Definite Integration Ques 91
- The integral $\int _1^{e} \frac{x}{e}^{2 x}-\frac{e}{x}^{x} \log _e x d x$ is equal to
(a) $\frac{3}{2}-e-\frac{1}{2 e^{2}}$
(b) $-\frac{1}{2}+\frac{1}{e}-\frac{1}{2 e^{2}}$
(c) $\frac{1}{2}-e-\frac{1}{e^{2}}$
(d) $\frac{3}{2}-\frac{1}{e}-\frac{1}{2 e^{2}}$
(2019 Main, 12 Jan II)
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Solution:
- Let $I=\int _1^{e} \frac{x}{e}^{2 x}-\frac{e}{x}^{x} \log _e x d x$
Now, put $\frac{x^{x}}{e}=t \Rightarrow \quad x \log _e \frac{x}{e}=\log t$
$\Rightarrow \quad x\left(\log _e x-\log _e e\right)=\log t$
$\Rightarrow \quad x \frac{1}{x}+\left(\log _e x-\log _e e\right) d x=\frac{1}{t} d t$
$\Rightarrow\left(1+\log _e x-1\right) d x=\frac{1}{t} d t \Rightarrow\left(\log _e x\right) d x=\frac{1}{t} d t$ Also, upper limit $x=e \Rightarrow t=1$ and lower limit $x=1 \Rightarrow$ $t=\frac{1}{e}$
$\therefore I=\int _{1 / e}^{1} t^{2}-\frac{1}{t} \cdot \frac{1}{t} d t \Rightarrow I=\int _{1 / e}^{1}\left(t-t^{-2}\right) d t$
$$ I=\frac{t^{2}}{2}+\frac{1}{t}{ } _{\frac{1}{e}}^{1}=\frac{1}{2}+1-\frac{1}{2 e^{2}}+e \quad=\frac{3}{2}-e-\frac{1}{2 e^{2}} $$
BETA
