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
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
Updated On: Apr 8, 2024
- $\frac {40}{3}$
- $\frac {16}{3}$
- $\frac {32}{3}$
- $\frac {8}{3}$
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The Correct Option is A
Solution and Explanation
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
\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
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