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Q. The area of the region enclosed by the curves $y=x,$ $x=e, y=\frac{1}{x}$ and the positive $x$ -axis is

AIEEEAIEEE 2011Application of Integrals

Solution:

image
Required area
$=\frac{1}{2} \times 1 \times 1+\int\limits_{1}^{e} \frac{1}{x} d x$
$=\frac{1}{2}+[\ln x]_{1}^{e}$
$=\frac{1}{2}+1-0=\frac{3}{2}$ sq units