evaluate the following integral given (f(x)=\begin{cases}2x^{4}&\text{if }-pi<x < 0\\7sin(x)&\text{if…

Explanation:

Step1: Split the integral by piece - wise function

Use the property $\int_{-a}^{a}f(x)dx=\int_{-a}^{0}f(x)dx+\int_{0}^{a}f(x)dx$. Here, $\int_{-\pi}^{\pi}f(x)dx=\int_{-\pi}^{0}2x^{4}dx+\int_{0}^{\pi}7\sin(x)dx$.

Step2: Integrate $2x^{4}$ from $-\pi$ to $0$

The antiderivative of $2x^{4}$ is $\frac{2}{5}x^{5}$. Using the fundamental theorem of calculus, $\int_{-\pi}^{0}2x^{4}dx=\left[\frac{2}{5}x^{5}\right]_{-\pi}^{0}=0 - \frac{2}{5}(-\pi)^{5}=\frac{2\pi^{5}}{5}$.

Step3: Integrate $7\sin(x)$ from $0$ to $\pi$

The antiderivative of $7\sin(x)$ is $- 7\cos(x)$. Using the fundamental theorem of calculus, $\int_{0}^{\pi}7\sin(x)dx=[-7\cos(x)]_{0}^{\pi}=-7\cos(\pi)+7\cos(0)=-7(-1)+7(1)=14$.

Step4: Sum the two results

$\int_{-\pi}^{\pi}f(x)dx=\frac{2\pi^{5}}{5}+14$.

Answer:

$\frac{2\pi^{5}}{5}+14$