Definite Integration Question 28
Question 28
- The value of the integral
(2000, 2M)
(a)
Show Answer
Solution:
- $\int_{e^{-1}}^{e^{2}} \frac{\log {e} x}{x} d x=\int{e^{-1}}^{1} \frac{\log {e} x}{x} d x-\int{1}^{e^{2}} \frac{\log _{e} x}{x} d x
$$ \begin{aligned} & =-\int_{e^{-1}}^{1} \frac{\log {e} x}{x} d x+\int{1}^{e^{2}} \frac{\log _{e} x}{x} d x \ & =-\frac{1}{2}\left[\left(\log {e} x\right)^{2}\right]{e^{-1}}^{1}+\frac{1}{2}\left[\left(\log {e} x\right)^{2}\right]{1}^{e^{2}} \ & =-\frac{1}{2}\left{0-(-1)^{2}\right}+\frac{1}{2}\left(2^{2}-0\right)=\frac{5}{2} \end{aligned} $$
