The area (in sq. units) of the region enclose | vedclass

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

Question

(B) To find the area enclosed by the curves $y=x^{2}-1$ and $y=1-x^{2}$,we first find their points of intersection by setting $x^{2}-1 = 1-x^{2}$.This

Vedclass · Accepted Answer

$\frac{8}{3}$

Vedclass · Answer

(B) To find the area enclosed by the curves $y=x^{2}-1$ and $y=1-x^{2}$,we first find their points of intersection by setting $x^{2}-1 = 1-x^{2}$.This gives $2x^{2} = 2$,so $x^{2} = 1$,which means $x = -1$ and $x = 1$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 1$.$A = \int_{-1}^{1} ((1-x^{2}) - (x^{2}-1)) dx$$A = \int_{-1}^{1} (2 - 2x^{2}) dx$Since the function $f(x) = 2 - 2x^{2}$ is an even function,we can write:$A = 2 \int_{0}^{1} (2 - 2x^{2}) dx = 4 \int_{0}^{1} (1 - x^{2}) dx$$A = 4 [x - \frac{x^{3}}{3}]_{0}^{1}$$A = 4 (1 - \frac{1}{3}) = 4 (\frac{2}{3}) = \frac{8}{3}$ sq. units.

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

Similar Questions

The area (in sq. units) of the region $A = \{(x, y) : x^2 \le y \le x + 2\}$ 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 $\sin \left(a+\frac{\pi}{4}\right)=$

Find the area enclosed by the parabola $4y = 3x^{2}$ and the line $2y = 3x + 12$.

The sine and cosine curves intersect infinitely many times,creating bounded regions of equal areas. The area of one such region is

The area (in square units) of the region bounded by the parabola $y^2=4(x-2)$ and the line $y=2x-8$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module