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

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

Question

(B) Given equations are $y = x^2 + 2$ and $y = 2|x| - \cos(\pi x)$.Equating the two,we get $x^2 + 2 = 2|x| - \cos(\pi x)$.This can be rewritten as $x^

Vedclass · Accepted Answer

$\frac{8}{3}$

Vedclass · Answer

(B) Given equations are $y = x^2 + 2$ and $y = 2|x| - \cos(\pi x)$.Equating the two,we get $x^2 + 2 = 2|x| - \cos(\pi x)$.This can be rewritten as $x^2 - 2|x| + 1 + 1 = -\cos(\pi x)$,which simplifies to $(|x| - 1)^2 + 1 = -\cos(\pi x)$.Since $(|x| - 1)^2 \ge 0$,the left side is $\ge 1$. Also,$-\cos(\pi x) \le 1$. Thus,equality holds only when $(|x| - 1)^2 = 0$ and $-\cos(\pi x) = 1$,which gives $x = \pm 1$.The area is given by $\int_{-1}^{1} ((x^2 + 2) - (2|x| - \cos(\pi x))) dx$.Since the integrand is an even function,the area is $2 \int_{0}^{1} (x^2 - 2x + 2 + \cos(\pi x)) dx$.Evaluating the integral: $2 [\frac{x^3}{3} - x^2 + 2x + \frac{\sin(\pi x)}{\pi}]_{0}^{1}$.$= 2 [(\frac{1}{3} - 1 + 2 + 0) - (0)] = 2 [\frac{4}{3}] = \frac{8}{3}$.

The area bounded by $y = x^2 + 2$ and $y = 2|x| - \cos(\pi x)$ is equal to

Similar Questions

The area enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ and the lines $x=0, x=\frac{\pi}{2}$ is:

The area of the shorter region bounded by $|y| = 4 - x^2$ and $|y| = 3x$ is given by $\left( 3K + \frac{1}{3} \right)$ sq-unit,where $K$ is equal to:

The area of the figure bounded by the curves $y = |x - 1|$ and $y = 3 - |x|$ is ....... $sq. \text{ unit}$.

The area included between the two curves $y^2 = 4ax$ and $x^2 = 4ay$ is

The area (in sq. units) of the region consisting of points $(x,y)$ on the $X-Y$ plane which satisfy $|x| \le 1 + |y|$ and $|y| \le 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module