MCQ
The area bounded by the parabolas $y=x^2$ and $y=1-x^2$ equals
- A$\frac{\sqrt{2}}{3}$
- ✓$\frac{2 \sqrt{2}}{3}$
- C$\frac{1}{3}$
- D$\frac{2}{3}$
✓
Answer
Correct option: B.
$\frac{2 \sqrt{2}}{3}$
b
(b)
(b)
We have, $y=x^2$ and $y=1-x^2$
Intersection point of $y=x^2$ and $y=1-x^2$ is $A\left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right)$ and $C\left(\frac{-1}{\sqrt{2}}, \frac{1}{2}\right)$
Area of shaded region
$=2 \int \limits_0^{1 / \sqrt{2}}\left[\left(1-x^2\right)-\left(x^2\right)\right] d x$
$=2 \int \limits_0^{1 / \sqrt{2}}\left(1-2 x^2\right) d x$
$=2\left[x-\frac{2 x^3}{3}\right]_0^{1 / \sqrt{2}}$
$=2\left[\frac{1}{\sqrt{2}}-\frac{1}{3 \sqrt{2}}\right]$
$=\frac{2}{\sqrt{2}}\left[\frac{3-1}{3}\right]=\frac{2 \sqrt{2}}{3}$ sq units
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