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Choose the correct option: Area bounded by the curve $^{y=x|x|}, x$-axis and the ordinates $x=-1$ and $x=1$ is given by [Hint: $y=x^{2}$ if $x>0$ and $y=-x^{2}$ if $x
Choose the correct option: Area bounded by the curve $^{y=x|x|}, x$-axis and the ordinates $x=-1$ and $x=1$ is given by [Hint: $y=x^{2}$ if $x>0$ and $y=-x^{2}$ if $x<0$ ]
A. 0
B. $^{\frac{1}{3}}$
C. $\frac{2}{3}$
D. $\frac{4}{3}$
Applications of Integrals | CBSE | Class 12 | Maths | Miscellaneous | NCERT
Choose the correct option: Area bounded by the curve $^{y=x|x|}, x$-axis and the ordinates $x=-1$ and $x=1$ is given by [Hint: $y=x^{2}$ if $x>0$ and $y=-x^{2}$ if $x<0$ ]
A. 0
B. $^{\frac{1}{3}}$
C. $\frac{2}{3}$
D. $\frac{4}{3}$

Solution:


$\text { Required area }=\int_{-1}^{1} y d x$
$\begin{array}{l}
=\int_{-1}^{1} x|x| d x \\
=\int_{-1}^{0} x^{2} d x+\int_{0}^{1} x^{2} d x
\end{array}$
$\begin{array}{l}
=\left[\frac{x^{3}}{3}\right]_{-1}^{0}+\left[\frac{x^{3}}{3}\right]_{0}^{1} \\
=-\left(-\frac{1}{3}\right)+\frac{1}{3} \\
=\frac{2}{3} \text { sq.units }
\end{array}$
As a result, the correct answer is C.

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