Giải bài 4 trang 113 – SGK môn Giải tích lớp 12
Sử dụng phương pháp tính tích phân từng phần, hãy tính:
a) \(\int\limits_{0}^{\frac{\pi }{2}}{\left( 1+x \right)\sin xdx}\);
b) \(\int\limits_{1}^{e}{{{x}^{2}}\ln xdx}\);
c) \(\int\limits_{0}^{1}{\ln \left( 1+x \right)dx}\);
d) \(\int\limits_{0}^{1}{\left( {{x}^{2}}-2x-1 \right){{e}^{-x}}dx}\)
Phương pháp:
Nếu u=u(x) và v=v(x) là hai hàm số có đạo hàm liên tục trên đoạn [a;b] thì
\(\int\limits_{{a}}^{{b}}{u\left( x \right)v'\left( x \right)dx}=\left( u\left( x \right)v\left( x \right) \right)| _{a}^{b}-\int\limits_{a}^{b}{u'\left( x \right)v\left( x \right)dx} \)
hay \(\int\limits_{a}^{b}{udv}=uv\left| _{a}^{b} \right.-\int\limits_{a}^{b}{vdu}\)
a) Đặt \( \left\{ \begin{aligned} & 1+x=u \\ & \sin xdx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=dx \\ & v=-\cos x \\ \end{aligned} \right. \)
\(\begin{aligned} \int\limits_{0}^{\frac{\pi }{2}}{\left( 1+x \right)\sin xdx}&=-\left( 1+x \right)\cos x\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\frac{\pi }{2}} \right.+\int\limits_{0}^{\frac{\pi }{2}}{\cos xdx} \\ & =1+\sin x\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\frac{\pi }{2}} \right. \\ & =2 \\ \end{aligned} \)
b) Đặt \( \left\{ \begin{aligned} & \ln x=u \\ & {{x}^{2}}dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=\dfrac{1}{x}dx \\ & v=\dfrac{{{x}^{3}}}{3} \\ \end{aligned} \right. \)
\(\begin{aligned} \int\limits_{1}^{e}{{{x}^{2}}\ln xdx}&=\dfrac{{{x}^{3}}}{3}\ln x\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} e \\ \end{smallmatrix}} \right.-\int\limits_{1}^{e}{\dfrac{{{x}^{3}}}{3}.\dfrac{1}{x}dx} \\ & =\dfrac{{{e}^{3}}}{3}-\dfrac{{{x}^{3}}}{9}\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} e \\ \end{smallmatrix}} \right. \\ & =\dfrac{{{e}^{3}}}{3}-\dfrac{{{e}^{3}}}{9}+\dfrac{1}{9} \\ & =\dfrac{2{{e}^{3}}+1}{9} \\ \end{aligned} \)
c) Đặt \( \left\{ \begin{aligned} & ln\left( 1+x \right)=u \\ & dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=\dfrac{1}{1+x}dx \\ & v=x \\ \end{aligned} \right. \)
\(\begin{aligned} \int\limits_{0}^{1}{\ln \left( 1+x \right)dx}&=x\ln \left( 1+x \right)\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right.-\int\limits_{0}^{1}{\dfrac{x}{x+1}dx} \\ & =\ln 2-\left( x-\ln \left| x+1 \right| \right)\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right. \\ & =\ln 2-1+\ln 2 \\ & =2\ln 2-1 \\ \end{aligned} \)
d) Đặt \(\left\{ \begin{aligned} & {{x}^{2}}-2x-1=u \\ & {{e}^{-x}}dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=2\left( x-1 \right)dx \\ & v=-{{e}^{-x}} \\ \end{aligned} \right. \)
\(\begin{aligned} \int\limits_{0}^{1}{\left( {{x}^{2}}-2x-1 \right){{e}^{-x}}dx}&=-\left( {{x}^{2}}-2x-1 \right){{e}^{-x}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right.+2\int\limits_{0}^{1}{\left( x-1 \right)}{{e}^{-x}}dx \\ & =\dfrac{2}{e}-1+2\int\limits_{0}^{1}{\left( x-1 \right)}{{e}^{-x}}dx \\ \end{aligned} \)
Đặt \( \left\{ \begin{aligned} & x-1=u \\ & {{e}^{-x}}dx=dx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & dx=du \\ & v=-{{e}^{-x}} \\ \end{aligned} \right. \)
\(\begin{aligned} \int\limits_{0}^{1}{\left( x-1 \right)}{{e}^{-x}}dx&=-\left( x-1 \right){{e}^{-x}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right.+\int\limits_{0}^{1}{{{e}^{-x}}dx} \\ & =-1-{{e}^{-x}}\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} 1 \\ \end{smallmatrix}} \right. \\ & =-1-\dfrac{1}{e}+1 \\ & =-\dfrac{1}{e} \\ \end{aligned}\\ \Rightarrow \int\limits_{0}^{1}{\left( {{x}^{2}}-2x-1 \right){{e}^{-x}}dx}=\dfrac{2}{e}-1-\dfrac{2}{e}=-1 \)