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 (in sq. units) of the region bounded by the curves $y = x \log x$ and $y = 2x - 2x^2$ is

The area (in sq. units) of the region $\{(x,y):y^2 \geq 2x, x^2+y^2 \leq 4x, x \geq 0, y \leq 0 \}$ is

Area of the region (in square units) enclosed by the curves $y^2=8(x+2)$,$y^2=4(1-x)$ and the $Y$-axis is

The area of the region common to the parabolas $4y^2 = 9x$ and $3x^2 = 16y$ is...

The area of the region bounded by the parabola $y^2 = 8x$ and the line $x + y = 0$ is . . . . . . sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module