The area bounded by the curves y = 1 - cos(pi | vedclass

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

Question

(D) The area $A$ is bounded by $y_1 = 1 - \cos(\pi x)$ and $y_2 = -x^2$ between $x = -\frac{1}{2}$ and $x = \frac{1}{2}$.Since both curves are symmetr

Vedclass · Accepted Answer

$\frac{13}{12} - \frac{2}{\pi}$

Vedclass · Answer

(D) The area $A$ is bounded by $y_1 = 1 - \cos(\pi x)$ and $y_2 = -x^2$ between $x = -\frac{1}{2}$ and $x = \frac{1}{2}$.Since both curves are symmetric about the $y$-axis,the total area is $2 	imes \int_{0}^{1/2} (y_1 - y_2) dx$.$A = 2 \int_{0}^{1/2} (1 - \cos(\pi x) - (-x^2)) dx$$A = 2 \int_{0}^{1/2} (1 - \cos(\pi x) + x^2) dx$$A = 2 \left[ x - \frac{\sin(\pi x)}{\pi} + \frac{x^3}{3} ight]_{0}^{1/2}$$A = 2 \left[ (\frac{1}{2} - \frac{\sin(\pi/2)}{\pi} + \frac{(1/2)^3}{3}) - (0 - 0 + 0) ight]$$A = 2 \left[ \frac{1}{2} - \frac{1}{\pi} + \frac{1}{24} ight]$$A = 2 \left[ \frac{12 + 1}{24} - \frac{1}{\pi} ight] = 2 \left[ \frac{13}{24} - \frac{1}{\pi} ight]$$A = \frac{13}{12} - \frac{2}{\pi}$.

The area bounded by the curves $y = 1 - \cos(\pi x)$,$y = -x^2$ and the lines $x = \frac{1}{2}$ and $x = -\frac{1}{2}$ is:

Similar Questions

The area of the region $\{(x, y) \in R^2: y \geq \sqrt{|x+3|}, 5y \leq x+9 \leq 15\}$ is equal to

The area (in sq. units) of the region enclosed by the set of points ${(x, y) | y \leq x^2, xy \leq 8, y \geq 1}$ is...

The area of the smaller region enclosed by the curves $y^{2}=8x+4$ and $x^{2}+y^{2}+4\sqrt{3}x-4=0$ is equal to.

If the area of the region $S=\{(x, y): 2y - y^2 \leq x^2 \leq 2y, x \geq y\}$ is equal to $\frac{n+2}{n+1} - \frac{\pi}{n-1}$,then the natural number $n$ is equal to $...............$.

The area of the region given by $A = \{(x, y) : x^{2} \leq y \leq \min \{x+2, 4-3x\}\}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module