# Evaluate the following integral. integral from -pi to pi {\sin x + x} dx

## Question:

Evaluate the following integral.

{eq}\displaystyle \int_{-\pi}^{\pi} (\sin x + x) \ dx {/eq}

## Integral of even or odd function:

**Test for even or odd function:**

Let us consider the function {eq}f(x) {/eq},

If {eq}f(-x) = f(x) {/eq}, the the function {eq}f(x) {/eq} is even.

If {eq}f(-x) = - f(x) {/eq}, the the function {eq}f(x) {/eq} is odd.

**Integral of even or odd function:**

If the function is even, then,

{eq}\displaystyle \int_{-a}^{a} f(x) \ dx = 2 \int_{0}^{a} f(x) \ dx {/eq}

If the function is odd, then,

{eq}\displaystyle \int_{-a}^{a} f(x) \ dx = 0 {/eq}

## Answer and Explanation:

We have,

$$\displaystyle \int_{-\pi}^{\pi} (\sin x + x) \ dx $$

Let us test the function for odd or even,

$$\displaystyle \begin{align*} f(x) &= \sin x + x \\ \Rightarrow f(-x) &= \sin (-x) + (-x) \\ &= - \sin x - x \\ &= - (\sin x + x) \\ f(-x) &= - f(x) \\ \end{align*} $$

The function {eq}f(x) = \sin x + x {/eq} is an **odd** function.

Therefore,

$$\displaystyle \Rightarrow \int_{-\pi}^{\pi} (\sin x + x) \ dx = 0 $$

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from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13