The area bounded by the parabolas y=x^2 and y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the parabolas $y=x^2$ and $y=1-x^2$.To find the intersection points,set $x^2 = 1-x^2$,which gives $2x^2 = 1$,so $x^2 = \frac{1}{2}$,implying

Vedclass · Accepted Answer

$\frac{2 \sqrt{2}}{3}$

Vedclass · Answer

(B) Given the parabolas $y=x^2$ and $y=1-x^2$.To find the intersection points,set $x^2 = 1-x^2$,which gives $2x^2 = 1$,so $x^2 = \frac{1}{2}$,implying $x = \pm \frac{1}{\sqrt{2}}$.The intersection points are $A\left(\frac{1}{\sqrt{2}}, \frac{1}{2}ight)$ and $C\left(-\frac{1}{\sqrt{2}}, \frac{1}{2}ight)$.The area of the shaded region is given by the integral of the upper curve minus the lower curve from $x = -\frac{1}{\sqrt{2}}$ to $x = \frac{1}{\sqrt{2}}$.Area $= \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} [(1-x^2) - x^2] dx = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (1-2x^2) dx$.Since the function is even,Area $= 2 \int_{0}^{\frac{1}{\sqrt{2}}} (1-2x^2) dx$.$= 2 \left[ x - \frac{2x^3}{3} ight]_{0}^{\frac{1}{\sqrt{2}}} = 2 \left[ \frac{1}{\sqrt{2}} - \frac{2}{3} \left( \frac{1}{\sqrt{2}} ight)^3 ight] = 2 \left[ \frac{1}{\sqrt{2}} - \frac{2}{3 \cdot 2 \sqrt{2}} ight] = 2 \left[ \frac{1}{\sqrt{2}} - \frac{1}{3 \sqrt{2}} ight]$.$= 2 \left[ \frac{3-1}{3 \sqrt{2}} ight] = 2 \left[ \frac{2}{3 \sqrt{2}} ight] = \frac{4}{3 \sqrt{2}} = \frac{2 \sqrt{2}}{3} 	ext{ sq units}$.

The area bounded by the parabolas $y=x^2$ and $y=1-x^2$ is equal to

Similar Questions

The area (in sq. units) bounded by the curves $y=\sqrt{x}$,$2y-x+3=0$,$X$-axis and lying in the first quadrant,is

The area (in square units) bounded by the curves $y=2x^2$ and $y=\max \{x-[x], x+|x|\}$ in between the lines $x=0$ and $x=2$ is

The area (in sq. units) of the region enclosed by the curves $y=x^{2}-1$ and $y=1-x^{2}$ is equal to

The area of the region bounded by the parabola $y=x^{2}-4x+5$ and the straight line $y=x+1$ is

If the area of the region bounded by $y=\cos x$,$y=\sin x$,$x=\frac{\pi}{4}$ and $x=\pi$ is bisected by the line $x=a$,then $\sin \left(a+\frac{\pi}{4}\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module