Mathematics | TS EAMCET 2019 | Discussion Page

Area of the region (in sq units) bounded by the curve y = $\sqrt{x}$ , x= $\sqrt{y}$ and the lines x=1, x=4 , is

$\frac{8}{3}$

$\frac{49}{3}$

$\frac{16}{3}$

$\frac{14}{3}$

Option B

We have ,

y = $\sqrt{x}$ , x= $\sqrt{y}$ ,x=1 ,x=4

The graph of curves are

Area of shaded region

$=\int_{1}^{4} (x^{2}-\sqrt{x})dx=\left[\frac{x^{3}}{3}-\frac{2}{3}(x)^{3/2}\right]_{1}^{4}$

= $\left(\frac{64}{3}-\frac{16}{3}\right)-\left(\frac{1}{3}-\frac{2}{3}\right)$

= $\frac{64-16-1+2}{3}=\frac{49}{3}$

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