# Calculus 1 Using properties of integrals and geometry, find: \int_{-2}^{2} 2-\left | x \right |...

## Question:

Calculus 1

Using properties of integrals and geometry, find:

{eq}\int_{-2}^{2} 2-\left | x \right | + \left | \sin(pix) \right | dx {/eq}

## Integrating Even Functions:

An even function is a function {eq}f(x) {/eq} with the property

{eq}f(x)=f(-x) {/eq}

for all values of {eq}x {/eq}. Even functions have a special property involving integration. For any real value {eq}a {/eq},

{eq}\int_{-a}^a f(x) \, dx=2\int_0^a f(x) \, dx {/eq}

whenever {eq}f(x) {/eq} is even. Using this property can make it much easier to find integrals of certain even functions.

## Answer and Explanation:

This integral can be broken up into three separate integrals:

{eq}\int_{-2}^2 2-|x|+|\sin(\pi x)| \, dx=\int_{-2}^2 2\, dx-\int_{-2}^2 |x|\, dx+\int_{-2}^2 |\sin(\pi x)| \, dx {/eq}

Take these integrals one at a time:

{eq}\begin{align} \int_{-2}^2 2 \, dx&=2x \Big|_{-2}^2\\ &=2(2)-2(-2)\\ &=8 \end{align} {/eq}

For the second integral, notice that {eq}|x| {/eq} is an even function. This means we can compute it using the following property of even functions:

{eq}\int_{-2}^2 |x|\, dx=2\int_0^2 |x| \, dx {/eq}

But on a positive interval, {eq}|x| {/eq} is equivalent to just {eq}x {/eq}. So we can solve the integral this way:

{eq}\begin{align} \int_{-2}^2 |x|\, dx&=2\int_0^2 |x| \, dx\\ &=2\int_0^2 x \, dx\\ &=x^2 \Big|_0^2 \\ &=2^2-0^2\\ &=4 \end{align} {/eq}

For the third interval, we again have an even function because of the absolute value sign, so

{eq}\int_{-2}^2 |\sin(\pi x)| \, dx=2\int_0^2 |\sin(\pi x)| \, dx {/eq}

Note that {eq}\sin(\pi x) {/eq} is positive on the interval {eq}x=(0, 1) {/eq} and negative on the interval {eq}x=(1, 2) {/eq}.

This means that

{eq}\int_0^2 |\sin(\pi x)|\, dx=\int_0^1 \sin(\pi x) \, dx-\int_1^2 \sin(\pi x) \, dx {/eq}

Solve each of these integrals:

{eq}\begin{align} \int_0^1 \sin(\pi x) \, dx&=-\frac{1}{\pi} \cos(\pi x) \Big|_0^1\\ &=-\frac{1}{\pi}(\cos(\pi)-\cos(0))\\ &=-\frac{1}{\pi}(-1-1)\\ &=\frac{2}{\pi} \end{align} {/eq}

{eq}\begin{align} \int_1^2 \sin(\pi x) \, dx&=-\frac{1}{\pi} \cos(\pi x) \Big|_1^2\\ &=-\frac{1}{\pi}(\cos(2\pi)-\cos(\pi))\\ &=-\frac{1}{\pi}(1+1)\\ &=-\frac{2}{\pi} \end{align} {/eq}

So we have

{eq}\begin{align} \int_{-2}^2 |\sin(\pi x)| \, dx&=2\int_0^2 |\sin(\pi x)| \, dx\\ &=2\left(\frac{2}{\pi}+\frac{2}{\pi}\right)\\ &=\frac{8}{\pi} \end{align} {/eq}

Now we have all the integrals we need, and we just have to add them together:

{eq}\begin{align} \int_{-2}^2 2-|x|+|\sin(\pi x)| \, dx&=\int_{-2}^2 2\, dx-\int_{-2}^2 |x|\, dx+\int_{-2}^2 |\sin(\pi x)| \, dx\\ &=8-4+\frac{8}{\pi}\\ &=4+\frac{8}{\pi} \end{align} {/eq}

#### Learn more about this topic:

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2