Một bài tích phân nữa

Question

Tính tích phân :
a) $I=\int\limits_{0}^{2}xe^{-\frac{x}{3} }dx $
b) $I=\int\limits_{1}^{e}(1+x)\ln xdx $

score 3 · Answer 1

b)

Đặt : $\left\{ \begin{array}{l} x=\ln x \Rightarrow du=\frac{dx}{x} \\ dv=(1+x)dx \Rightarrow v=x+\frac{x^2}{2} \end{array} \right. $

Lúc đó : $I=(x+\frac{x^2}{2} )\ln x \left|\begin{array}{l}e\\1\end{array}\right. -\int\limits_{1}^{e}(x^2+\frac{x^2}{2} )(\frac{1}{x} )dx$

$=(e^2+\frac{e^2}{2} )\ln e-0-\int\limits_{1}^{e}dx-\int\limits_{1}^{e}\frac{xdx}{2} $

$=(e+\frac{e^2}{2} )-x \left|\begin{array}{l}e\\1\end{array}\right. -\frac{x^2}{4} \left|\begin{array}{l}e\\1\end{array}\right. $

$=e+\frac{e^2}{2}-e(e-1)-\frac{1}{4}(e^2-1)=\frac{1}{4}(e^2+5)$

score 3 · Answer 2

Bình chọn tăng 3 Bình chọn giảm

a)

Đặt : $\left\{ \begin{array}{l} u=x \Rightarrow du=dx\\ dv=e^{-\frac{x}{3} }dx \Rightarrow v= \frac{1}{-\frac{1}{3} }e^{-\frac{x}{3} }=-3e^{-\frac{x}{3} } \end{array} \right. $

$\Rightarrow I=-3xe^{-\frac{x}{3} }\left|\begin{array}{l}3\\0\end{array}\right.+3 \int\limits_{0}^{3}e^{-\frac{x}{3} }dx=-9e^{-1}+0+3 \frac{1}{-\frac{1}{3} }e^{-\frac{x}{3} } \left|\begin{array}{l}3\\0\end{array}\right. =-\frac{18}{e}+9=\frac{9e-18}{e}$