Bài tập tính tích phân áp dụng bất đẳng thức tích phân có đáp án chi tiết

Bài tập tính tích phân áp dụng bất đẳng thức tích phân có đáp án

Dưới đây là một số bài tập áp dụng bất đẳng thức để tính tích phân có Lời giải chi tiết

Lời giải chi tiết

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

Suy ra $\frac{5}{2}=\frac{f\left( 1 \right)}{2}-\int\limits_{0}^{1}{\frac{{{x}^{2}}}{2}.{f}'\left( x \right)dx}\Rightarrow \int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)dx}=1$

Ta chọn k sao cho: $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+k{{x}^{2}} \right]}^{2}}dx=\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx+2k\int\limits_{0}^{1}{{f}'\left( x \right){{x}^{2}}dx+{{k}^{2}}\int\limits_{0}^{1}{{{x}^{4}}dx}}}}=0$

$=5+2k+\frac{{{k}^{2}}}{2}=0\Rightarrow k=-5\Rightarrow \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)-5{{x}^{2}} \right]}^{2}}dx=0\Rightarrow {f}'\left( x \right)=5{{x}^{2}}}\Rightarrow f\left( x \right)=\frac{5{{x}^{3}}}{3}+C$

Do $f\left( 1 \right)=6\Rightarrow C=\frac{13}{3}\Rightarrow f\left( x \right)=\frac{5{{x}^{3}}}{3}+\frac{13}{3}\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\frac{19}{4}$. Chọn D.

Lời giải chi tiết

Đặt $\left\{ \begin{array}  {} u=f\left( x \right) \\  {} dv=xdx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du={f}'\left( x \right)dx \\  {} v=\frac{{{x}^{2}}}{2} \\ \end{array} \right.$

Do đó $\int\limits_{0}^{1}{x.f\left( x \right)dx=\left. \frac{{{x}^{2}}}{2}.f\left( x \right) \right|_{0}^{1}-\frac{1}{2}\int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)dx}}=\frac{1}{2}-\frac{1}{2}\int\limits_{0}^{1}{{{x}^{2}}{f}'\left( x \right)dx}=\frac{1}{5}$

Suy ra $\int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)dx}=\frac{3}{5};\int\limits_{0}^{1}{{{x}^{4}}}dx=\frac{1}{5}$

Chọn k sao cho: $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+k{{x}^{2}} \right]}^{2}}dx=\frac{9}{5}+\frac{6k}{5}+\frac{{{k}^{2}}}{5}=0\Rightarrow k=-3}$

Như vậy $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)-3{{x}^{2}} \right]}^{2}}dx=0\Rightarrow {f}'\left( x \right)=3{{x}^{2}}}\Rightarrow f\left( x \right)={{x}^{3}}+C$

Do $f\left( 1 \right)=1\Rightarrow C=0\Rightarrow I=\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{{{x}^{3}}dx}=\frac{1}{4}$. Chọn B.

Lời giải chi tiết

Đặt $\left\{ \begin{array}  {} u=f\left( x \right) \\  {} dv={{x}^{3}}dx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=f'\left( x \right)dx \\  {} v=\frac{{{x}^{4}}}{4} \\ \end{array} \right.$

Do đó $\int\limits_{0}^{1}{{{x}^{3}}f\left( x \right)dx=\left. \frac{{{x}^{4}}}{4}.f\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\frac{{{x}^{4}}}{4}.{f}'\left( x \right)dx}}=\frac{3}{20}-\int\limits_{0}^{1}{\frac{{{x}^{4}}{f}'\left( x \right)}{4}dx}\Rightarrow \int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)dx}=-\frac{2}{9}$

Lại có: $\int\limits_{0}^{1}{{{x}^{8}}dx}=\frac{1}{9}$ ta chọn k sao cho: $\int\limits_{0}^{1}{\left[ {f}'\left( x \right)+k{{x}^{4}} \right]dx=\frac{4}{9}+2k.\frac{-2}{9}+\frac{{{k}^{2}}}{9}=0\Rightarrow k=2}$

Như vậy $\int\limits_{0}^{1}{\left[ {f}'\left( x \right)+2{{x}^{4}} \right]dx=0\Rightarrow {f}'\left( x \right)=-2{{x}^{4}}}\Rightarrow f\left( x \right)=\frac{-2{{x}^{5}}}{5}+C$

Do $f\left( 1 \right)=\frac{3}{5}\Rightarrow \frac{3}{5}=\frac{-2}{5}+C\Leftrightarrow C=1\Leftrightarrow f\left( x \right)-1=\frac{-2}{5}{{x}^{5}}\Rightarrow \int\limits_{0}^{1}{\left[ f\left( x \right)-1 \right]dx}=\frac{-1}{15}$. Chọn B.

Lời giải chi tiết

Đặt $\left\{ \begin{array}  {} u=f\left( x \right) \\  {} dv={{x}^{3}}dx \\ \end{array} \right.\Rightarrow \left\{ \begin{array}  {} du=f'\left( x \right)dx \\  {} v=\frac{{{x}^{4}}}{4} \\ \end{array} \right.$khi đó $\int\limits_{0}^{1}{{{x}^{3}}f\left( x \right)dx=\left. \frac{{{x}^{4}}}{4}.f\left( x \right) \right|_{0}^{3}-\int\limits_{0}^{3}{\frac{{{x}^{4}}}{4}{f}'\left( x \right)dx}}$

Suy ra $\frac{45}{2}=\frac{81f\left( 3 \right)}{4}-\int\limits_{0}^{3}{\frac{{{x}^{4}}}{4}.{f}'\left( x \right)dx}\Rightarrow \int\limits_{0}^{3}{{{x}^{4}}}{f}'\left( x \right)dx=-9$

Ta chọn k sao cho: $\int\limits_{0}^{3}{{{\left[ {f}'\left( x \right)+k{{x}^{4}} \right]}^{2}}dx=\int\limits_{0}^{3}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx+2k\int\limits_{0}^{3}{{f}'\left( x \right){{x}^{4}}dx+{{k}^{2}}\int\limits_{0}^{3}{{{x}^{8}}dx}}}}$

$=\frac{1}{27}-2.9k+2187{{k}^{2}}=0\Rightarrow k=\frac{1}{243}\Rightarrow {f}'\left( x \right)=-\frac{1}{243}{{x}^{4}}\Rightarrow f\left( x \right)=\frac{-{{x}^{5}}}{1215}+\frac{6}{5}\Rightarrow \int\limits_{0}^{3}{f\left( x \right)dx=}\frac{7}{2}$. Chọn C.

Lời giải chi tiết

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

Suy ra $I=f\left( 1 \right)-\int\limits_{0}^{1}{{{x}^{3}}.{f}'\left( x \right)dx\Rightarrow }\int\limits_{0}^{1}{{{x}^{3}}.{f}'\left( x \right)dx=-1\Leftrightarrow \int\limits_{0}^{1}{14{{x}^{3}}.{f}'\left( x \right)dx=-7}}$. Mà $\int\limits_{0}^{1}{49{{x}^{6}}}dx=7$ suy ra  $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx+\int\limits_{0}^{1}{7{{x}^{3}}}{f}'\left( x \right)dx+\int\limits_{0}^{1}{49{{x}^{6}}dx}=0}\Leftrightarrow \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+7{{x}^{3}} \right]}^{2}}dx}=0$

Vậy ${f}'\left( x \right)+7{{x}^{3}}=0\Rightarrow f\left( x \right)=-\frac{7}{4}{{x}^{4}}+C$

Mà lại có: $f\left( 1 \right)=0\Rightarrow f\left( x \right)=\frac{7}{4}\left( 1-{{x}^{4}} \right)\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\frac{7}{5}$. Chọn A.