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

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(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^

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$\frac{8}{3}$

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(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

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