Definite Integrals: Odd and Even Functions (Exercise 7.10)

This lesson works through Exercise 7.10, evaluating definite integrals over symmetric limits by applying property P7: the rule for integrals of odd and even functions. The key idea is to test the symmetry of the function first, which often turns a hard integral into a simple one.

Property P7: odd and even functions

For limits running from $-a$ to $a$, the symmetry of $f$ decides the answer.

If $f$ is even, meaning $f(-x) = f(x)$, then

$\int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx.$

If $f$ is odd, meaning $f(-x) = -f(x)$, then

$\int_{-a}^{a} f(x)\,dx = 0.$

So whenever the limits have the form $-a$ to $a$, first check whether the function is odd or even, and only fall back to direct integration if neither applies.

Example 1: $\int_{-\pi/2}^{\pi/2} \sin^2 x\,dx$

Let $I = \displaystyle\int_{-\pi/2}^{\pi/2} \sin^2 x\,dx$, with $f(x) = \sin^2 x$.

Check the symmetry. Replacing $x$ with $-x$:

$f(-x) = \big(\sin(-x)\big)^2 = \big(-\sin x\big)^2 = \sin^2 x = f(x).$

So $f$ is even, and by P7

$I = 2\int_{0}^{\pi/2} \sin^2 x\,dx.$

Finishing with the double-angle identity

Write $\sin^2 x$ in terms of a multiple angle using $\sin^2 x = \tfrac{1 - \cos 2x}{2}$:

$I = 2\int_{0}^{\pi/2} \frac{1 - \cos 2x}{2}\,dx = 2 \cdot \frac{1}{2}\left[\int_{0}^{\pi/2} 1\,dx - \int_{0}^{\pi/2} \cos 2x\,dx\right].$

The factor of $2$ and the $\tfrac{1}{2}$ cancel. Integrating, $\int 1\,dx = x$ and $\int \cos 2x\,dx = \tfrac{\sin 2x}{2}$:

$I = \left[x\right]_{0}^{\pi/2} - \frac{1}{2}\left[\sin 2x\right]_{0}^{\pi/2}.$

Applying the limits:

$I = \left(\frac{\pi}{2} - 0\right) - \frac{1}{2}\big(\sin \pi - \sin 0\big) = \frac{\pi}{2} - \frac{1}{2}(0 - 0) = \frac{\pi}{2}.$

Example 2: $\int_{-1}^{1} \sin^5 x \, \cos^4 x\,dx$

The limits are again of the form $-a$ to $a$, so test the symmetry of $f(x) = \sin^5 x \, \cos^4 x$.

$f(-x) = \big(\sin(-x)\big)^5 \big(\cos(-x)\big)^4 = \big(-\sin x\big)^5 \big(\cos x\big)^4 = -\sin^5 x \, \cos^4 x = -f(x).$

Since $f(-x) = -f(x)$, the function is odd, so by P7

$\int_{-1}^{1} \sin^5 x \, \cos^4 x\,dx = 0.$

Example 3: $\int_{0}^{4} |x - 1|\,dx$

Here the limits are not symmetric, so integrate directly. Using $\int |u|\,du = \tfrac{u\,|u|}{2}$ with $u = x - 1$:

$I = \left[\frac{(x-1)\,|x-1|}{2}\right]_{0}^{4}.$

Applying the limits:

$I = \frac{1}{2}\Big[(4-1)\,|4-1| - (0-1)\,|0-1|\Big] = \frac{1}{2}\big[3 \cdot |3| - (-1)\cdot|-1|\big].$

$I = \frac{1}{2}\big[3 \cdot 3 - (-1)\cdot 1\big] = \frac{1}{2}(9 + 1) = \frac{1}{2}\cdot 10 = 5.$

Key takeaways

• For limits $-a$ to $a$, test symmetry first: an even function gives $2\int_{0}^{a} f(x)\,dx$, an odd function gives $0$.
• $\sin^2 x$ is even and integrates cleanly once rewritten as $\tfrac{1-\cos 2x}{2}$, giving $\tfrac{\pi}{2}$ over $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$.
• A modulus integral over non-symmetric limits is evaluated directly using $\int |u|\,du = \tfrac{u\,|u|}{2}$.