The area (in $s q$. units) bounded by the curve $y=x|x|, \mathrm{X}$-axis and the lines $x=-1$ and $x=1$ is
The area (in $s q$. units) bounded by the curve $y=x|x|, \mathrm{X}$-axis and the lines $x=-1$ and $x=1$ is
- $\frac{2}{3}$
- $\frac{1}{3}$
- 1
- $\frac{4}{3}$
Solution
$y=x|x| ...[Given]$
Required area
$\begin{aligned}
& =\int_{-1}^1 x|x| \mathrm{d} x \\
& =2 \int_0^1 x^2 \mathrm{~d} x \quad \ldots[\because \text { Area is always positive }] \\
& =2 \times\left[\frac{x^3}{3}\right]_0^1 \\
& =2 \times\left(\frac{1}{3}-0\right)=\frac{2}{3} \text { sq.units }
\end{aligned}$
Asked in: MHT CET 2023 (09 May Shift 2)
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