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Application of Integrals

Multiple Choice Questions

151.

$The area of the region described by \{(x, y) / x^{2} + y \leq 1 and y \leq 1 - x\} is$

$\frac{π}{2} - \frac{2}{3}$
$\frac{π}{2} + \frac{2}{3}$
$\frac{π}{2} + \frac{4}{3}$
$\frac{π}{2} - \frac{4}{3}$

152.
Consider a region R = {(x, y) $\in$ R²: x^{2 $\leq y \leq 2 x$}}. If a line y = $α$ divides the area of region R into two equal parts, then which of the following is true ?

$3 α^{2} - 8 α + 8 = 0$

$α^{3} - 6 α^{\frac{3}{2}} - 16 = 0$

$α^{3} - 6 α^{2} + 16 = 0$

${3α}^{2} - 8 α^{\frac{3}{2}} + 8 = 0$

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${3α}^{2} - 8 α^{\frac{3}{2}} + 8 = 0$

$\int_{0}^{4} (\sqrt{y} - \frac{y}{20}) dy = 2 \int_{0}^{\infty} (\sqrt{y} - \frac{y}{2}) dy \Rightarrow {(\frac{2 y^{\frac{3}{2}}}{3} - \frac{y^{2}}{2})}_{0}^{4} = 2 {(\frac{2 y^{\frac{3}{2}}}{3} - \frac{y^{2}}{4})}_{0}^{α} \Rightarrow \frac{16}{3} - 4 = 2 (\frac{2}{3} α^{\frac{3}{2}} - \frac{α^{2}}{4}) \Rightarrow \frac{4}{3} = 2 (\frac{2}{3} α^{\frac{3}{2}} - \frac{α^{2}}{4}) \Rightarrow \frac{2}{3} = \frac{8 α^{\frac{3}{2}} - 3 α^{2}}{12} \Rightarrow 8 α^{\frac{3}{2}} - 3 α^{2} = 8$

153.

$The area (in sq unit) of the region \{(x, y); \frac{1}{2} \leq y \leq \sin (x), 0 \leq x \leq π\} is equal to$

$3 - 2 π$
$\sqrt{3} - \frac{π}{6}$
3 - $\frac{π}{3}$
$\sqrt{3} - \frac{π}{3}$

154.

The area (in sq. units) of the region A = {x, y) : (x –1) [x] $\leq y \leq 2 \sqrt{x}, 0 \leq x \leq 2$ ,where [t] denotes the greatest integer function, is :