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The area (in square units) of the region enclosed by the curves $y = x, x =e, y=\frac{1}{x}$ and the positive x-axis is |
$\frac{1}{2}$ 1 $\frac{3}{2}$ $\frac{5}{2}$ |
$\frac{3}{2}$ |
Required area = $\int\limits_0^1x\,dx+\int\limits_1^e\frac{1}{x}dx$ $=\left[\frac{x^2}{2}\right]_0^1+[\log x]_1^e=\frac{1}{2}+1=\frac{3}{2}$
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