Question

The area bounded by the curve $y= \begin{cases} x^2,x<0 & \quad\\\ x,x \geq 0 & \quad \\\ \end{cases}$ and the line $y = 4$ is

• A. $\frac {40}{3}$
• B. $\frac {16}{3}$
• C. $\frac {32}{3}$
• D. $\frac {8}{3}$

Answer 1

Answer: (A)

$\frac {40}{3}$

Answer 2

Given, curve $y= \begin{cases} x^2,x<0 \\\ x, x \ge 0 \end{cases}$
and line $y=4$
Area of OABO,

A_{1} =\int_\limits{y=0}^{4} \sqrt{y} d y
$=\left[\frac{2}{3} y^{3 / 2}\right]_{0}^{4}$
$t=\frac{2}{3}\left[(4)^{3 / 2}-0\right]$
$=\frac{2}{3} \times(2)^{3}$
$=\frac{16}{3}$
and area of $OBCO$,
A_{2} =\int_\limits{0}^{4} y d y=\left[\frac{y^{2}}{2}\right]_{0}^{4}
$=\frac{1}{2}\left[(4)^{2}-0\right]=\frac{16}{2}=8$
Hence, area of $O A B O=A_{1}+A_{2}$
$=\frac{16}{3}+8=\frac{40}{3}$ sq unit