Một số bài tập chọn lọc về tích phân có đáp án chi tiết

Một số bài tập chọn lọc về tích phân có đáp án chi tiết

Bài tập trắc nghiệm tích phân vận dụng cao có Lời giải chi tiết

Lời giải chi tiết:

Ta có $f\left( x \right)+x.{f}'\left( x \right)={{\left[ x.f\left( x \right)+1 \right]}^{2}}\Leftrightarrow \frac{f\left( x \right)+x.{f}'\left( x \right)}{{{\left[ x.f\left( x \right)+1 \right]}^{2}}}=1\Leftrightarrow \frac{{{\left[ x.f\left( x \right) \right]}^{\prime }}}{{{\left[ x.f\left( x \right)+1 \right]}^{2}}}=1$

Lấy nguyên hàm 2 vế ta được: $\frac{-1}{x.f\left( x \right)+1}=x+C$

Thay $x=1\Rightarrow \frac{-1}{-2+1}=1+C\Rightarrow C=0\Rightarrow f\left( x \right)=\frac{-1}{{{x}^{2}}}-\frac{1}{x}\Rightarrow f\left( 2 \right)=-\frac{3}{4}$. Chọn D.

Lời giải chi tiết:

Ta có: ${{x}^{2}}.f\left( x \right)+{{x}^{3}}.{f}'\left( x \right)={{\left[ xf\left( x \right)+1 \right]}^{2}}\Leftrightarrow {{x}^{2}}\left[ f\left( x \right)+x.{f}'\left( x \right) \right]={{\left[ xf\left( x \right)+1 \right]}^{2}}$

$\Leftrightarrow \frac{f\left( x \right)+x.{f}'\left( x \right)}{{{\left[ xf\left( x \right)+1 \right]}^{2}}}=\frac{1}{{{x}^{2}}}\Leftrightarrow \frac{{{\left[ x.f\left( x \right) \right]}^{\prime }}}{{{\left[ xf\left( x \right)+1 \right]}^{2}}}=\frac{1}{{{x}^{2}}}$

Lấy nguyên hàm 2 vế ta được: $\int{\frac{d\left[ xf\left( x \right)+1 \right]}{{{\left[ xf\left( x \right)+1 \right]}^{2}}}}=\int{\frac{1}{{{x}^{2}}}}dx\Rightarrow \frac{-1}{xf\left( x \right)+1}=\frac{-1}{x}+C$

Lại có: $f\left( 1 \right)=\frac{-2}{3}\Rightarrow \frac{-1}{1.f\left( 1 \right)+1}=-1+C\Rightarrow C=\frac{-1}{\frac{-2}{3}+1}+1=-2$

Do đó $\frac{1}{x.f\left( x \right)+1}=\frac{1}{x}+2$, thay $x=3\Rightarrow \frac{1}{3.f\left( 3 \right)+1}=\frac{1}{3}+2\Rightarrow f\left( 3 \right)=\frac{-4}{21}$. Chọn D.

Lời giải chi tiết:

Ta có: ${{\left[ {f}'\left( x \right) \right]}^{2}}=\left( {{x}^{4}}+3{{x}^{2}} \right)f\left( x \right)\Rightarrow \frac{{{\left[ {f}'\left( x \right) \right]}^{2}}}{f\left( x \right)}={{x}^{2}}\left( {{x}^{2}}+3 \right)$

$\Leftrightarrow \frac{{f}'\left( x \right)}{\sqrt{f\left( x \right)}}=x\sqrt{{{x}^{2}}+3}$  (do $f\left( x \right)>0\,\,\forall x\in \left[ 1;2 \right]$)

Lấy nguyên hàm 2 vế ta được: $\int{\frac{d\left[ f\left( x \right) \right]}{\sqrt{f\left( x \right)}}}=\int{x\sqrt{{{x}^{2}}+3}dx\Leftrightarrow 2}\sqrt{f\left( x \right)}=\frac{1}{2}\int{\sqrt{{{x}^{2}}+3}dx}\left( {{x}^{2}}+3 \right)$

$\Leftrightarrow \sqrt{f\left( x \right)}=\frac{1}{4}.\frac{2}{3}\sqrt{{{\left( {{x}^{2}}+3 \right)}^{3}}}+C=\frac{1}{6}\sqrt{{{\left( {{x}^{2}}+3 \right)}^{3}}}+C$

Do $f\left( 1 \right)=4\Rightarrow 2=\frac{8}{6}+C\Rightarrow C=\frac{2}{3}\Rightarrow \sqrt{f\left( x \right)}=\frac{\sqrt{{{\left( {{x}^{2}}+3 \right)}^{3}}}}{6}+\frac{2}{3}.$

$\Rightarrow f\left( x \right)={{\left[ \frac{\sqrt{{{\left( {{x}^{2}}+3 \right)}^{3}}}}{6}+\frac{2}{3} \right]}^{2}}\Rightarrow f\left( 2 \right)\approx 14,1$. Chọn D.

Lời giải chi tiết:

Ta có: ${{\left[ {f}'\left( x \right) \right]}^{2}}={{f}'}'\left( x \right)\Rightarrow \frac{{{f}'}'\left( x \right)}{{{\left[ {f}'\left( x \right) \right]}^{2}}}=1$

Lấy nguyên hàm 2 vế ta có: $\int{\frac{d\left[ {f}'\left( x \right) \right]}{{{\left[ {f}'\left( x \right) \right]}^{2}}}}=\int{dx\Leftrightarrow \frac{-1}{{f}'\left( x \right)}}=x+C\Rightarrow {f}'\left( x \right)=\frac{-1}{x+C}$

Do ${f}'\left( 0 \right)=-1\Rightarrow C=1$

Suy ra $\int\limits_{0}^{1}{{f}'\left( x \right)dx=\int\limits_{0}^{1}{\frac{-1}{x+1}}}dx\Leftrightarrow f\left( 1 \right)-f\left( 0 \right)=-\ln 2$. Chọn B.

Lời giải chi tiết:

Ta có: ${{\left[ {f}'\left( x \right)+2x \right]}^{2}}=9{{x}^{3}}+9x.f\left( x \right)\Leftrightarrow \frac{{{\left[ {f}'\left( x \right)+2x \right]}^{2}}}{f\left( x \right)+{{x}^{2}}}=9x\Leftrightarrow \frac{{f}'\left( x \right)+2x}{\sqrt{f\left( x \right)+{{x}^{2}}}}=3\sqrt{x}$

Lấy nguyên hàm 2 vế ta được $\int{\frac{d\left[ f\left( x \right)+{{x}^{2}} \right]}{\sqrt{f\left( x \right)+{{x}^{2}}}}}=\int{3\sqrt{x}dx}$

$\Leftrightarrow 2\sqrt{f\left( x \right)+{{x}^{2}}}=2\sqrt{{{x}^{3}}}+2C\Leftrightarrow \sqrt{f\left( x \right)+{{x}^{2}}}=\sqrt{{{x}^{3}}}+C$

Thay $x=0\Rightarrow C=1\Rightarrow \sqrt{f\left( x \right)+{{x}^{2}}}=\sqrt{{{x}^{3}}}+1.$

Suy ra $f\left( x \right)={{\left( \sqrt{{{x}^{3}}}+1 \right)}^{2}}-{{x}^{2}}\Rightarrow f\left( 1 \right)=3.$ Chọn A.

Lời giải chi tiết:

Ta có: ${{\left[ f\left( x \right).{f}'\left( x \right) \right]}^{\prime }}={{\left[ {f}'\left( x \right) \right]}^{2}}+f\left( x \right).{{f}'}'\left( x \right)=15{{x}^{4}}+12x$

Nguyên hàm 2 vế ta được $f\left( x \right).{f}'\left( x \right)=\frac{15{{x}^{5}}}{5}+6{{x}^{2}}+C=3{{x}^{5}}+6{{x}^{2}}+C$

Do $f\left( 0 \right)={f}'\left( 0 \right)=1\Rightarrow C=1$

Tiếp tục nguyên hàm 2 vế ta được: $\int{f\left( x \right)df\left( x \right)=\int{\left( 3{{x}^{5}}+6{{x}^{2}}+1 \right)}}dx$

$\Rightarrow \frac{{{f}^{2}}\left( x \right)}{2}=\frac{3{{x}^{6}}}{6}+\frac{6{{x}^{3}}}{3}+x+D=\frac{1}{2}{{x}^{6}}+2{{x}^{3}}+x+D$. Do $f\left( 0 \right)=1\Rightarrow D=\frac{1}{2}\Rightarrow {{f}^{2}}\left( 1 \right)=8$. Chọn A.

Lời giải chi tiết:

Ta có: ${{f}'}'\left( x \right).f\left( x \right)={{{f}'}^{2}}\left( x \right)-{{x}^{2}}.{{f}^{2}}\left( x \right)\Leftrightarrow {{f}'}'\left( x \right).f\left( x \right)-{{{f}'}^{2}}\left( x \right)={{x}^{2}}{{f}^{2}}\left( x \right)$

$\Leftrightarrow \frac{{{f}'}'\left( x \right).f\left( x \right)-{{{{f}'}}^{2}}\left( x \right)}{{{f}^{2}}\left( x \right)}={{x}^{2}}\,\,\,\left( * \right)$

Mặt khác ${{\left[ \frac{{f}'\left( x \right)}{f\left( x \right)} \right]}^{\prime }}=\frac{{{f}'}'\left( x \right).f\left( x \right)-{{{{f}'}}^{2}}\left( x \right)}{{{f}^{2}}\left( x \right)}$, lấy nguyên hàm 2 vế của $\left( * \right)$ ta được:

$\frac{{f}'\left( x \right)}{f\left( x \right)}=\frac{{{x}^{3}}}{3}+C$

Do $f\left( 1 \right)={f}'\left( 1 \right)=1\Rightarrow C=\frac{2}{3}$. Tiếp tục nguyên hàm 2 vế ta được: $\ln f\left( x \right)=\left( \frac{{{x}^{4}}}{12}+\frac{2x}{3} \right)+D$

Do $f\left( 1 \right)=1\Rightarrow D=-\frac{3}{4}\Rightarrow \ln f\left( x \right)=\frac{{{x}^{4}}}{12}+\frac{2x}{3}-\frac{3}{4}\Rightarrow \ln \left[ f\left( 3 \right) \right]=8$. Chọn B.

Lời giải chi tiết:

Đặt $t=\sqrt{x}\Leftrightarrow {{t}^{2}}=x\Leftrightarrow dx=2tdt$ và $\left\{ \begin{array}  {} x=0\Rightarrow t=0 \\  {} x=1\Rightarrow t=1 \\ \end{array} \right..$

Khi đó: $\int\limits_{0}^{1}{f\left( \sqrt{x} \right)dx}=\int\limits_{0}^{1}{2t.f\left( t \right)dt=2\int\limits_{0}^{1}{x.}}f\left( x \right)dx=\frac{2}{5}\Leftrightarrow \int\limits_{0}^{1}{x.f\left( x \right)dx=\frac{1}{5}}.$

Đặt $\left\{ \begin{array}  {} u=f\left( x \right) \\  {} dv=xdx \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}  {} du={f}'\left( x \right)dx \\  {} v=\frac{{{x}^{2}}}{2} \\ \end{array} \right.\Rightarrow \int\limits_{0}^{1}{x.f\left( x \right)dx=\left. \frac{{{x}^{2}}.f\left( x \right)}{2} \right|}_{0}^{1}-\frac{1}{2}\int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)dx}\Rightarrow \int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)dx}=\frac{3}{5}.$

Xét ${{\int\limits_{0}^{1}{\left[ {f}'\left( x \right)+k{{x}^{2}} \right]}}^{2}}dx=\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx+2k}\int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)dx}+{{k}^{2}}\int\limits_{0}^{1}{{{x}^{4}}dx=\frac{9}{5}+\frac{6}{5}k+\frac{1}{5}{{k}^{2}}=0\Leftrightarrow k=-3.}$

Do đó ${f}'\left( x \right)-3{{x}^{2}}=0\Leftrightarrow {f}'\left( x \right)=3{{x}^{2}}\Rightarrow f\left( x \right)=\int{{f}'\left( x \right)dx={{x}^{3}}+C}$ mà $f\left( 1 \right)=1\Rightarrow C=0.$

Vậy $f\left( x \right)={{x}^{3}}\xrightarrow{{}}I=\int\limits_{0}^{1}{{{x}^{3}}dx=\left. \frac{{{x}^{4}}}{4} \right|}_{0}^{1}=\frac{1}{4}$. Chọn B.              $$

Lời giải chi tiết:

Ta có $\int\limits_{0}^{1}{{f}'\left( x \right).cos\pi xdx=\int\limits_{0}^{1}{\cos \pi xd\left( f\left( x \right) \right)}}=\left. f\left( x \right).cos\pi x \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( x \right)}.{{\left( \cos \pi x \right)}^{\prime }}dx$

$=-\left[ f\left( 1 \right)+f\left( 0 \right) \right]+\pi \int\limits_{0}^{1}{f\left( x \right).sin\pi xdx}=\frac{\pi }{2}\Rightarrow \int\limits_{0}^{1}{f\left( x \right).sin\pi xdx}=\frac{1}{2}.$

Xét $\int\limits_{0}^{1}{{{\left[ f\left( x \right)+k.\sin \pi x \right]}^{2}}}dx=0\Leftrightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)}dx+2k.\int\limits_{0}^{1}{f\left( x \right).sin\pi xdx}+{{k}^{2}}.\int\limits_{0}^{1}{{{\sin }^{2}}}\left( \pi x \right)dx=0$

$\Leftrightarrow \frac{1}{2}{{k}^{2}}+2k.\frac{1}{2}+\frac{1}{2}=0\Leftrightarrow {{\left( k+1 \right)}^{2}}=0\Leftrightarrow k=-1.$ Suy ra $\int\limits_{0}^{1}{{{\left[ f\left( x \right)-\sin \pi x \right]}^{2}}}dx=0.$

Vậy $f\left( x \right)=\sin \pi x\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx=}\int\limits_{0}^{1}{sin\pi xdx}=\frac{2}{\pi }$. Chọn B.

Lời giải chi tiết:

Giả thuyết tương đương với $\int\limits_{0}^{1}{{{\left[ {{f}^{2}}'\left( x \right)f\left( x \right)-x \right]}^{2}}}dx=0$

$\Leftrightarrow {{f}^{2}}'\left( x \right)f\left( x \right)-x=0\Leftrightarrow {{f}^{2}}'\left( x \right)f\left( x \right)=x\Leftrightarrow {f}'\left( x \right).\sqrt{f\left( x \right)}=\sqrt{x}$

Nguyên hàm 2 vế ta được: $\frac{2}{3}\sqrt{{{f}^{3}}\left( x \right)}=\frac{2}{3}\sqrt{{{x}^{3}}}+C$

Mặt khác $f\left( 0 \right)=1\Rightarrow C=\frac{2}{3}\Rightarrow \sqrt{{{f}^{3}}\left( x \right)}=\sqrt{{{x}^{3}}}+1$

Vậy $f\left( 1 \right)=\sqrt[3]{4}.$ Chọn A.

Lời giải chi tiết:

Ta có $\int{{f}'\left( x \right)dx=\ln \left| 2x-1 \right|+}C=\left\{ \begin{array}  {} \ln \left( 2x-1 \right)+{{C}_{1}}\text{ khi  }x>\frac{1}{2} \\  {} \ln \left( 1-2x \right)+{{C}_{2}}\text{ khi }x<\frac{1}{2} \\ \end{array} \right..$

Do $f\left( 0 \right)=1$ và $f\left( 1 \right)=2\Rightarrow \left\{ \begin{array}  {} {{C}_{1}}=1 \\  {} {{C}_{2}}=2 \\ \end{array} \right.\Rightarrow f\left( -1 \right)+f\left( 3 \right)=\ln 3+\ln 5+{{C}_{1}}+{{C}_{2}}=3+\ln 15$. Chọn C.

Lời giải chi tiết:

Ta có: ${f}'\left( x \right)=\frac{4}{{{x}^{2}}-4}\Rightarrow \int{{f}'\left( x \right)dx=\int{\frac{4dx}{{{x}^{2}}-4}}}=\int{\frac{4dx}{\left( x-2 \right)\left( x+2 \right)}=\int{\left( \frac{1}{x-2}-\frac{1}{x+2} \right)}}dx$

$\Rightarrow f\left( x \right)=\ln \left| \frac{x-2}{x+2} \right|+C=\left\{ \begin{array}  {} \ln \frac{x-2}{x+2}+{{C}_{1}}\text{     khi }x>2\text{ } \\  {} \ln \left( \frac{2-x}{x+2} \right)+{{C}_{2}}\text{  khi  }-2<x<2 \\  {} \ln \frac{x-2}{x+2}+{{C}_{3}}\text{      khi }x<-2 \\ \end{array} \right..$

Lại có: $f\left( -3 \right)=0\Rightarrow {{C}_{3}}=-\ln 5$; $f\left( 0 \right)=1\Rightarrow {{C}_{2}}=1$; $f\left( 3 \right)=2\Rightarrow {{C}_{1}}=2-\ln \frac{1}{5}$.

Do đó $P=f\left( -4 \right)+f\left( -1 \right)+f\left( 4 \right)=\ln 3+\ln 3+\ln \frac{1}{3}+{{C}_{1}}+{{C}_{2}}+{{C}_{3}}=3+\ln 3$. Chọn B.

Lời giải chi tiết:

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

$\Rightarrow f\left( x \right)=\frac{1}{2}\ln \left| \frac{x-1}{x+1} \right|+C=\left\{ \begin{array}  {} \frac{1}{2}\ln \frac{x-1}{x+1}+{{C}_{1}}\text{     khi }x>1\text{ } \\  {} \frac{1}{2}\ln \left( \frac{1-x}{x+1} \right)+{{C}_{2}}\text{  khi  }-1<x<1 \\  {} \frac{1}{2}\ln \frac{x-1}{x+1}+{{C}_{3}}\text{      khi }x<-1 \\ \end{array} \right..$

Theo bài ra ta có: $\left\{ \begin{array}  {} f\left( -3 \right)+f\left( 3 \right)=0 \\  {} f\left( -\frac{1}{2} \right)+f\left( \frac{1}{2} \right)=2 \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}  {} {{C}_{1}}+{{C}_{3}}=0 \\  {} 2{{C}_{2}}=2\Rightarrow {{C}_{2}}=1 \\ \end{array} \right.$

Do đó $T=f\left( -2 \right)+f\left( 0 \right)+f\left( 4 \right)=\left[ f\left( -2 \right)+f\left( 4 \right) \right]+f\left( 0 \right)$

$=\frac{1}{2}\ln 3+{{C}_{1}}+\frac{1}{2}\ln \frac{3}{5}+{{C}_{3}}+{{C}_{2}}=1+\frac{1}{2}\ln \frac{9}{5}$. Chọn B.

Lời giải chi tiết:

Ta có ${{\left( \int\limits_{0}^{{{x}^{2}}}{f\left( t \right)dt} \right)}^{\prime }}={{\left( x.cos\pi x \right)}^{\prime }}\Leftrightarrow {{\left( {{x}^{2}} \right)}^{\prime }}.f\left( {{x}^{2}} \right)=cos\pi x-\pi x.\sin \pi x$

$\Leftrightarrow 2x.f\left( {{x}^{2}} \right)=cos\pi x-\pi x.\sin \pi x.$ Thay $x=2$ vào 2 vế, ta được $4f\left( 4 \right)=1\Leftrightarrow f\left( 4 \right)=\frac{1}{4}$. Chọn B.

Lời giải chi tiết:

Gọi $F\left( t \right)$ là nguyên hàm của hàm số $f\left( t \right)=\cos \sqrt{t}.$

Ta có $G\left( x \right)=\int\limits_{0}^{{{x}^{2}}}{\cos \sqrt{t}dt=F\left( {{x}^{2}} \right)-F\left( 0 \right)\xrightarrow{{}}{G}'\left( x \right)={{\left[ F\left( {{x}^{2}} \right) \right]}^{\prime }}}=2x.{F}'\left( {{x}^{2}} \right)=2x.f\left( {{x}^{2}} \right).$

Lại có $f\left( {{x}^{2}} \right)=\cos \sqrt{{{x}^{2}}}=\cos x$ nên suy ra ${G}'\left( x \right)=2x.cosx.$ Chọn B.