Area Question 12
Question 12
- The area of the region $\mathrm{A}={(x, y) ; 0 \leq y \leq x|x|+1$ and $-1 \leq x \leq 1}$ in sq. units, is
(2019 Main, 9 Jan II) (a) 2 (b) $\frac{4}{3}$ (c) $\frac{1}{3}$ (d) $\frac{2}{3}$
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Solution:
- We have,
$A={(x, y): 0 \leq y \leq x|x|+1$ and $-1 \leq x \leq 1}$
When $x \geq 0$, then $0 \leq y \leq x^{2}+1$
and when $x<0$, then $0 \leq y \leq-x^{2}+1$
Now, the required region is the shaded region.
$\left[\because y=x^{2}+1 \Rightarrow x^{2}=(y-1)\right.$, parabola with vertex $(0,1)$ and $y=-x^{2}+1 \Rightarrow x^{2}=-(y-1)$,
parabola with vertex $(0,1)$ but open downward] We need to calculate the shaded area, which is equal to
$$ \begin{aligned} \int_{-1}^{0}\left(-x^{2}+1\right) & d x+\int_{0}^{1}\left(x^{2}+1\right) d x \ & =-\frac{x^{3}}{3}+x_{-1}^{0}+\frac{x^{3}}{3}+x_{0}^{1} \ & =0–\frac{(-1)^{3}}{3}+(-1)+\frac{1}{3}+1-0 \ & =-\frac{1}{3}-1+\frac{4}{3} \ & =\frac{2}{3}+\frac{4}{3}=2 \end{aligned} $$
