Question

The area bounded by the curve $y=\{\begin{array}{ll}{x}^2,x<0& \\ x,x\ge 0& \end{array}$ 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)

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

$A_1=\underset{y=0}{\overset{4}{\int }}\sqrt{y}dy$
$={\left[\frac{2}{3}{y}^{3/2}\right]}_0^4$
$t=\frac{2}{3}\left[(4{)}^{3/2}-0\right]$
$=\frac{2}{3}\times (2{)}^3$
$=\frac{16}{3}$
and area of $OBCO$,
$A_2=\underset{0}{\overset{4}{\int }}ydy={\left[\frac{y^2}{2}\right]}_0^4$
$=\frac{1}{2}\left[(4{)}^2-0\right]=\frac{16}{2}=8$
Hence, area of $OABO=A_1+A_2$
$=\frac{16}{3}+8=\frac{40}{3}$ sq unit