm{d}x\). Suy ra

\(I=\displaystyle\frac{1}{2}\displaystyle\int\limits_{0}^{2} f(t)\textrm{d}t=8.\)

\[ \begin{aligned} I=&\ \int\limits_0^2\left[2x + xf(x^2 + 1)\right]\, \mathrm{d}x = x^2 \bigg|_0^2 + \displaystyle\frac{1}{2}\int\limits_0^2 f(x^2 + 1)\, \mathrm{d}(x^2 + 1)\\ =\ & 4 + \displaystyle\frac{1}{2}\int\limits_0^5 f(t) \, \mathrm{d}t \text{ (ta đặt } x^2 + 1 = t\text{) } = 4 + \displaystyle\frac{1}{2}\cdot 12 = 10. \end{aligned} \]

Xét tích phân \(I=\displaystyle\int\limits_{\frac{\pi}{3}}^{\frac{2\pi}{3}}f(2\cos x)\sin x\mathrm{\,d}x\).

Đặt \(t=2\cos x\Rightarrow \mathrm{\,d}t=-2\sin x\mathrm{\,d}x\) hay \(\sin x\mathrm{\,d}x=-\displaystyle\frac{1}{2}\mathrm{\,d}t\).

Đổi cận: \(x=\displaystyle\frac{\pi}{3}\Rightarrow t=1\), \(x=\displaystyle\frac{2\pi}{3}\Rightarrow t=-1\).

\[ \begin{aligned} I=&\ -\displaystyle\frac{1}{2}\displaystyle\int\limits_{1}^{-1}f(t)\mathrm{\,d}t=\displaystyle\frac{1}{2}\displaystyle\int\limits_{-1}^1f(x)\mathrm{\,d}x=\displaystyle\frac{1}{2}\cdot 12=6. \end{aligned} \]

Đặt \(u= \sqrt{x^3 + 1} \Rightarrow u^2=x^3+1\Rightarrow 2u\mathrm{\, d}u = 3x^2 \mathrm{\, d}x.\)

\[ \begin{aligned} I=&\ \displaystyle\frac{2}{3} \displaystyle\int\limits_{1}^{3}u^2 \mathrm{\, d}u = \displaystyle\frac{2}{9}\cdot u^3\Big|_1^3=\displaystyle\frac{52}{9}. \end{aligned} \]

\[ \begin{aligned} I= &\ \displaystyle \int \limits^4_0 \displaystyle\frac{\mathrm{\, d} x}{3 + \sqrt{2x+1} } = \int \limits^4_0 \left (1 - \displaystyle\frac{3}{3+ \sqrt{2x+1}} \right )\mathrm{\, d }(\sqrt{2x+1})\\ =\ & \left [ \sqrt{2x+1} -3\ln (\sqrt{2x+1} +3) \right ] \Big |^4_0 = 2 + 3 \cdot \ln \displaystyle\frac{2}{3}. \end{aligned} \]

Vậy \( a+ b = 5 \).

Đặt \(t= \sqrt[3]{1-x}\Rightarrow t^3 = 1 - x \Rightarrow 3t^2 \mathrm{\,d}t = - \mathrm{\,d}x\).

Ta có \(x=0 \Rightarrow t=1\), \(x=1 \Rightarrow t = 0\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits_0^1 x\sqrt[3]{1-x} \mathrm{\,d}x = 3\displaystyle\int\limits_0^1 \left(1 -t^3 \right)t^3 \mathrm{\,d}t =3\left( \displaystyle\frac{t^4}{4} - \displaystyle\frac{t^7}{7} \right) \bigg|_0^1 = \displaystyle\frac{9}{28}. \end{aligned} \]

Do đó \(M= 9\), \(N=28\). Suy ra \(M + N = 37\).

Đặt \(u=\sqrt{3+\ln x}\Rightarrow u^2 = 3+ \ln x \Rightarrow 2u\mathrm{d}u = \displaystyle\frac{\mathrm{d}x}{x}\).

Đổi cận: \(x=1\Rightarrow u = \sqrt{3}\); \(x=\mathrm{e} \Rightarrow u = 2 \).

\[ \begin{aligned} I=&\ \displaystyle \int_{1}^{e} \displaystyle\frac{\sqrt{3+\ln x}}{x} \mathrm{\, d}x=2\displaystyle \int_{\sqrt{3}}^{2} u^2 \mathrm{\, d}x = \displaystyle\frac{2u^3}{3}\biggr|_{\sqrt{3}}^2=\displaystyle\frac{16-6\sqrt{3}}{16}. \end{aligned} \]

Khi đó \(\). Vậy \(a=16, b= 6 \Rightarrow a- b = 10\).

Với \(u=\sin x\), ta có \(x=0 \Rightarrow u=0\); \(x=\displaystyle\frac{\pi}{2} \Rightarrow u=1\).

\[ \begin{aligned} I=&\ \displaystyle \int \limits_0^{\frac{\pi}{2}} \sin^2 x \cos x \mathrm{d}x=\displaystyle \int \limits_0^{\frac{\pi}{2}} \sin^2 x \mathrm{d}(\sin x)=\displaystyle \int \limits_0^1 u^2 \mathrm{d}u. \end{aligned} \]

Đặt \(u=\cos 2x\), ta có \(\mathrm{\,d}u=-2\sin2x\mathrm{\,d}x\).

Đổi cận: với \(x=0\) ta có \(u=1\) và \(x=a\), ta có \(u=\cos 2a\).

\[ \begin{aligned} I=&\ -4\displaystyle\displaystyle\int\limits_1^{\cos2a}\mathrm{e}^u\mathrm{\,d}u=-4\mathrm{e}^u\Big|_1^{\cos2a}=4\left(\mathrm{e}-\mathrm{e}^{\cos2a}\right). \end{aligned} \]

Đặt \(t=\sqrt{x^2+1}\Rightarrow t^2=x^2+1\Rightarrow t\mathrm{\,d}t=x\mathrm{\,d}x\).

Với \(x=1\Rightarrow t=\sqrt{2}\), \(x=\sqrt{3}\Rightarrow t=2\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits_1^{\sqrt{3}}x\sqrt{x^2+1}\mathrm{\,d}x=\displaystyle\int\limits_{\sqrt{2}}^2t^2\mathrm{\,d}t=\displaystyle\frac{t^3}{3}\Bigg|_{\sqrt{2}}^2=\displaystyle\frac{8-2\sqrt{2}}{3}=\displaystyle\frac{2}{3}\left(4-\sqrt{2}\right). \end{aligned} \]

Vậy \(a=4\) và \(b=2\) nên ta suy ra \(a=2b\).

Ta có \(u=\sin x\Rightarrow\mathrm{\,d}u=\cos x\mathrm{\,d}x\), với \(x=0\Rightarrow u=0\), \(x=\displaystyle\frac{\pi}{2}\Rightarrow u=1\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits_0^{\frac{\pi}{2}}\sin^2x\cos x\mathrm{\,d}x=\displaystyle\int\limits_0^1u^2\mathrm{\,d}u. \end{aligned} \]

Đặt \(t=\sqrt{1+3\ln x} \Rightarrow t^2=1+3\ln x \Rightarrow 2t \mathrm{\,d}t=\displaystyle\frac{3}{x} \mathrm{\,d}x \Rightarrow \displaystyle\frac{\mathrm{\,d}x}{x}=\displaystyle\frac{2}{3}t \mathrm{\,d}t\).

Đổi cận: \(x=1 \Rightarrow t=1\); \(x=\mathrm{e} \Rightarrow t=2\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits_1^2 t\cdot\displaystyle\frac{2}{3}t \mathrm{\,d}t=\displaystyle\frac{2}{3}\displaystyle\int\limits_1^2 t^2\mathrm{\,d}t=\displaystyle\frac{2}{9}t^3\bigg|_1^2=\displaystyle\frac{14}{9}. \end{aligned} \]

Đặt \(t=2+\cos x\Rightarrow \mathrm{\,d} t=-\sin x\mathrm{\,d} x\Rightarrow \sin x\mathrm{\,d} x=-\mathrm{\,d} t\).

Khi \(x=0\Rightarrow t=3\), khi \(x=\displaystyle\frac{\pi}{2}\Rightarrow t=2\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{3}^{2}\sqrt{t}(-\mathrm{\,d} t)=\displaystyle\int\limits_{2}^{3}\sqrt{t}\mathrm{\,d} t. \end{aligned} \]

Ta có \(t=\sqrt{1+\ln x}\Rightarrow t^2=1+\ln x \Rightarrow 2t\mathrm{\,d}t=\displaystyle\frac{\mathrm{\,d}x}{x}.\)

Khi \(x=1\Rightarrow t=1,\) \(x=\mathrm{e}\Rightarrow t=\sqrt{2}.\)

\[ \begin{aligned} I=&\ \displaystyle\int\limits_1^{\sqrt{2}}t\cdot2t\mathrm{\,d}t=2\displaystyle\int\limits_1^{\sqrt{2}} t^2\mathrm{\,d}t. \end{aligned} \]

Đặt \(t=\cos x\) suy ra \(\mathrm{\,d}t=-\sin x\mathrm{\,d}x\). Đổi cận \(x=0 \Rightarrow t=1\), \(x=\pi\Rightarrow t=-1.\)

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{1}^{-1}t^2\cdot (-1)\mathrm{\,d}t=\displaystyle\int\limits_{-1}^{1}t^2\mathrm{\,d}t=\displaystyle\frac{t^3}{3}\bigg|_{-1}^1=\displaystyle\frac{2}{3}. \end{aligned} \]

\[ \begin{aligned} & F\left(\displaystyle\frac{\pi}{2}\right) - F(0) = \displaystyle \int \limits _{0}^{\frac{\pi}{2}} \sin^3x \cos x \mathrm{\,d}x = \left(\displaystyle\frac{\sin^4x}{4}\right) \bigg|_{0}^{\frac{\pi}{2}} = \displaystyle\frac{1}{4}\\ \Leftrightarrow\ & F\left(\displaystyle\frac{\pi}{2}\right) = \displaystyle\frac{1}{4} + F(0) = \displaystyle\frac{1}{4} + \pi. \end{aligned} \]

Tích phân \(J=\displaystyle\int\limits_{1}^{9}{\displaystyle\frac{f(\sqrt{x})}{\sqrt{x}}}\mathrm{\,d}x=4\)

Đặt \(t=\sqrt{x}\Rightarrow dt=\displaystyle\frac{1}{2\sqrt{x}}\mathrm{\,d}x\Rightarrow \displaystyle\frac{1}{2}\mathrm{\,d}t=\displaystyle\frac{1}{\sqrt{x}}\mathrm{\,d}x\).

Khi \(x=1\Rightarrow t=1\), khi \(x=9\Rightarrow t=3\).

\[ \begin{aligned} J=&\ 4\Leftrightarrow \displaystyle\int\limits_{1}^{3}{f(t)\displaystyle\frac{1}{2}}\mathrm{\,d}t=4 \Leftrightarrow \displaystyle\int\limits_{1}^{3}{f(t)}\mathrm{\,d}t=8 \Leftrightarrow \displaystyle\int\limits_{1}^{3}{f(x)}\mathrm{\,d}x=8. \end{aligned} \]

Tích phân \(K=\displaystyle\int\limits_{0}^{\frac{\pi}{2}}{f(\sin x)}\cos x\mathrm{\,d}x=2\)

Đặt \(t=\sin x\Rightarrow \mathrm{\,d}t = \cos x \mathrm{\,d}x\).

Khi \(x=0\Rightarrow t=0\), khi \(x=\dfrac{\pi}{2}\Rightarrow t=1\).

\(K=\displaystyle\int\limits_{0}^{1}{f(t)}\mathrm{\,d}t=2 \Leftrightarrow \displaystyle\int\limits_{0}^{1}{f(x)}\mathrm{\,d}x=2\).

Như vậy ta có: \(I=\displaystyle\int\limits_{0}^{3}{f(x)}\mathrm{\,d}x = \displaystyle\int\limits_{0}^{1}{f(x)}\mathrm{\,d}x + \displaystyle\int\limits_{1}^{3}{f(x)}\mathrm{\,d}x = 2+8=10\).

Đặt \(t=\sqrt{5x^2+4}\Rightarrow t^2=5x^2+4\Rightarrow t\mathrm{\,d}t=5x\mathrm{\,d}x\). Do đó

\[ \begin{aligned} I=&\ \displaystyle\int\limits_0^1 {\displaystyle\frac{x\mathrm{\,d}x}{\sqrt{5x^2+4}}}=\displaystyle\frac{1}{5}\displaystyle\int\limits_2^3\mathrm{\,d}t =\displaystyle\frac{1}{5} \Rightarrow a=1;\ b=5. \end{aligned} \]

Vậy \(T=1^2+5^2=26\).

Đặt \(t=\sqrt{x}\Rightarrow \mathrm{\, d}t=\displaystyle\frac{1}{2\sqrt{x}}\mathrm{\, d}x\Rightarrow \displaystyle\frac{\mathrm{d}x}{\sqrt{x}}=2\mathrm{\, d}t\).

Với \(x=4\) thì \(t=2\), với \(x=1\) thì \(t=1\).

\[ \begin{aligned} I=&\ 2\displaystyle\int\limits_1^2\mathrm{e}^t\mathrm{\, d}t. \end{aligned} \]

Đặt \(t=\sqrt{3x+1}\Rightarrow t^2=3x+1\Rightarrow 2t\mathrm{d}t=3\mathrm{d}x\).

Đổi cận: \(x=1\Rightarrow t=2;x=5\Rightarrow t=4\).

\[ \begin{aligned} I=&\ \int\limits_1^5{\displaystyle\frac{\mathrm{d}x}{x\sqrt{3x+1}}}=\displaystyle\frac{2}{3}\int\limits_2^4{\displaystyle\frac{t\mathrm{d}t}{\displaystyle\frac{t^2-1}{3}\cdot t}}=2\int\limits_2^4{\displaystyle\frac{\mathrm{d}t}{t^2-1}}\\ =\ & \int\limits_2^4{\left(\displaystyle\frac{1}{t-1}-\displaystyle\frac{1}{t+1}\right)}\mathrm{d}t=\left. \ln \left| \displaystyle\frac{t-1}{t+1}\right|\right|_2^4=2\ln 3-\ln 5. \end{aligned} \]

Khi đó \(a=2,b=-1\Rightarrow a^2+ab+3b^2=4-2+3=5\).

Nhân cả tử và mẫu với lượng liên hợp của \(3x+\sqrt{9x^2-1}\) ta được

\[ \begin{aligned} I=&\ \int\limits _{\tfrac{1}{3}}^1\displaystyle\frac{x}{3x+\sqrt{9x^2-1}}\mathrm{\, d}x=\int\limits _{\tfrac{1}{3}}^1x(3x-\sqrt{9x^2-1})\mathrm{\, d}x\\ =\ & 3\int\limits _{\tfrac{1}{3}}^1 x^2\mathrm{\, d}x-\int\limits _{\tfrac{1}{3}}^1 x\sqrt{9x^2-1}\mathrm{\, d}x. \end{aligned} \]

Đặt \(u=9x^2-1\Rightarrow \mathrm{\, d}u=18x \mathrm{\, d}x\) và đổi cận, ta được

\(\begin{aligned} I=\ & x^3\bigg |^1_{\tfrac{1}{3}}-\displaystyle\frac{1}{18}\int\limits^1_{\tfrac{1}{3}}\sqrt{9x^2-1}\cdot 18x\mathrm{\, d}x\\ =\ &x^3\bigg |^1_{\tfrac{1}{3}}-\displaystyle\frac{1}{18}\int\limits_0^8\sqrt{u}\mathrm{\, d}u=\displaystyle\frac{26}{27}+\left (-\displaystyle\frac{u^{\tfrac{3}{2}}}{27}\right )\Bigg |_0^8=\displaystyle\frac{26}{27}-\displaystyle\frac{16\sqrt{2}}{27}\end{aligned}\).

Xét tích phân: \(I=\displaystyle\int\limits_1^{\mathrm{e}}\displaystyle\frac{\sqrt{3+\ln x}}{x}\mathrm{\,d}x\).

Đặt \(u=\sqrt{3+\ln x}\Rightarrow u^2=3+\ln x \Rightarrow 2u\mathrm{\,d}u=\displaystyle\frac{1}{x}\mathrm{\,d}x\).

Khi \(x=1\) thì \(u=\sqrt{3}\), khi \(x=\mathrm{e}\) thì \(u=2\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{\sqrt{3}}^{2}u\cdot 2u\mathrm{\,d}u=2\displaystyle\int\limits_{\sqrt{3}}^{2}u^2\mathrm{\,d}u=\left.\displaystyle\frac{2}{3}u^3\right|_{\sqrt{3}}^2=\displaystyle\frac{16-6\sqrt{3}}{3}. \end{aligned} \]

Suy ra \(a=16\), \(b=6\), \(c=3\). Do đó \(S=a+b+c=16+6+3=25\).

\[ \begin{aligned} a\sqrt{5}+b\sqrt{2}+c =\ & \int\limits_{1}^{2}\displaystyle\frac{x^3\mathrm{\,d}x}{\sqrt{x^2+1}-1}\\ =\ & \int\limits_{1}^{2}\displaystyle\frac{x^3(\sqrt{x^2}+1)\mathrm{\,d}x}{x^2+1-1}\\ =\ & \int\limits_{1}^{2}x(\sqrt{x^2+1}+1)\mathrm{\,d}x\\ =\ & \int\limits_{1}^{2}(\sqrt{x^2+1})^2\mathrm{\,d}(\sqrt{x^2+1})+\int\limits_{1}^{2}x\mathrm{\,d}x\\ =\ & \displaystyle\frac{(\sqrt{x^2+1})^3}{3}\bigg|_{1}^{2}+\displaystyle\frac{x^2}{2}\bigg|_{1}^{2}\\ =\ & \displaystyle\frac{5\sqrt{5}-2\sqrt{3}}{3}+\displaystyle\frac{3}{2}. \end{aligned} \]

Do đó \(a=\displaystyle\frac{5}{3}\), \(b=-\displaystyle\frac{2}{3}\) và \(c=\displaystyle\frac{3}{2}\) hay \(P=\displaystyle\frac{5}{2}\).

Gọi \(I=\displaystyle \int \limits_0^1 (2x-2)f'(x)\mathrm {\,d}x\).

Đặt \(\left\{\begin{aligned} & u=2x-2\\ & \mathrm{\,d}v=f'(x)\mathrm{\,d}x\end{aligned} \right.\Rightarrow\) \(\left\{\begin{aligned} & \mathrm {\,d}u=2\mathrm{\,d}x\\ & v=f(x)\end{aligned}\right.\)

\[ \begin{aligned} I=&\ (2x-2)f(x) \bigg|_{0}^{1}-\displaystyle \int \limits_0^1 2f(x) \mathrm {\,d}x\\ \Leftrightarrow\ & 6=2f(0)-2\displaystyle \int \limits_0^1 f(x) \mathrm {\,d}x \Rightarrow \displaystyle \int \limits_0^1 f(x) \mathrm {\,d}x=f(0)-3=3. \end{aligned} \]

\[ \begin{aligned} I=&\ \displaystyle\int\limits_1^{\mathrm{e}} (1+x\ln x) \mathrm{\,d}x=\displaystyle\int\limits_1^{\mathrm{e}} 1 \mathrm{\,d}x+\displaystyle\int\limits_1^{\mathrm{e}} x\ln x \mathrm{\,d}x\\ =\ & \mathrm{e}-1+\displaystyle\int\limits_1^{\mathrm{e}} x\ln x \mathrm{\,d}x. \end{aligned} \]

Đặt \(\left\{\begin{aligned} & u=\ln x\\ & \mathrm{d}v=x\mathrm{\,d}x \end{aligned}\right. \Rightarrow\left\{\begin{aligned} & \mathrm{d}u=\displaystyle\frac{1}{x}\mathrm{\,d}x\\ & v=\displaystyle\frac{x^2}{2}.\end{aligned}\right.\)

Khi đó

\(\displaystyle\int\limits_1^{\mathrm{e}} x\ln x \mathrm{\,d}x=\displaystyle\frac{x^2}{2}\ln x \bigg|_1^{\mathrm{e}} -\displaystyle\frac{1}{2}\displaystyle\int\limits_1^{\mathrm{e}} x \mathrm{\,d}x=\displaystyle\frac{\mathrm{e}^2}{2}-\displaystyle\frac{1}{4}x^2\bigg|_1^{\mathrm{e}} =\displaystyle\frac{\mathrm{e}^2}{2}-\displaystyle\frac{\mathrm{e}^2}{4}+\displaystyle\frac{1}{4}=\displaystyle\frac{\mathrm{e}^2}{4}+\displaystyle\frac{1}{4}\).

Suy ra

\(\displaystyle\int\limits_1^{\mathrm{e}} (1+x\ln x) \mathrm{\,d}x= \mathrm{e}-1+\displaystyle\frac{\mathrm{e}^2}{4}+\displaystyle\frac{1}{4}=\displaystyle\frac{\mathrm{e}^2}{4}+ \mathrm{e}-\displaystyle\frac{3}{4}\) nên \(a=\displaystyle\frac{1}{4}\), \(b=1\), \(c=-\displaystyle\frac{3}{4}\).

Vậy \(a-b=c\).

Đặt \(I=\displaystyle\int\limits_0^{\frac{\pi}{4}}{\left( x+1 \right)\cos 2x}\mathrm{\, d}x\).

Đặt \(\left\{\begin{aligned}& u=x+1\\ & \mathrm{\, d}v=\cos 2x \mathrm{\, d}x \end{aligned}\right. \Rightarrow \left\{\begin{aligned}& \mathrm{\, d}u =\mathrm{ d}x\\ & v=\displaystyle\frac{1}{2}\sin 2x.\end{aligned}\right.\).

\[ \begin{aligned} I=&\ \left. \displaystyle\frac{1}{2}(x+1) \sin 2x \right|_0^{\frac{\pi}{4}}- \displaystyle\frac{1}{2}\displaystyle\int\limits_0^{\frac{\pi}{4}}{\sin 2x}\mathrm{\, d}x= \displaystyle\frac{1}{2} \left( \displaystyle\frac{\pi}{4}+1\right)+\displaystyle\frac{1}{4} \cos 2x \Bigg|_0^{\tfrac{\pi}{4}}= \displaystyle\frac{\pi}{8}+\displaystyle\frac{1}{4}. \end{aligned} \]

Vậy \(a \cdot b=8\cdot4=32\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{1}^{2}\displaystyle\frac{\ln x}{x^{2}}\mathrm{\,d}x=-\displaystyle\int\limits_{1}^{2}\ln x\mathrm{\,d}\left(\displaystyle\frac{1}{x}\right)=-\displaystyle\frac{\ln x}{x}\Big |_{1}^{2}+\displaystyle\int\limits_{1}^{2}\displaystyle\frac{1}{x^{2}}\mathrm{\,d}x\\ =\ &-\displaystyle\frac{\ln 2}{2}-\displaystyle\frac{1}{x}\Big |_{1}^{2}=\displaystyle\frac{1}{2}-\displaystyle\frac{\ln 2}{2}\\ \Rightarrow\ & a=-\displaystyle\frac{1}{2},\ b=1,\ c=2\Rightarrow T=4. \end{aligned} \]

\[ \begin{aligned} \displaystyle\int\limits_{0}^{\pi} (3x+2)\cos^2 x\mathrm{\,d}x =\ & \displaystyle\int\limits_{0}^{\pi} (3x+2)\cdot \displaystyle\frac{1+\cos 2x}{2}\mathrm{\,d}x\\ =\ &\displaystyle\frac{1}{2} \displaystyle\int\limits_{0}^{\pi} (3x+2)\mathrm{\,d}x + \displaystyle\int\limits_{0}^{\pi}\cos 2x\mathrm{\,d}x+ \displaystyle\frac{3}{2}\displaystyle\int\limits_{0}^{\pi} x\cos 2x\mathrm{\,d}x\\ =\ &\displaystyle\frac{1}{2}\left( \displaystyle\frac{3x^2}{2}+2x \right)\Bigg|_0^{\pi} + \left(\displaystyle\frac{\sin 2x}{2} \right)\Bigg|_0^{\pi} + \left(\displaystyle\frac{x\sin 2x}{2}\right)\Bigg|_0^{\pi} -\displaystyle\int\limits_{0}^{\pi} \displaystyle\frac{\sin 2x}{2}\mathrm{\,d}x\\ =\ & \displaystyle\frac{3\pi^2}{4}+\pi + \displaystyle\frac{\cos 2x}{4}\Bigg|_0^\pi\\ =\ &\displaystyle\frac{3\pi^2}{4}+\pi. \end{aligned} \]

Đặt \( \left\{\begin{aligned}&u=\ln x\\&v =(4x-1)\mathrm{\,d}x\end{aligned}\right. \Rightarrow \left\{\begin{aligned} &\mathrm{\,d}u =\displaystyle\frac{1}{x}\mathrm{\,d}x\\ &v=2x^2-x.\end{aligned}\right.\)

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{1}^{2}\left(4x-1\right)\ln x \mathrm{\,d}x =\left. (2x^2-x)\ln x\right|_1^2 -\displaystyle\int\limits_{1}^{2}(2x-1)\mathrm{dx}\\ =\ &6\ln 2-\left. (x^2-x)\right|_1^2 =6\ln 2-2\Rightarrow a=6,b=-2. \end{aligned} \]

Vậy \( 2a+b=10. \)

Đặt \(\left\{\begin{aligned} &u = x\\ & \mathrm{d}v = \cos 2x\, \mathrm{d} x\end{aligned}\right. \Rightarrow \left\{\begin{aligned} & \mathrm{d} u = \mathrm{d} x\\ &v = \displaystyle\frac{1}{2}\sin 2x\end{aligned}\right.\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{0}^{1}x\cos 2x\, \mathrm{d}x = \displaystyle\frac{x}{2}\sin 2x\Big\vert_{0}^{1} - \displaystyle\frac{1}{2}\displaystyle\int\limits_{0}^{1}\sin 2x\, \mathrm{d}x\\ =\ & \displaystyle\frac{x}{2}\sin 2x\Big\vert_{0}^{1} + \displaystyle\frac{1}{4}\cos 2x\Big\vert_{0}^{1} = \displaystyle\frac{1}{2}\sin 2 + \displaystyle\frac{1}{4}\cos 2 - \displaystyle\frac{1}{4}. \end{aligned} \]

Suy ra \(a = 2\), \(b = 1\) và \(c = - 1\) nên \(a - b + c = 0\).

\[ \begin{aligned} J=&\ \displaystyle \int \limits_1^3 (4x-6) \cdot \mathrm{e}^{2x} \mathrm{d}x=\displaystyle \int_1^3(2x-3)\mathrm{d}(\mathrm{e}^{2x})=(2x-3)\mathrm{e}^{2x}\bigg|_1^3-\displaystyle \int_1^3 2\mathrm{e}^{2x} \mathrm{d}x\\ =\ &(2x-4)\mathrm{e}^{2x}\bigg|_1^3=2\mathrm{e}^6+2\mathrm{e}^2. \end{aligned} \]

Suy ra \(m=n=2\Rightarrow J=0.\)

Đặt \(\left\{ \begin{aligned} &u=\ln x \\ &\mathrm{\,d}v=(2x-1)\mathrm{\,d}x \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} &\mathrm{\,d}u=\displaystyle\frac{1}{x}\mathrm{\,d}x \\ &v=x^2-x \end{aligned} \right.\).

\[ \begin{aligned} I=&\ \displaystyle \int\limits_1^2 (2x-1)\ln x\mathrm{\,d}x=(x^2-x)\ln x\Bigg|_1^2-\displaystyle \int\limits_1^2 (x-1)\mathrm{\,d}x\\ =\ & (x^2-x)\ln x\Bigg|_1^2-\left(\displaystyle\frac{x^2}{2}-x\right)\Bigg|_1^2=2\ln 2-\displaystyle\frac{1}{2}. \end{aligned} \]

Khi đó \(a=2\), \(b=-\displaystyle\frac{1}{2}\) suy ra \(a+b=\displaystyle\frac{3}{2}\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits_{0}^{\pi}x\sin x\, \mathrm{d}x = - x\cdot \cos x\Big\vert^{\pi}_{0} + \displaystyle\int\limits_{0}^{\pi}\cos x\, \mathrm{d}x = - x\cdot \cos x\Big\vert^{\pi}_{0} + \sin x\Big\vert^{\pi}_{0} = \pi. \end{aligned} \]

Suy ra \(a = 1\) và \(b = 0\) nên \(a + b = 1\).

Đặt \(\left\{\begin{aligned}&u=4x+3 \\& \mathrm{d} v=\mathrm{e}^x \mathrm{d} x\end{aligned}\right. \Rightarrow \left\{\begin{aligned}&\mathrm{d} u = 4 \mathrm{d} x \\ & v = \mathrm{e}^x\end{aligned}\right.\).

\[ \begin{aligned} I=&\ \displaystyle (4x+3) \mathrm{e}^x \bigg|_0^1 - \int \limits_{0}^{1} 4\mathrm{e}^x \mathrm{d} x = 7\mathrm{e} -3 - 4 \mathrm{e} + 4 = 3 \mathrm{e} +1. \end{aligned} \]

Đặt \(\left\{\begin{aligned}& u=2x+1\\ \mathrm{d}&v=\sin x\mathrm{d}x\end{aligned}\right. \Rightarrow \left\{\begin{aligned}\mathrm{d}u&=2\mathrm{d}x\\ v&=-\cos x.\end{aligned}\right.\)

\[ \begin{aligned} I=&\ -(2x+1)\cos x\bigg|_0^\pi +2\displaystyle \int \limits_0^\pi \cos x \mathrm{d}x=2\pi +2+2\sin x\bigg|_0^\pi=2\pi+2. \end{aligned} \]

Đặt \(\left\{\begin{aligned} &u=\ln x\\ \mathrm{d}&v=x\mathrm{\,d}x\end{aligned}\right. \Rightarrow \left\{\begin{aligned}\mathrm{d}u &=\displaystyle\frac{1}{x}\mathrm{\,d}x\\ v& =\displaystyle\frac{x^2}{2}.\end{aligned}\right.\)

\[ \begin{aligned} I=&\ \displaystyle \int \limits_1^{\mathrm{e}}x\ln x \mathrm{\,d}x=\left(\displaystyle\frac{x^2}{2}\ln x\right) \Bigr|^{\mathrm{e}}_1 - \displaystyle \int \limits_1^{\mathrm{e}}\displaystyle\frac{x}{2} \mathrm{\,d}x= \displaystyle\frac{\mathrm{e}^2}{2}-\displaystyle\frac{x^2}{4} \Bigr|^{\mathrm{e}}_1=\displaystyle\frac{\mathrm{e}^2}{2}-\left(\displaystyle\frac{\mathrm{e}^2}{4}-\displaystyle\frac{1}{4}\right)\\ =\ &\displaystyle\frac{\mathrm{e}^2}{4}+\displaystyle\frac{1}{4}=\displaystyle\frac{1}{4}(\mathrm{e}^2+1) \end{aligned} \]

Đặt \(\left\{\begin{aligned}&u=2x+1\\&\mathrm{d}v=\mathrm{e}^x\mathrm{\,d}x\end{aligned}\right. \Rightarrow\left\{\begin{aligned}&\mathrm{d}u=2\mathrm{d}x\\ &v=\mathrm{e}^x\end{aligned}\right.\).

\[ \begin{aligned} I=&\ \displaystyle\int\limits_0^1(2x+1)\mathrm{e}^x\mathrm{\,d}x=\left((2x+1)\mathrm{e}^x\right)\biggr|_0^1-2\displaystyle\int\limits_0^1\mathrm{e}^x\mathrm{\,d}x\\ =\ &3\mathrm{e}-1-2\mathrm{e}^x\biggr|_0^1=3\mathrm{e}-1-2\mathrm{e}+2=\mathrm{e}+1. \end{aligned} \]

Suy ra \(a=1,b=1\). Vậy \(ab=1\).

\[ \begin{aligned} I =\ & \displaystyle\int\limits_{1}^{\mathrm{e}} x\ln x\mathrm{\, d} x=\displaystyle\int\limits_{1}^{\mathrm{e}} \ln x\mathrm{\, d} \left(\displaystyle\frac{x^2}{2}\right) \\ =\ & \left.\displaystyle\frac{x^2}{2}\ln x\right |_1^{\mathrm{e}}-\displaystyle\int\limits_{1}^{\mathrm{e}} \displaystyle\frac{x^2}{2x}\mathrm{\, d} x=\left.\left(\displaystyle\frac{x^2}{2}\ln x-\displaystyle\frac{x^2}{4}\right)\right|_1^{\mathrm{e}}\\ =\ &\displaystyle\frac{\mathrm{e}^2}{2}-\displaystyle\frac{\mathrm{e}^2}{4}+\displaystyle\frac{1}{4}=\displaystyle\frac{\mathrm{e}^2+1}{4}. \end{aligned} \]

Suy ra \(a=1, b=1\). Vậy \(a+b=2\).

Đặt \(\left\{\begin{aligned}&u=\ln (x^2+16)\\& \mathrm{\,d}v=x\mathrm{\,d}x\end{aligned}\right. \Rightarrow \left\{\begin{aligned}&\mathrm{\,d}u=\displaystyle\frac{2x}{x^2+16}\mathrm{\,d}x\\ & v=\displaystyle\frac{1}{2}(x^2+16).\end{aligned}\right.\)

\[ \begin{aligned} \displaystyle \int\limits_{0}^{3}x\ln (x^2+16)\mathrm{\,d}x =\ &\left[\displaystyle\frac{1}{2}(x^2+16)\cdot \ln (x^2+16)\right] \bigg|_0^3-\int\limits_{0}^{3}\displaystyle\frac{2x}{x^2+16}\cdot\displaystyle\frac{1}{2}(x^2+16) \mathrm{\,d}x\\ =\ &\displaystyle\frac{1}{2}\cdot 25\cdot \ln 25-\displaystyle\frac{1}{2}\cdot 16\cdot \ln 16-\displaystyle\int\limits_{0}^{3}x\mathrm{\,d}x\\ =\ &25\ln 5-16\ln 4-\displaystyle\frac{x^2}{2}\bigg|_0^3\\&=25\ln 5-32\ln 2-\displaystyle\frac{9}{2}. \end{aligned} \]

Suy ra \(a=25,b=-32,c=-9 \Rightarrow T=a+b+c=25-32-9=-16\).

Theo giả thiết \( f(x) +x\cdot f'(x)=3x^2 +2x, \,\, \forall x \in \mathbb{R} \).
Ta có \(\left (xf(x)\right )'=3x^2+2x\Rightarrow \displaystyle\int\limits_{0}^{1}\left(x f(x)\right)'\mathrm{\,d}x =\displaystyle\int\limits_{0}^{1} (3x^2+2x)\mathrm{\,d}x=2\Rightarrow \left(xf(x)\right) \bigg|_0^1 =2 \Rightarrow f(1) =2.\)

Từ \(\displaystyle\int\limits_0^1 f(2x) \mathrm{\,d}x=2\) ta đặt \(t=2x\) được \[\displaystyle\frac{1}{2}\displaystyle\int\limits_0^2 f(t) \mathrm{\,d}t=2 \Rightarrow \displaystyle\int\limits_0^2 f(x) \mathrm{\,d}x=4 \quad (1).\] Từ \[I=\displaystyle\int\limits_0^2 x\cdot f'(x) \mathrm{\,d}x=[x\cdot f(x)]\bigg|_0^2-\displaystyle\int\limits_0^2 f(x) \mathrm{\,d}x=28.\]

Ta có \[f'(x)=2x{\left[f(x)\right]}^2\overset{f(x)\ne 0}{\mathop{\Leftrightarrow}}\displaystyle\frac{f'(x)}{{\left[f(x)\right]}^2}=2x \Leftrightarrow {\left[\displaystyle\frac{1}{f(x)}\right]}'=-2x \Leftrightarrow \displaystyle\frac{1}{f(x)}=-x^2+C.\] Từ \(f(2)=-\displaystyle\frac{2}{9}\) suy ra \(C=-\displaystyle\frac{1}{2}\).
Do đó \(f(1)=\displaystyle\frac{1}{-1^2+\left(-\displaystyle\frac{1}{2}\right)}=-\displaystyle\frac{2}{3}\).

Ta có \[f'(x)=4x^3[f(x)]^2 \Rightarrow -\displaystyle\frac{f'(x)}{{\left[f(x)\right]}^2}=-4x^3 \Rightarrow {\left[\displaystyle\frac{1}{f(x)}\right]}'=-4x^3 \Rightarrow \displaystyle\frac{1}{f(x)}=-x^4+C.\] Do \(f(2)=-\displaystyle\frac{1}{25}\) nên ta có \(C=-9\). Do đó \[f(x)=-\displaystyle\frac{1}{x^4+9} \Rightarrow f(1)=-\displaystyle\frac{1}{10}.\]

Xét tích phân \(I=\displaystyle\int\limits_0^1 xf(x)\mathrm{\,d}x\).
Đặt \(\left\{\begin{aligned} &u=f(x)\\ &\mathrm{\,d}v=x\mathrm{\,d}x\end{aligned}\right. \Rightarrow \left\{\begin{aligned} &\mathrm{\,d}u=f'(x)\mathrm{\,d}x\\ &v=\displaystyle\frac{x^2}{2}.\end{aligned}\right.\)
Khi đó \[I=\left.\displaystyle\frac{x^2}{2}f(x)\right|_0^1-\displaystyle\int\limits_0^1 \displaystyle\frac{x^2}{2}f'(x)\mathrm{\,d}x\Rightarrow \displaystyle\int\limits_0^1 x^2f'(x)\mathrm{\,d}x=4\Rightarrow \displaystyle\int\limits_0^1 40x^2f'(x)\mathrm{\,d}x=160.\]

Ta có: \(\displaystyle\int\limits_0^1 400x^4\mathrm{\,d}x=80\), \(\displaystyle\int\limits_0^1\left[f'(x)\right]^2\mathrm{\,d}x=80\Rightarrow \displaystyle\int\limits_0^1\left[f'(x)-20x^2\right]^2\mathrm{\,d}x=0\Rightarrow f'(x)=20x^2\).
Nên \(f(x)=\displaystyle\frac{20x^3}{3}+C\) và \(f(1)=0\Rightarrow f(x)=\displaystyle\frac{20x^3}{3}-\displaystyle\frac{20}{3}\).
Do đó: \(\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x=\left.\left(\displaystyle\frac{5}{3}x^4-\displaystyle\frac{20}{3}x\right)\right|_0^1=-5\).

Ta có \( f(x) = f'(x)\cdot \sqrt{3x+1} \Leftrightarrow \displaystyle\frac{f'(x)}{f(x)} = \displaystyle\frac{1}{ \sqrt{3x+1} }\).
Khi đó \( \displaystyle \int \limits^5_1 \displaystyle\frac{f'(x)}{f(x)} \mathrm{d\, }x= \int \limits^5_1 \displaystyle\frac{1}{\sqrt{3x+1}} \mathrm{d\, }x = \displaystyle\frac{4}{3} \Rightarrow \ln \left [ f(5) \right ] = \displaystyle\frac{4}{3} \Leftrightarrow f(5) = e^{ \frac{4}{3} } \approx 3{ , }79\).

Thay \(x\) bởi \(1-x\) vào đẳng thức đề bài ta có \[2f(1-x)+3f(1-1+x)=\sqrt{1-1+x} \Leftrightarrow 2f(1-x)+3f(x)=\sqrt{x}.\] Ta xét hệ phương trình \(\heva{&2f(1-x)+3f(x)=\sqrt{x}\\&3f(1-x)+2f(x)=\sqrt{1-x}.}\Rightarrow f(x)=\displaystyle\frac{1}{5}\left(3\sqrt{x}-2\sqrt{1-x}\right)\).
Vậy \(I=\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x=\displaystyle\frac{1}{5}\displaystyle\int\limits_0^1 \left(3\sqrt{x}-2\sqrt{1-x}\right)\mathrm{\,d}x=\displaystyle\frac{2}{3}\).

Ta có \[\begin{aligned} 2=\ &\displaystyle\int\limits_0^{\frac{\pi}{4}}{(\sin x \tan x f(x)) \mathrm{\,d}x}=\displaystyle\int\limits_0^{\frac{\pi}{4}}{\displaystyle\frac{\sin^2x}{\cos x} f(x) \mathrm{\,d}x} \\ =\ &\displaystyle\int\limits_0^{\frac{\pi}{4}}{\displaystyle\frac{f(x)}{\cos x} \mathrm{\,d}x}-\displaystyle\int\limits_0^{\frac{\pi}{4}}{\cos x f(x) \mathrm{\,d}x}=1-\displaystyle\int\limits_0^{\frac{\pi}{4}}{\cos x f(x) \mathrm{\,d}x}\\ \Rightarrow\ & \displaystyle\int\limits_0^{\frac{\pi}{4}}{\cos x f(x) \mathrm{\,d}x}=-1. \end{aligned}\] Đặt \(\left\{\begin{aligned} & u=f(x) \\ & \mathrm{\,d}v=\cos x \mathrm{\,d}x\end{aligned}\right.\Rightarrow \left\{\begin{aligned} & \mathrm{\,d}u=f'(x) \mathrm{\,d}x \\ & v=\sin x. \end{aligned}\right.\) \[\begin{aligned} \Rightarrow-1=\ &\displaystyle\int\limits_0^{\frac{\pi}{4}}{\cos x f(x) \mathrm{\,d}x}=(f(x) \sin x)\bigg|_0^{\frac{\pi}{4}}-\displaystyle\int\limits_0^{\frac{\pi}{4}}{\sin x f'(x) \mathrm{\,d}x}\\ =\ &f\left(\displaystyle\frac{\pi}{4}\right) \displaystyle\frac{\sqrt{2}}{2}-\displaystyle\int\limits_0^{\frac{\pi}{4}}{\sin x f'(x) \mathrm{\,d}x} =\displaystyle\frac{3\sqrt{2}}{2}-\displaystyle\int\limits_0^{\frac{\pi}{4}}{\sin x f'(x) \mathrm{\,d}x}\\ \Rightarrow\ & \displaystyle\int\limits_0^{\frac{\pi}{4}}{\sin x f'(x) \mathrm{\,d}x}=\displaystyle\frac{2+3\sqrt{2}}{2}. \end{aligned}\]

Xét \(K=\displaystyle\int\limits_0^{\frac{\pi}{4}}f\left(\tan x\right) \mathrm{\,d}x=3\).
Đặt \(t=\tan x \Rightarrow \mathrm{d}x=\displaystyle\frac{\mathrm{d}t}{t^2+1}\).
Suy ra \(K=\displaystyle\int\limits_0^1 \displaystyle\frac{f(t)}{t^2+1}\mathrm{\,d}t=3\).
Xét \(J=\displaystyle\int\limits_0^1 \displaystyle\frac{x^2 f(x)}{x^2+1}\mathrm{\,d}x=\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x-\displaystyle\int\limits_0^1 \displaystyle\frac{f(x)}{x^2+1}\mathrm{\,d}x =1 \Rightarrow \displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x-3=1\).
Vậy \(I=\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x=4\).

Ta có: \[\begin{aligned} &f'\left( x \right)\sqrt{x^2+1}=2x\sqrt{f\left( x \right)+1}\\ \Leftrightarrow\ & \displaystyle\frac{f'\left( x \right)}{\sqrt{f\left( x \right)+1}}=\displaystyle\frac{2x}{\sqrt{x^2+1}} \\ \Leftrightarrow\ & \displaystyle\int\limits_0^{\sqrt{3}} \displaystyle\frac{f'\left( x \right)}{\sqrt{f\left( x \right)+1}}\mathrm{\,d}x=\displaystyle\int\limits_{0}^{\sqrt{3}}\displaystyle\frac{2x}{\sqrt{x^2+1}}\mathrm{\,d}x \\ \Leftrightarrow\ & 2\sqrt{f\left( x \right)+1}\Big|_0^{\sqrt{3}} = 2\sqrt{x^2+1}\big|_0^{\sqrt{3}} \\ \Leftrightarrow\ & 2\sqrt{f\left( \sqrt{3} \right)+1}-2 =4-2\\ \Leftrightarrow\ & f\left( \sqrt{3} \right)=3. \end{aligned}\]

Ta có \(T=f(1)-f(0)=\displaystyle\int\limits_0^1{f'(x) \mathrm{\, d}x}\)
Lại có \(\left[f'(x)\right]^2=f''(x)\Leftrightarrow -1=-\displaystyle\frac{f''(x)}{\left[f'(x)\right]^2}\Leftrightarrow -1=\left[\displaystyle\frac{1}{f'(x)}\right]'\Leftrightarrow -x+c=\displaystyle\frac{1}{f'(x)}\).
Mà \(f'(0)=-1\) nên \(c=-1\).
Vậy \(T=\displaystyle\int\limits_0^1{f'(x) \mathrm{\, d}x}=\displaystyle\int\limits_0^1{\displaystyle\frac{1}{-x-1} \mathrm{\, d}x}=-\ln \left|-x-1\right|\bigg|_0^1 =-\ln 2\).

Ta có \( 3f'(x)+2f^{2}(x)=0\Leftrightarrow \displaystyle\frac{f'(x)}{f^{2}(x)}=-\displaystyle\frac{2}{3} \). Lấy tích phân hai vế ta được \[ \displaystyle\int\limits_{0}^{1}\displaystyle\frac{f'(x)}{f^{2}(x)}\mathrm{d}x=-\displaystyle\int\limits_{0}^{1}\displaystyle\frac{2}{3}\mathrm{d}x\Leftrightarrow -\displaystyle\frac{1}{f(x)}\Big |_{0}^{1}=-\displaystyle\frac{2}{3}x\Big |_{0}^{1}\Leftrightarrow \displaystyle\frac{1}{f(1)}-1=\displaystyle\frac{2}{3}\Leftrightarrow \displaystyle\frac{1}{f(1)}=\displaystyle\frac{5}{3}\Leftrightarrow f(1)=\displaystyle\frac{3}{5}. \]

Đặt \(t=4-x \Rightarrow \mathrm{\,d}x=-\mathrm{\,d}t\).
Với \(x=1 \Rightarrow t=3\), \(x=3 \Rightarrow t=1\).
\(\displaystyle\int\limits_1^3 x\cdot f(x)\mathrm{\,d}x=-\displaystyle\int\limits_3^1 (4-t)\cdot f(4-t)\mathrm{\,d}t=\displaystyle\int\limits_1^3 (4-x)\cdot f(x)\mathrm{\,d}x \).
Suy ra \[2\displaystyle\int\limits_1^3 x\cdot f(x)\mathrm{\,d}x=4\displaystyle\int\limits_1^3 f(x)\mathrm{\,d}x\] hay \[\displaystyle\int\limits_1^3 f(x)\mathrm{\,d}x=-1.\]

Xét \(A=\displaystyle\int\limits_0^{\frac{\pi}{4}} f(\tan x)\mathrm{\,d}x\).
Đặt \(t=\tan x \Rightarrow \mathrm{\,d}t=\left(1+\tan^2x\right)\mathrm{\,d}x \Rightarrow \mathrm{\,d}x =\displaystyle\frac{\mathrm{\,d}t}{1+t^2}.\)
Khi \(x=0\) thì \(t=0\); khi \(x= \displaystyle\frac{\pi}{4}\) thì \(t=1\).
Ta có \(A= \displaystyle\int\limits_0^1 \displaystyle\frac{f(t)}{t^2+1}\mathrm{\,d}t = \displaystyle\int\limits_0^1 \displaystyle\frac{f(x)}{x^2+1}\mathrm{\,d}x \Rightarrow \displaystyle\int\limits_0^1 \displaystyle\frac{f(x)}{x^2+1}\mathrm{\,d}x =3\quad (1).\)
Mà theo giả thiết, ta có \(\displaystyle\int\limits_0^1 \displaystyle\frac{x^2 f(x)}{x^2+1}\mathrm{\,d}x=1 \quad (2).\)
Lấy \((1)\) cộng \((2)\) vế với vế, ta được \(\displaystyle\int\limits_0^1 \displaystyle\frac{x^2 f(x)+f(x)}{x^2+1}\mathrm{\,d}x=4 \Leftrightarrow \displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x=4\).

Từ giả thiết, ta có \[\displaystyle\frac{f'(x)}{f^2(x)}=-(2x+4) \Rightarrow \displaystyle\int\limits \displaystyle\frac{f'(x)}{f^2(x)}\mathrm{\,d}x=- \displaystyle\int\limits (2x+4)\mathrm{\,d}x \Rightarrow \displaystyle\int\limits \displaystyle\frac{\mathrm{\,d}f(x)}{f^2(x)}=-\displaystyle\int\limits (2x+4)\mathrm{\,d}x. \] Suy ra \(\displaystyle\frac{1}{f(x)}=x^2+4x+C\). Vì \(f(2)=\displaystyle\frac{1}{15} \Rightarrow C= 3\) nên \(f(x)=\displaystyle\frac{1}{x^2+4x+3}\).
Do đó \(S=f(1)+f(2)+f(3)=\displaystyle\frac{1}{8}+\displaystyle\frac{1}{15}+\displaystyle\frac{1}{24}=\displaystyle\frac{7}{30}.\)

Từ giả thiết \(2\cdot f(3x)+3\cdot f\left(\displaystyle\frac{2}{x}\right)=-\displaystyle\frac{15x}{2}\), suy ra \[2\displaystyle\int\limits_{1}^{3}f(3x)\mathrm{\,d}x+3\displaystyle\int\limits_{1}^{3}f\left(\displaystyle\frac{2}{x}\right)\mathrm{\,d}x=\displaystyle\int\limits_{1}^{3}\left(-\displaystyle\frac{15x}{2}\right)\mathrm{\,d}x=-30.\] Xét tích phân \(K=\displaystyle\int\limits_{1}^{3}f(3x)\mathrm{\,d}x\).
Đặt \(t=3x\Rightarrow \mathrm{\, d}x=\displaystyle\frac{1}{3}\mathrm{\,d}t\). Với \(x=1\Rightarrow t=3\); \(x=3\Rightarrow t=9\). Suy ra \[K=\displaystyle\int\limits_{3}^{9}f(t)\displaystyle\frac{1}{3}\mathrm{\,d}t=\displaystyle\frac{k}{3}.\] Xét tích phân \(L=\displaystyle\int\limits_{1}^{3}f\left(\displaystyle\frac{2}{x}\right)\mathrm{\,d}x\).
Đặt \(\displaystyle\frac{1}{t}=\displaystyle\frac{2}{x}\Leftrightarrow x=2t\Rightarrow \mathrm{\,d}x=2\mathrm{\,d}t\). Với \(x=1\Rightarrow t=\displaystyle\frac{1}{2}\); \(x=3\Rightarrow t=\displaystyle\frac{3}{2}\). Suy ra \[L=\displaystyle\int\limits_{\tfrac{1}{2}}^{\tfrac{3}{2}}f\left(\displaystyle\frac{1}{t}\right)2\mathrm{\,d}t=2I.\] Vậy ta có \[2\cdot\displaystyle\frac{k}{3}+3\cdot 2I=-30\Leftrightarrow I=-\displaystyle\frac{45+k}{9}.\]

Do \(f(x)\) đồng biến trên \([1;4]\) nên \(f'(x) \ge 0\), \(\forall x \in [1;4]\).
Suy ra \(f'(x) = \sqrt{x+2xf(x)} = \sqrt{x} \cdot \sqrt{1+2f(x)} \Rightarrow \displaystyle\frac{f'(x)}{\sqrt{1+2f(x)}} = \sqrt{x}\). Ta có \[\begin{aligned} & \displaystyle \int\limits_{1}^{4} \displaystyle\frac{f'(x)}{\sqrt{1+2f(x)}} \mathrm{\,d}x = \displaystyle \int\limits_{1}^{4} \sqrt{x} \mathrm{\,d}x\\ \Leftrightarrow\ & \displaystyle\frac{1}{2} \displaystyle \int\limits_{1}^{4} \displaystyle\frac{1}{\sqrt{1+2f(x)}} \mathrm{\,d}\left(2f(x)\right) = \displaystyle \int\limits_{1}^{4} \sqrt{x} \mathrm{\,d}x\\ \Leftrightarrow\ & \left(\sqrt{1+2f(x)}\right) \bigg|_{1}^{4} = \displaystyle\frac{2}{3} \sqrt{x^3} \bigg|_{1}^{4} \\ \Leftrightarrow\ & \sqrt{1+2f(4)} - \sqrt{1+2f(1)} = \displaystyle\frac{14}{3} \\ \Leftrightarrow\ & f(4) = \displaystyle\frac{391}{18}. \end{aligned}\]

Từ giả thiết ta có \[\mathrm{e}^{-x^3}f'(x)-3x^2\mathrm{e}^{-x^3}f(x)=2x \Rightarrow \left( \mathrm{e}^{-x^3}f(x) \right)'=2x.\] Suy ra \[1=\int\limits_0^1 2x\mathrm{\,d}x=\int\limits_0^1 \left( \mathrm{e}^{-x^3}f(x) \right)'\mathrm{\,d}x=\mathrm{e}^{-x^3}f(x)\Big|_0^1= \displaystyle\frac{f(1)}{\mathrm{e}}-1.\] Vậy \(f(1)=2\mathrm{e}\).

Đặt \(\left\{\begin{aligned} &u = x\\ &\mathrm{d}v = f'(2x)\mathrm{\,d}x\end{aligned}\right. \Rightarrow \left\{\begin{aligned} &\mathrm{d}u = \mathrm{d}x\\ &v = \displaystyle\frac{1}{2}f(2x).\end{aligned}\right.\)
Do đó \(\displaystyle \int \limits_{0}^{1}xf'(2x) \mathrm{\,d}x = \displaystyle\frac{x}{2}f(2x)\bigg|_0^1 - \displaystyle\frac{1}{2} \displaystyle \int \limits_{0}^{1}f(2x)\mathrm{\,d}x = \displaystyle\frac{1}{2} - \displaystyle\frac{1}{2}\cdot \displaystyle\frac{3}{2} = -\displaystyle\frac{1}{4}\).

Đặt \(t=\displaystyle\frac{1}{x}\), thay vào biểu thức \(f(x)+2f\left(\displaystyle\frac{1}{x}\right)=3x\), ta được \(f\left(\displaystyle\frac{1}{t}\right)+2f(t)=\displaystyle\frac{3}{t}\) hay \(f\left(\displaystyle\frac{1}{x}\right)+2f(x)=\displaystyle\frac{3}{x}\).
Suy ra \(f(x)=\displaystyle\frac{2}{x}-x\).
Vậy \(I=\displaystyle\int\limits_{\tfrac{1}{2}}^2\displaystyle\frac{f(x)}{x}\mathrm{d}x = \displaystyle\int\limits_{\tfrac{1}{2}}^2\left(\displaystyle\frac{2}{x^2}-1\right)\mathrm{d}x=\displaystyle\frac{3}{2}\).

Ta có \(I=\displaystyle\int\limits_{-2}^2 \displaystyle\frac{f(x)}{3^x+1} \mathrm{\,d}x=\displaystyle\int\limits_{-2}^0 \displaystyle\frac{f(x)}{3^x+1} \mathrm{\,d}x+\displaystyle\int\limits_0^2 \displaystyle\frac{f(x)}{3^x+1} \mathrm{\,d}x\).
Đặt \(x=-t \Rightarrow \mathrm{\,d}x=-\mathrm{d}t\); \(x=-2 \Rightarrow t=2;x=0 \Rightarrow t=0\)
\(\Rightarrow \displaystyle\int\limits_{-2}^0 \displaystyle\frac{f(x)}{3^x+1}\mathrm{\,d}x=\displaystyle\int\limits_2^0 \displaystyle\frac{f(-t)}{3^{-t}+1}(-\mathrm{d}t)=\displaystyle\int\limits_0^2 \displaystyle\frac{3^tf(-t)}{3^t+1}\mathrm{d}t=\displaystyle\int\limits_0^2 \displaystyle\frac{3^tf(t)}{3^t+1}\mathrm{d}t=\displaystyle\int\limits_0^2 \displaystyle\frac{3^xf(x)}{3^x+1}\mathrm{\,d}x\).
\( \Rightarrow I=\displaystyle\int\limits_0^2 \displaystyle\frac{3^xf(x)}{3^x+1}\mathrm{\,d}x+\displaystyle\int\limits_0^2 \displaystyle\frac{f(x)}{3^x+1}\mathrm{\,d}x=\displaystyle\int\limits_0^2 f(x)\mathrm{\,d}x=\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x+\displaystyle\int\limits_1^2 f(x)\mathrm{\,d}x=3\).

\(\bullet\) Ta có \(f'(x)\mathrm{e}^{f(x)} = 2x + 3 \Rightarrow \displaystyle \int \mathrm{e}^{f(x)}f'(x) \mathrm{\,d}x = \int (2x + 3) \mathrm{\,d}x \Rightarrow \mathrm{e}^{f(x)} = x^2 + 3x + C\).
Do \(f(0) = \ln 2\) nên \(\mathrm{e}^{\ln 2} = 0^2 + 3 \cdot 0 + C \Rightarrow C = 2\). Suy ra \(f(x) = \ln (x^2 + 3x + 2)\).
\[\begin{aligned} I =\ & \int\limits_{1}^{2} f(x) \mathrm{\,d}x = \int\limits_{1}^{2} \ln (x^2 + 3x + 2) \mathrm{\,d}x\\ =\ & x\ln(x^2+3x+2)\Big|_1^2 - \int\limits_{1}^{2} \displaystyle\frac{x(2x + 3)}{x^2+3x+2} \mathrm{\,d}x\\ =\ &x\ln(x^2+3x+2)\Big|_1^2 - \int\limits_{1}^{2} \left(2 - \displaystyle\frac{2}{x + 2} - \displaystyle\frac{1}{x + 1}\right) \mathrm{\,d}x\\ =\ & x\ln(x^2+3x+2)\Big|_1^2 - 2x\Big|_1^2 + 2\ln|x+ 2|\Big|_1^2 + \ln|x + 1|\Big|_1^2 = 6\ln 2 - 2. \end{aligned}\]

Đặt \(\left\{\begin{aligned} & u=f(x) \\ & \mathrm{\,d}v=(x+1)\mathrm{e}^x \mathrm{d}x\end{aligned}\right. \Rightarrow \left\{\begin{aligned} & \mathrm{\,d}u=f'(x)\mathrm{\,d}x \\ & v=x\mathrm{e}^x.\end{aligned}\right.\)
Khi đó \(\displaystyle\frac{e^2-1}{4}=x.\mathrm{e}^xf(x)\bigg|_0^1-\displaystyle\int\limits_0^1 x.e^xf'(x)\mathrm{\,d}x\Rightarrow \displaystyle\int\limits_0^1x.e^x f'(x)\mathrm{\,d}x=-\displaystyle\frac{e^2-1}{4}\).
Xét \[\displaystyle\int\limits_0^1 [f'(x)+x\mathrm{e}^x]^2\mathrm{\,d}x=\displaystyle\int\limits_0^1 \left[f'(x)\right]^2\mathrm{\,d}x+2\displaystyle\int\limits_0^1 x\cdot \mathrm{e}^xf'(x)\mathrm{\,d}x+\displaystyle\int\limits_0^1 x^2\mathrm{e}^{2x}\mathrm{\,d}x=0.\] \(\Rightarrow f'(x)=-x\mathrm{e}^x\Rightarrow f(x)=-\displaystyle\int\limits {x\mathrm{e}^x\mathrm{\, d}x}=(1-x)\mathrm{e}^x+C.\)
Mà \(f(1)=0\) nên \(C=0\).
Do đó \[\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x=\displaystyle\int\limits_0^1 \left(1-x\right)\mathrm{e}^x\mathrm{\,d}x=(2-x)\mathrm{e}^x\bigg|_0^1=2-\mathrm{e}.\]

Ta có \[\begin{aligned} (x+2)f(x)+(x+1)f'(x)=\mathrm{e}^x \Leftrightarrow\ & \mathrm{e}^x(x+2)f(x)+\mathrm{e}^x(x+1)f'(x)=\mathrm{e}^{2x}\\ \Leftrightarrow\ & \left(\mathrm{e}^x(x+1) \right)'f(x)+ \mathrm{e}^x(x+1)f'(x)=\mathrm{e}^{2x}. \end{aligned}\] Từ đó suy ra \[\begin{aligned} &\displaystyle\int\limits_0^{2}\left[ \left(\mathrm{e}^x(x+1) \right)'f(x)+ \mathrm{e}^x(x+1)f'(x)\right] \mathrm{\,d}x= \displaystyle\int\limits_0^{2}\mathrm{e}^{2x} \mathrm{\,d}x\\ \Leftrightarrow\ & (x+1)\mathrm{e}^x f(x)\bigg|_{0}^{2}=\displaystyle\frac{1}{2}\left(\mathrm{e}^4-1 \right) \\ \Leftrightarrow\ & 3\mathrm{e}^2f(2)-f(0)=\displaystyle\frac{1}{2}\left(\mathrm{e}^4-1 \right)\\ \Rightarrow\ & f(2)=\displaystyle\frac{\mathrm{e}^2}{6}. \end{aligned}\]

Cách 1: Chọn hàm
Chọn \(f(x)=1, \forall x \) ta được \(I=\displaystyle\int\limits_0^2 f(x) \mathrm{\,d}x = 2\) nên \(I\in [1;+\infty).\)
Cách 2: Tự luận
\[I=\displaystyle\int\limits_0^2 f(x) \mathrm{\,d}x =\displaystyle\int\limits_0^1 f(x) \mathrm{\,d}x +\displaystyle\int\limits_1^2 f(x) \mathrm{\,d}x=A+B\] Đặt \(\left\{\begin{aligned} &u=f(x)\\ &\mathrm{\,d}v=\mathrm{\,d}x\end{aligned}\right.\Rightarrow \left\{\begin{aligned} &\mathrm{\,d}u=f'(x)\mathrm{\,d}x\\ &v=x-1.\end{aligned}\right.\)
Suy ra \[A=(x-1)f(x)\Big|_0^1-\displaystyle\int\limits_0^1 (x-1)f'(x)\mathrm{\,d}x=1+\displaystyle\int\limits_0^1 (1-x)f'(x)\mathrm{\,d}x\geq 1-\displaystyle\int\limits_0^1 (1-x)\mathrm{\,d}x=\displaystyle\frac{1}{2}.\] \[B=(x-1)f(x)\Big|_1^2-\displaystyle\int\limits_1^2 (x-1)f'(x)\mathrm{\,d}x=1-\displaystyle\int\limits_1^2 (x-1)f'(x)\mathrm{\,d}x\geq 1-\displaystyle\int\limits_1^2 (x-1)\mathrm{\,d}x=\displaystyle\frac{1}{2}.\] Vậy \(I\geq 1\).

Ta có \[\begin{aligned} \displaystyle\frac{1}{2}=\displaystyle\int\limits_0^1 x^3f(x) \mathrm{\,d}x=\displaystyle\frac{x^4f(x)}{4}\Bigg|_0^1-\displaystyle\frac{1}{4}\displaystyle\int\limits_0^1 x^4f'(x) \mathrm{\,d}x=\displaystyle\frac{1}{4}-\displaystyle\frac{1}{4}\displaystyle\int\limits_0^1 x^4f'(x) \mathrm{\,d}x. \end{aligned}\] Suy ra \(\displaystyle\int\limits_0^1 x^4f'(x) \mathrm{\,d}x=-1\), do đó ta có \[\begin{aligned} \displaystyle\int\limits_{0}^{1} \left[ f'(x)+9x^4 \right]^2\mathrm{\,d}x &= \displaystyle\int\limits_0^1 \left[ f'(x) \right]^2 \mathrm{\,d}x +18\displaystyle\int\limits_0^1 x^4f'(x) \mathrm{\,d}x+81\displaystyle\int\limits_{0}^{1}x^8\mathrm{\,d}x\\ &= 9-18+9=0. \end{aligned}\] Do đó \(f'(x)=-9x^4\), kết hợp với \(f(1)=1\), suy ra \(f(x)=-\displaystyle\frac{9x^5}{5}+\displaystyle\frac{14}{5}\). Vậy \(\displaystyle\int\limits_0^1 f(x)\mathrm{\,d}x=\displaystyle\frac{5}{2}\).

Đặt \(t=x-1\Rightarrow 3=\displaystyle\int\limits_{1}^{2}f(x-1)\mathrm{\,d}x=\displaystyle\int\limits_{0}^{1}f(t)\mathrm{\,d}t\).
Xét \(\displaystyle\int\limits_{0}^{1}x^{3}f'(x^{2})\mathrm{\,d}x=\displaystyle\frac{1}{2}\displaystyle\int\limits_{0}^{1}x^{2}f'(x^{2})\mathrm{\,d}x^{2}=\displaystyle\frac{1}{2}\displaystyle\int\limits_{0}^{1}tf'(t)\mathrm{\,d}t=\displaystyle\frac{1}{2}\left[tf(t)\Big |_{0}^{1}-\displaystyle\int\limits_{0}^{1}f(t)\mathrm{\,d}t \right]=\displaystyle\frac{1}{2}(4-3)=\displaystyle\frac{1}{2}\).

Xét \(\displaystyle\int\limits_0^1x^4f(x)\mathrm{\, d}x=\displaystyle\frac{7}{11}\).
Đặt \(\left\{\begin{aligned} &u=x^4f(x)\\ &\mathrm{\, d}v=\mathrm{\, d}x\end{aligned}\right. \Rightarrow \left\{\begin{aligned} &\mathrm{\, d}u=[4x^3f(x)+x^4f'(x)]\mathrm{\, d}x\\ &v=x.\end{aligned}\right.\)
Suy ra \[\displaystyle\int\limits_0^1x^4f(x)\mathrm{\, d}x=x^5f(x)\Big|_0^1-\displaystyle\int\limits_0^1 4x^4f(x)\mathrm{\, d}x -\displaystyle\int\limits_0^1x^5f'(x)\mathrm{\, d}x.\] Suy ra \[\displaystyle\int\limits_0^1x^5f'(x)\mathrm{\, d}x=-\displaystyle\frac{2}{11}.\] Khi đó \(\displaystyle\int\limits_0^1[f'(x)+2x^5]^2\mathrm{\, d}x=\displaystyle\int\limits_0^1[f'(x)]^2\mathrm{\, d}x +4\displaystyle\int\limits_0^1x^5f'(x)\mathrm{\, d}x+4\displaystyle\int\limits_0^1x^{10}\mathrm{\, d}x=0\).
Suy ra \(f'(x)=-2x^5\Rightarrow f(x)=-\displaystyle\frac{x^6}{3}+C\). Mà \(f(1)=3\) nên \(C=\displaystyle\frac{10}{3}\).
Vậy \(\displaystyle\int\limits_0^1f(x)\mathrm{\, d}x=\displaystyle\int\limits_0^1\left(-\displaystyle\frac{x^6}{3}+\displaystyle\frac{10}{3}\right)\mathrm{\, d}x=\displaystyle\frac{23}{7}\).

Ta có \[\begin{aligned} f'(x) - \sqrt{x^2 + 1} \cdot f(x) = 0 \Leftrightarrow\ & \displaystyle\frac{f'(x)}{f(x)} = \sqrt{x^2 + 1}\\ \Rightarrow\ & \displaystyle \int \displaystyle\frac{f'(x)}{f(x)} \mathrm{\,d}x = \int \sqrt{x^2 + 1} \mathrm{\,d}x\\ \Rightarrow\ & \ln f(x) = \displaystyle\frac{x}{2} \sqrt{x^2 + 1} +\displaystyle\frac{1}{2} \ln \left(x + \sqrt{x^2+1}\right)+C. \end{aligned}\] Ta có \(f\left(\sqrt{3}\right) = \mathrm{e}^3 \Rightarrow C = 3 - \sqrt{3} - \displaystyle\frac{1}{2}\ln \left(\sqrt{3} + 2\right) \).
Do đó \[\begin{aligned} \displaystyle \int\limits_{0}^{\sqrt{3}} \ln f(x) \mathrm{\,d}x =\ & \int\limits_{0}^{\sqrt{3}} \left[\displaystyle\frac{x}{2} \sqrt{x^2+1} + \displaystyle \displaystyle\frac{1}{2} \ln \left(x + \sqrt{x^2 + 1}\right) + \displaystyle \left(3-\sqrt{3} -\displaystyle\frac{1}{2}\ln\left(\sqrt{3}+2\right)\right)\right] \mathrm{\,d}x\\ =\ & 3 \sqrt{3} - \displaystyle\frac{7}{3}. \end{aligned}\]

Ta có \[\begin{aligned} &f'(x)=x^3\left[f(x)\right]^2\Rightarrow\displaystyle\frac{f'(x)}{f^2(x)}=x^3\Rightarrow\displaystyle\int\limits_{1}^{2} \displaystyle\frac{f'(x)}{f^2(x)}\mathrm{\,d}x =\displaystyle\int\limits_{1}^{2} x^3\mathrm{\,d}x\\ \Leftrightarrow\ & -\displaystyle\frac{1}{f(x)}\Big|_{1}^{2}=\displaystyle\frac{15}{4} \Leftrightarrow -\displaystyle\frac{1}{f(2)}+\displaystyle\frac{1}{f(1)} =\displaystyle\frac{15}{4}\Leftrightarrow f(1)=-\displaystyle\frac{4}{5}. \end{aligned}\]

Từ giả thiết ta có \(f'(x)\mathrm{e}^{f(x)} = 2x + 3 \Rightarrow \left(\mathrm{e}^{f(x)}\right)'=2x+3 \Rightarrow \mathrm{e}^{f(x)}=x^2+3x+C\).
Từ \(f(0)=\ln2\), suy ra \(C = 2\). Vậy \(f(x) = \ln (x^2 + 3x + 2)\). \[\begin{aligned} I =\ & \int\limits_{1}^{2} f(x) \mathrm{\,d}x = \int\limits_{1}^{2} \ln (x^2 + 3x + 2) \mathrm{\,d}x\\ =\ & x\ln(x^2+3x+2)\Big|_1^2 - \int\limits_{1}^{2} \displaystyle\frac{x(2x + 3)}{x^2+3x+2} \mathrm{\,d}x\\ =\ &x\ln(x^2+3x+2)\Big|_1^2 - \int\limits_{1}^{2} \left(2 - \displaystyle\frac{2}{x + 2} - \displaystyle\frac{1}{x + 1}\right) \mathrm{\,d}x\\ =\ & x\ln(x^2+3x+2)\Big|_1^2 - 2x\Big|_1^2 + 2\ln|x+ 2|\Big|_1^2 + \ln|x + 1|\Big|_1^2 = 6\ln 2 - 2. \end{aligned}\]

Với \(x \in[1;2]\), ta có \(xf'(x) - f(x) = x^2 \mathrm{e}^x \Leftrightarrow \displaystyle\frac{f'(x)}{x} - \displaystyle\frac{f(x)}{x^2} = \mathrm{e}^x\Leftrightarrow \left(\displaystyle\frac{f(x)}{x}\right)'=\mathrm{e}^x\).
Từ đó suy ra \(\displaystyle\frac{f(x)}{x} = \mathrm{e}^x + C\). Do \(f(1) = \mathrm{e}\) nên \(C = 0\).
Suy ra, \[f(x) = x \mathrm{e}^x \Rightarrow\displaystyle I = \int\limits_1^2 f(x)\,\mathrm{d}x = \int\limits_1^2 x\mathrm{e}^x\,\mathrm{d}x = x\mathrm{e}^x\bigg|_1^2 - \mathrm{e}^x\bigg|_1^2 = \mathrm{e}^2.\]

Từ giả thiết \(f'(x)+2xf(x)=2x\mathrm{e}^{-x^2}\), ta suy ra \[\begin{aligned} &\mathrm{e}^{x^2}f'(x)+2x\mathrm{e}^{x^2}f(x)=2x\Leftrightarrow \left[\mathrm{e}^{x^2}f(x)\right]'=2x\\ \Leftrightarrow\ & \displaystyle\int\limits_0^1\left[\mathrm{e}^{x^2}f(x)\right]' \mathrm{\,d}x=\displaystyle\int\limits_0^1 2x\mathrm{\,d}x\\ \Leftrightarrow\ &\left. \mathrm{e}^{x^2}f(x)\right|_0^1=\left. x^2\right|_0^1\\ \Leftrightarrow\ &\mathrm{e}f(1)-f(0)=1\Rightarrow f(1)=\displaystyle\frac{2}{\mathrm{e}}. \end{aligned}\]