# Find the area of the region enclosed by the given curves. y = 4\cos \pi x,\ y= 8x^2 - 2

## Question:

Find the area of the region enclosed by the given curves.

{eq}y = 4\cos \pi x,\ y= 8x^2 - 2 {/eq}

## Area of Region:

The form of the definite integral used to find the area bounded between the given curves is:

{eq}\displaystyle A = \int_{a}^{b} f(x) - g(x) \, dx {/eq}

Where, f(x) is the upper bound and g(x) is the lower bound of the bounded region.

## Answer and Explanation:

The given curves are:

{eq}y = f(x) = 4\cos \pi x \\ y = g(x) = 8x^2 - 2 {/eq}

Finding the limits of the integral:

{eq}\begin{align} 4\cos \pi x &= 8x^2 - 2 \\ 8x^2 - 4\cos \pi x - 2 &= 0 \\ x &= \pm \dfrac{1}{2} \end{align} {/eq}

Implies that:

{eq}-\dfrac{1}{2} \leq x \leq \dfrac{1}{2} {/eq}

Evaluating the area of bounded region by using the formula:

{eq}\displaystyle A = \int_{a}^{b} f(x) - g(x) \, dx \\ \displaystyle A = \int_{-\frac{1}{2}}^{\frac{1}{2}} 4\cos \pi x - 8x^2 + 2 \, dx \\ \displaystyle A = 2\int_{0}^{\frac{1}{2}} 4\cos \pi x - 8x^2 + 2 \, dx {/eq}

Integrating the equation, we get:

{eq}\displaystyle A = 2\left[ \frac{4\sin \pi x}{\pi} - \frac{8x^3}{3} + 2x \right]_{0}^{\frac{1}{2}} \\ \displaystyle A = 2\left[ \frac{4\sin \frac{\pi}{2}}{\pi} - \frac{8(\frac{1}{2})^3}{3} + 2(\frac{1}{2}) \right] \\ \displaystyle A = 2\left[ \frac{4}{\pi} - \frac{1}{3} + 1 \right] \\ \displaystyle A = 2\left[ \frac{2}{3}+\frac{4}{\pi} \right] \\ \boxed{A = \dfrac{4\pi +24}{3\pi}} {/eq}