39 câu Trắc nghiệm Toán 12 Tích phân hàm ẩn có đáp án (Mới nhất)

Chọn D

Xét $I=\underset{e^2}{\overset{e^4}{\int }}\frac{f\left(\ln x\right) }{x}\text{d}x$

Đặt $t=\ln x\Rightarrow \text{d}t=\frac{1}{x}\text{d}x$

Đổi cận: $\begin{array}{l}x={\text{e}}^2\Rightarrow t=2\\ x={\text{e}}^4\Rightarrow t=4\end{array}$.

Chọn A

Xét $I=\underset{-7}{\overset{1}{\int }}f\left(\sqrt[3]{1-x}\right)\text{d}x$

Đặt $t=\sqrt[3]{1-x}\Rightarrow -3t^2\text{d}t=\text{d}x$

Đổi cận: $\begin{array}{l}x=-7\Rightarrow t=2\\ x=1\Rightarrow t=0\end{array}$.

$I=-3\underset{2}{\overset{0}{\int }}{t}^{2}f\left(t\right)\text{d}t=3\underset{0}{\overset{2}{\int }}{x}^{2}f\left(x\right)\text{d}x=3\left[\underset{0}{\overset{1}{\int }}{x}^2\left(x+1\right)\text{d}x+\underset{1}{\overset{2}{\int }}x\text{d}x\right]=\frac{25}{12}$.

Chọn C

Bảng khử dấu giá trị tuyệt đối:

Ta có:  $\underset{1}{\overset{2}{\int }}f\left(x\right)\text{d}x=F\left(2\right)-F\left(1\right)=F\left(2\right)-3$ mà $\underset{1}{\overset{2}{\int }}f\left(x\right)\text{d}x=\underset{1}{\overset{2}{\int }}2\text{d}x=2$ nên $F\left(2\right)=5$.

Ø$\underset{0}{\overset{1}{\int }}f\left(x\right)\text{d}x=F\left(1\right)-F\left(0\right)=3-F\left(0\right)$ mà $\underset{0}{\overset{1}{\int }}f\left(x\right)\text{d}x=\underset{0}{\overset{1}{\int }}2x\text{d}x=x^2\left|{}_0^1\right.=1$ nên $F\left(0\right)=2$.

Ø $\underset{-1}{\overset{0}{\int }}f\left(x\right)\text{d}x=F\left(0\right)-F\left(-1\right)=2-F\left(-1\right)$ mà $\underset{-1}{\overset{0}{\int }}f\left(x\right)\text{d}x=\underset{-1}{\overset{0}{\int }}2x\text{d}x=x^2\left|{}_{-1}^0=-1\right.$ nên $F\left(-1\right)=3$ .

Ø$\underset{-3}{\overset{-1}{\int }}f\left(x\right)\text{d}x=F\left(-1\right)-F\left(-3\right)=3-F\left(-3\right)$ mà $\underset{-3}{\overset{-1}{\int }}f\left(x\right)\text{d}x=\underset{-3}{\overset{-1}{\int }}-2\text{d}x=-4$ nên $F\left(-3\right)=7$ .

Vậy $F\left(0\right)+F\left(2\right)+F\left(-3\right)=2+5+7=14$.

Chọn C

Từ giả thiết ta có $f\left(x^3+3x+1\right)=3x+2$ nên suy ra $f\left(1\right)=2$, $f\left(5\right)=5$.

Suy ra $I=\underset{1}{\overset{5}{\int }}x{f}^{\text{'}}\left(x\right)\text{d}x={\left.xf\left(x\right)\right|}_1^5-\underset{1}{\overset{5}{\int }}f\left(x\right)\text{d}x=23-\underset{1}{\overset{5}{\int }}f\left(x\right)\text{d}x$.

Đặt $x=t^3+3t+1\Rightarrow \text{d}x=\left(3t^2+3\right)\text{d}t$.

Với $x=1\Rightarrow t=0;x=5\Rightarrow t=1$

Do đó $\underset{1}{\overset{5}{\int }}f\left(x\right)\text{d}x=\underset{0}{\overset{1}{\int }}f\left(t^3+3t+1\right)\left(3t^2+3\right)\text{d}t=\underset{0}{\overset{1}{\int }}\left(3t+2\right)\left(3t^2+3\right)\text{d}t=\frac{59}{4}$ .

Vậy $I=23-\frac{59}{4}=\frac{33}{4}$.

Chọn B

Đặt $x=t^5+4t+3\Rightarrow dx=\left(5t^4+4\right)dt$.

Đổi cận: $\left\{\begin{array}{l}x=-2\Rightarrow t=-1\\ x=8\Rightarrow t=1\end{array}\right.$

Chọn C

Ta có $f^{\text{'}}\left(x\right)=\frac{2}{2x-1}$

 $\frac{2}{2x-1}dx=\ln \left|2x-1\right|+C=\left\{\begin{array}{l}\ln \left(1-2x\right)+C_1;x<\frac{1}{2}\\ \ln \left(2x-1\right)+C_2;x>\frac{1}{2}\end{array}\right.$

 $f\left(0\right)=1\Rightarrow C_1=1$ và $f\left(1\right)=2\Rightarrow C_2=2$ .

Do đó $f\left(x\right)=\left\{\begin{array}{l}\ln \left(1-2x\right)+1;x<\frac{1}{2}\\ \ln \left(2x-1\right)+2;x>\frac{1}{2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}f\left(-1\right)=\ln 3+1\\ f\left(3\right)=\ln 5+2\end{array}\right.$

$\Rightarrow f\left(-1\right)+f\left(3\right)=3+\ln 15.$

Chọn A

Đặt $t=\sin x\Rightarrow dt=\cos xdx$ . Đổi cận$\left\{\begin{array}{l}x=-\frac{\pi }{2}\Rightarrow t=-1\\ x=\frac{\pi }{2}\Rightarrow t=1\end{array}\right.$ .

$\Rightarrow I=\underset{-1}{\overset{1}{\int }}f\left(t\right)dt=\underset{-1}{\overset{1}{\int }}f\left(x\right)dx$

Do $f\left(x\right)=\left\{\begin{array}{ll}3x^2+2x& \text{ khi }x\ge 0\\ 5-x& \text{ khi }x<0\end{array}\right.$

$\Rightarrow I=\underset{-1}{\overset{0}{\int }}\left(5-x\right)dx+\underset{0}{\overset{1}{\int }}\left(3x^2+2x\right)dx=\frac{15}{2}$. 

Chọn C

          Đặt $t=3-2x\Rightarrow dt=-2dx\Rightarrow dx=-\frac{1}{2}dt$. Đổi cận $\left\{\begin{array}{l}x=0\Rightarrow t=3\\ x=1\Rightarrow t=1\end{array}\right.$.

Do $f\left(x\right)=\left\{\begin{array}{ll}{x}^2-2x+3& \text{ khi }x\ge 2\\ x+1& \text{ khi }x<2\end{array}\right.$

Chọn A

Đặt $t=\cos x+1\Rightarrow dt=-\sin xdx$. Đổi cận $\left\{\begin{array}{l}x=0\Rightarrow t=2\\ x=\frac{\pi }{2}\Rightarrow t=1\end{array}\right.$.

$\Rightarrow I=\underset{1}{\overset{2}{\int }}f\left(t\right)dt=\underset{1}{\overset{2}{\int }}f\left(x\right)dx$

Do $f\left(x\right)=\left\{\begin{array}{ll}{x}^2+2x& \text{ khi }x\ge \frac{3}{2}\\ x-2& \text{ khi }x<\frac{3}{2}\end{array}\right.$

$\Rightarrow I=\underset{1}{\overset{\frac{3}{2}}{\int }}\left(x-2\right)dx+\underset{\frac{3}{2}}{\overset{2}{\int }}\left(x^2+2x\right)dx=\frac{35}{12}$.

Chọn A

Đặt $t=\sin x\Rightarrow dt=\cos xdx$. Đổi cận $\left\{\begin{array}{l}x=-\frac{\pi }{2}\Rightarrow t=-1\\ x=\frac{\pi }{2}\Rightarrow t=1\end{array}\right.$.

$\Rightarrow I=\underset{-1}{\overset{1}{\int }}f\left(t\right)dt=\underset{-1}{\overset{1}{\int }}f\left(x\right)dx$

Do $f\left(x\right)=\left\{\begin{array}{ll}{x}^2-x& \text{ khi }x\ge 0\\ x& \text{ khi }x<0\end{array}\right.$

$\Rightarrow I=\underset{-1}{\overset{0}{\int }}xdx+\underset{0}{\overset{1}{\int }}\left(x^2-x\right)dx=-\frac{2}{3}$.

Chọn B

Đặt $t=x^2+1\Rightarrow dt=2xdx\Rightarrow xdx=\frac{1}{2}dt$ . Đổi cận $\left\{\begin{array}{l}x=0\Rightarrow t=1\\ x=2\Rightarrow t=5\end{array}\right.$.

$\Rightarrow I=\frac{1}{2}\underset{1}{\overset{5}{\int }}f\left(t\right)dt=\frac{1}{2}\underset{1}{\overset{5}{\int }}f\left(x\right)dx$

Do $f\left(x\right)=\left\{\begin{array}{ll}{x}^2+x+1& \text{ khi }x\ge 3\\ 2x-1& \text{ khi }x<3\end{array}\right.$

$\Rightarrow I=\frac{1}{2}\left(\underset{1}{\overset{3}{\int }}\left(2x-1\right)dx+\underset{3}{\overset{5}{\int }}\left(x^2+x+1\right)dx\right)=\frac{73}{3}$.

Chọn B

Xét $I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(\sin x\right)\cos x\text{d}x$

Đặt $\sin x=t\Rightarrow \cos x\text{d}x=\text{d}t$

Với $x=0\Rightarrow t=0$

$x=\frac{\pi }{2}\Rightarrow t=1$

$I=\underset{0}{\overset{1}{\int }}f\left(t\right)\text{d}t=\underset{0}{\overset{1}{\int }}f\left(x\right)\text{d}x=\underset{0}{\overset{\frac{1}{2}}{\int }}f\left(x\right)\text{d}x+\underset{\frac{1}{2}}{\overset{1}{\int }}f\left(x\right)\text{d}x=\underset{0}{\overset{\frac{1}{2}}{\int }}\left(3x+3\right)\text{d}x+\underset{\frac{1}{2}}{\overset{1}{\int }}\left(x+4\right)\text{d}x=\frac{17}{4}.$

Chọn D

Xét $I=\underset{0}{\overset{\frac{\pi }{3}}{\int }}f\left(3\cos x-2\right)\sin x\text{d}x$

Đặt $3\cos x-2=t\Rightarrow -3\sin x\text{d}x=\text{d}t\Rightarrow \sin x\text{d}x=-\frac{1}{3}\text{d}t$

Với $x=0\Rightarrow t=0$

    $x=\frac{\pi }{3}$$\Rightarrow t=-\frac{1}{2}$

$I=\frac{1}{3}\underset{-\frac{1}{2}}{\overset{1}{\int }}f\left(t\right)\text{d}t=\frac{1}{3}\underset{-\frac{1}{2}}{\overset{1}{\int }}f\left(x\right)\text{d}x=\frac{1}{3}\underset{-\frac{1}{2}}{\overset{0}{\int }}f\left(x\right)\text{d}x+\frac{1}{3}\underset{0}{\overset{1}{\int }}f\left(x\right)\text{d}x$

$=\frac{1}{3}\underset{-\frac{1}{2}}{\overset{0}{\int }}\left(2x^2-x+1\right)\text{d}x+\frac{1}{3}\underset{0}{\overset{1}{\int }}\left(2x^2+1\right)\text{d}x=\frac{19}{24}.$

Chọn C

Xét $I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{4}}{\int }}f\left(5\sin 2x-1\right)\cos 2x\text{d}x$

Đặt $5\sin 2x-1=t\Rightarrow 10\cos 2x\text{d}x=\text{d}t\Rightarrow \cos 2x\text{d}x=\frac{1}{10}\text{d}t$

Với $x=-\frac{\pi }{2}\Rightarrow t=-1$

$x=\frac{\pi }{4}\Rightarrow t=4$

$\Rightarrow I=\frac{1}{10}\underset{-1}{\overset{4}{\int }}f\left(t\right)\text{d}t=\frac{1}{10}\underset{-1}{\overset{4}{\int }}f\left(x\right)\text{d}x=\frac{1}{10}\underset{-1}{\overset{1}{\int }}f\left(x\right)\text{d}x+\frac{1}{10}\underset{1}{\overset{4}{\int }}f\left(x\right)\text{d}x$

$=\frac{1}{10}\underset{-1}{\overset{1}{\int }}\left(1-x^2\right)\text{d}x+\frac{1}{10}\underset{1}{\overset{4}{\int }}\left(2x-2\right)\text{d}x=\frac{31}{30}.$

Chọn A

Xét $I=\underset{1}{\overset{e}{\int }}f\left(2+\ln x\right)\frac{1}{x}\text{d}x$

Đặt $2+\ln x=t\Rightarrow \frac{1}{x}\text{d}x=\text{d}t$

Với $x=\frac{1}{e}\Rightarrow t=1$

$x=e\Rightarrow t=3$

$\Rightarrow I=\underset{1}{\overset{3}{\int }}f\left(t\right)\text{d}t=\underset{1}{\overset{3}{\int }}f\left(x\right)\text{d}x=\underset{1}{\overset{2}{\int }}f\left(x\right)\text{d}x+\underset{2}{\overset{3}{\int }}f\left(x\right)\text{d}x=\underset{1}{\overset{2}{\int }}\left(11-x\right)\text{d}x+\underset{2}{\overset{3}{\int }}\left(2x^3-x-5\right)\text{d}x=\frac{69}{2}.$

Chọn C

Xét $I=\underset{0}{\overset{\ln 2}{\int }}f\left(3e^x-1\right)e^x\text{d}x$

Đặt $3e^x-1=t\Rightarrow 3e^x\text{d}x=\text{d}t\Rightarrow e^x\text{d}x=\frac{1}{3}\text{d}t$

Với $x=0\Rightarrow t=2$

$x=\ln 2\Rightarrow t=5$

$\Rightarrow I=\frac{1}{3}\underset{2}{\overset{5}{\int }}f\left(t\right)\text{d}t=\frac{1}{3}\underset{2}{\overset{3}{\int }}f\left(x\right)\text{d}x+\frac{1}{3}\underset{3}{\overset{5}{\int }}f\left(x\right)\text{d}x=\frac{1}{3}\underset{2}{\overset{3}{\int }}\left(1-x^2\right)\text{d}x+\frac{1}{3}\underset{3}{\overset{5}{\int }}(7-5x)\text{d}x=-\frac{94}{9}.$

Chọn C

Ta có phương trình $\sin x-\cos x=0$ có một nghiệm trên đoạn $\left[0;\frac{\pi }{2}\right]$ là $x=\frac{\pi }{4}$.

Bảng xét dấu

ChọnB

Ta có   $x\left(x+1\right).f^{\text{'}}\left(x\right)+f\left(x\right)=x^2+x$       (1)

Chia cả 2 vế của biểu thức (1) cho ${\left(x+1\right)}^2$ ta được $\frac{x}{x+1}.f^{\text{'}}\left(x\right)+\frac{1}{{\left(x+1\right)}^2}f\left(x\right)=\frac{x}{x+1}$

$\Rightarrow {\left[\frac{x}{x+1}.f\left(x\right)\right]}^{\text{'}}=\frac{x}{x+1}$, với $\forall x\in ℝ\backslash \left\{0;-1\right\}$$\Rightarrow \frac{x}{x+1}.f\left(x\right)=\int \frac{x}{x+1}\text{d}x$.

Mặt khác, $f\left(1\right)=-2\ln 2\iff 2\left(1-\ln 2+C\right)=-2\ln 2\iff C=-1$.

Do đó $f\left(x\right)=\frac{x+1}{x}\left(x-\ln \left|x+1\right|-1\right)$.

Với $x=2$ thì $f\left(x\right)=\frac{3}{2}\left(1-\ln 3\right)=\frac{3}{2}-\frac{3}{2}\ln 3$. Suy ra $a=\frac{3}{2}$ và $b=-\frac{3}{2}$.

Vậy $a^2+b^2=\frac{9}{2}$.

ChọnC

Lấy đạo hàm theo hàm số

$f^{\text{'}}\left(x+y\right)=f^{\text{'}}\left(y\right)+3x^2+6xy$,$\forall x\in ℝ$ .

Cho $y=0\Rightarrow f^{\text{'}}\left(x\right)=f^{\text{'}}\left(0\right)+3x^2\Rightarrow f^{\text{'}}\left(x\right)=1+3x^2$

 $\Rightarrow f\left(x\right)=\int f^{\text{'}}\left(x\right)dx=x^3+x+C$ mà $f\left(0\right)=1\Rightarrow C=1$. Do đó $f\left(x\right)=x^3+x+1$.

Vậy $\underset{0}{\overset{1}{\int }}f\left(x-1\right)\text{d}x=\underset{-1}{\overset{0}{\int }}f\left(x\right)\text{\hspace{0.17em}d}x=\underset{-1}{\overset{0}{\int }}\left(x^3+x+1\right)\text{\hspace{0.17em}d}x=\frac{1}{4}$.

Chọn A

Ta có$\underset{0}{\overset{1}{\int }}{x}^{2}f\left(x\right)dx={\left.\left[\frac{x^3}{3}f\left(x\right)\right]\right|}_0^1-\underset{0}{\overset{1}{\int }}\frac{x^3}{3}{f}^{\text{'}}\left(x\right)dx$. Suy ra $\underset{0}{\overset{1}{\int }}\frac{x^3}{3}{f}^{\text{'}}\left(x\right)dx=-\frac{1}{3}$.

Hơn nữa ta dễ dàng tính được $\underset{0}{\overset{1}{\int }}\frac{x^6}{9}\text{\hspace{0.17em}d}x=\frac{1}{63}$.

Do đó$\underset{0}{\overset{1}{\int }}{\left[f^{\text{'}}\left(x\right)\right]}^2\text{d}x+2.21\underset{0}{\overset{1}{\int }}\frac{x^3}{3}{f}^{\text{'}}\left(x\right)\text{d}x+{21}^2\underset{0}{\overset{1}{\int }}\frac{x^6}{9}\text{d}x=0$$\iff \underset{0}{\overset{1}{\int }}{\left[f^{\text{'}}\left(x\right)+7x^3\right]}^2\text{d}x=0$ .

Suy ra $f^{\text{'}}\left(x\right)=-7x^3$, do đó $f\left(x\right)=-\frac{7}{4}{x}^4+C$. Vì $f\left(1\right)=0$ nên $C=\frac{7}{4}$ .

Vậy $\underset{0}{\overset{1}{\int }}f\left(x\right)\text{d}x=-\frac{7}{4}\underset{0}{\overset{1}{\int }}\left(x^4-1\right)\text{d}x=\frac{7}{5}$ .

Chọn D

Ta có $J=\underset{1}{\overset{2}{\int }}\left(\frac{f^{\text{'}}\left(x\right)+2}{x}-\frac{f\left(x\right)+1}{x^2}\right)\text{d}x=\underset{1}{\overset{2}{\int }}\frac{f^{\text{'}}\left(x\right)}{x}\text{d}x-\underset{1}{\overset{2}{\int }}\frac{f\left(x\right)}{x^2}\text{d}x+\underset{1}{\overset{2}{\int }}\left(\frac{2}{x}-\frac{1}{x^2}\right)\text{d}x$.

Đặt $\left\{\begin{array}{l}u=\frac{1}{x}\\ \text{d}v=f^{\text{'}}\left(x\right)\text{d}x\end{array}\right.\Rightarrow \left\{\begin{array}{l}\text{d}u=-\frac{1}{x^2}\text{d}x\\ v=f\left(x\right)\end{array}\right.$.

$J=\underset{1}{\overset{2}{\int }}\left(\frac{f^{\text{'}}\left(x\right)+2}{x}-\frac{f\left(x\right)+1}{x^2}\right)\text{d}x$$={\left.\frac{1}{x}.f\left(x\right)\right|}_1^2+\underset{1}{\overset{2}{\int }}\frac{f\left(x\right)}{x^2}\text{d}x-\underset{1}{\overset{2}{\int }}\frac{f\left(x\right)}{x^2}\text{d}x+\underset{1}{\overset{2}{\int }}\left(\frac{2}{x}-\frac{1}{x^2}\right)\text{d}x$.

$=\frac{1}{2}f\left(2\right)-f\left(1\right)+{\left.\left(2\ln x+\frac{1}{x}\right)\right|}_1^2=\frac{1}{2}+\ln 4$

Chọn B

Ta có: $f^{\text{'}}\left(x\right)=\frac{1}{x^2+x-2}=\frac{1}{3}\left(\frac{1}{x-1}-\frac{1}{x+2}\right)$

$f\left(x\right)=\frac{1}{3}\int \left(\frac{1}{x-1}-\frac{1}{x+2}\right)dx=\frac{1}{3}\ln \left|\frac{x-1}{x+2}\right|+C=\left\{\begin{array}{l}\frac{1}{3}\left[\ln \left(1-x\right)-\ln \left(-x-2\right)\right]+C_1;x\in \left(-\infty ;-2\right)\\ \frac{1}{3}\left[\ln \left(1-x\right)-\ln \left(x+2\right)\right]+C_2;x\in \left(-2;1\right)\\ \frac{1}{3}\left[\ln \left(x-1\right)-\ln \left(x+2\right)\right]+C_3;x\in \left(1;+\infty \right)\end{array}\right.$