Integrals of even/odd functions
We already know what integrals represent and how to solve indefinite and definite integrals, but have you ever wondered what happens if we have to find an integral of an even or odd function? Will these properties make calculating integrals easier?
Let’s find out.
What is an integral of an even/odd function?
Before explaining integrals of even and odd functions, let’s remind ourselves what even and odd functions are in general.
A function $$f$$ is even if the following is true for every $$x$$ in its domain:
A graph of an even function is symmetric about the $$y$$-axis.
A function $$f$$ is odd if the following is true for every $$x$$ in its domain:
A graph of an odd function is symmetric about the origin.
Now, let’s dig into integrals of even and odd functions!
Let $$f$$ be an integrable function on some closed interval that is symmetric about zero — for example $$[-a,a]$$, for $$a\geq0$$.
If $$f$$ is even, then:
If $$f$$ is odd, then:
One important thing to remember is that this only works if the integral is calculated on a closed interval symmetric about zero. Keep that in mind for later!
Further evaluation of these integrals is done with our usual suspects: special methods or properties of integrals.
Here’s our favorite chart to keep nearby when working with integrals:
| Constant multiple property of integrals | $$\int{(c\times f(x))}dx=c\times \int{f(x)}dx$$ |
| Sum rule for integrals | $$\int{(f(x) + g(x))}dx=\int{f(x)}dx + \int{g(x)}dx$$ |
| Difference rule for integrals | $$\int{(f(x) - g(x))}dx=\int{f(x)}dx - \int{g(x)}dx$$ |
| Substitution rule | $$\int{f(\varphi(t))}\varphi^{\prime}(t)dt=\int{f(x)}dx$$ |
| Integration by parts | $$\int{u}dv=uv-\int{v}du$$ |
Why are integrals of even/odd functions so useful?
When we have a symmetric interval, the knowledge that a function is even or odd can make calculating an integral much easier! For example, let’s say we want to calculate $$\int_{-1}^{1}x$$. Instead of going through the whole calculation, it’s easier just to conclude that the function $$f(x)=x$$ is odd, and, therefore, the result will be $$0$$.
How to solve integrals of even/odd functions
Let’s see some integrals of even and odd functions in action! We’ll go through a few examples together.
Example 1
Find the integral:
Determine if the function is even, odd, or neither:
$$f(x)=\sin(x)$$
Substitute $$-x$$ for $$x$$:
$$f(-x)=\sin(-x)$$
Simplify the expression by using the identity for the sine function, $$\sin(-x)=-\sin(x)$$:
$$f(-x)=-\sin(x)$$
Use the original equation $$f(x)=\sin(x)$$ to substitute $$f(x)$$ for $$\sin(x)$$:
$$f(-x)=-f(x)$$
Since $$f(-x)=-f(x)$$, we know the function is an odd function:
$$\text{Odd}$$
The function is odd, and the upper and lower limits are opposite values, so the integral equals $$0$$:
Not too bad, right? Let’s try another one.
Example 2
Find the integral:
Determine if the function is even, odd, or neither:
$$f(x)=x^2$$
Substitute $$-x$$ for $$x$$:
$$f(-x)=(-x)^2$$
A negative base raised to an even power equals a positive:
$$f(-x)=x^2$$
Use the original equation $$f(x)=x^2$$ to substitute $$f(x)$$ for $$x^2$$:
$$f(-x)=f(x)$$
Since $$f(-x)=f(x)$$, this function is an even function:
$$\text{Even}$$
The function is even, so the integral equals $$2\int_0^af(x)dx$$:
$$2\int_0^1x^2dx$$
To evaluate the definite integral, we first evaluate the indefinite integral:
$$2\int x^2dx$$
Use $$\int x^n dx=\frac{x^{n+1}}{n+1},n\neq-1$$ to evaluate the integral:
$$2\times \frac{x^3}{3}$$
Multiply the factors:
$$\frac{2x^3}{3}$$
To evaluate the definite integral, return the limits of integration:
$$\frac{2x^3}{3}\biggr|_0^1$$
Use $$F(x)\biggr|_a^b=F(b)-F(a)$$ to evaluate the expression:
$$\frac{2\times1^3}{3}-\frac{2\times0^3}{3}$$
Simplify the expression:
We did it!
Need a summary of our solving steps? Here it is:
Study summary
- Determine if the function is even, odd, or neither.
- If the function is odd and the upper and the lower limits are opposite values, the integral equals zero.
- If the function is even, use the property of even functions and evaluate the integral.
Do it yourself!
If you’d like a little more practice, you’re in luck — we’ve got some practice problems ready for you! Try these and see how it goes.
Evaluate the integral:
- $$\int_{-2}^2 x^4dx$$
- $$\int_{-2}^2 x^3dx$$
- $$\int_{-1}^1 \tan(x)dx$$
- $$\int_{-1}^1 |x| dx$$
Solutions:
- $$\frac{64}{5}$$
- $$0$$
- $$0$$
- $$1$$
If you’re still struggling through the solving process, that’s totally okay! Stumbling a few times is actually good for learning. If you get too stuck or lost, scan the problem using your Photomath app, and we’ll walk you through to the other side!
Here’s a sneak peek of what you’ll see:
