Why the Integral of an Odd Function Over Symmetric Limits Is Zero

An indefinite integral gives a family of functions; a definite integral gives a single number — the net accumulated change of a quantity between two limits, such as distance from speed or area under a curve. A handful of properties let many definite integrals be found with almost no calculation.

This lesson covers the fundamental theorem, reversing and splitting limits, reflection properties, and even, odd and doubled intervals.

**The fundamental theorem of calculus states that if F is an anti-derivative of a function f that is continuous on $[a,b]$, then ${\int }_a^{b}f(x)dx=F(b)-F(a)$ — the constant of integration cancels.

Worked example.**

$${\int }_1^3(3x^2+2x)dx={\left[x^3+x^2\right]}_1^3=(27+9)-(1+1)=34$$

Worked example 2. ${\int }_0^{\pi /2}\cos xdx=[\sin x{]}_0^{\pi /2}=1-0=1$.

Worked example 3. ${\int }_0^1\frac{dx}{1+x^2}=[{\tan }^{-1}x{]}_0^1=\frac{\pi }{4}\approx 0.785$.

With a substitution. Change the limits along with the variable: in ${\int }_0^{2}2x{e}^{x^2}dx$, put $t=x^2$, so the limits become 0 and 4 and the value is $e^4-1\approx 53.6$.

An everyday example. A car's speed during a 3-second burst on an expressway, integrated over that time, gives the distance covered, whatever its starting position.

The substance. The integrand must be continuous on the interval — ${\int }_{-1}^1\frac{dx}{x^2}$ cannot be found this way, because $\frac{1}{x^2}$ grows without bound at 0.

**Reversing the limits changes the sign, ${\int }_b^{a}f(x)dx=-{\int }_a^{b}f(x)dx$, and an integral can be split at any point c, ${\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$.

Other basic properties:**

- ${\int }_a^{a}f(x)dx=0$
- ${\int }_a^{b}f(x)dx={\int }_a^{b}f(t)dt$ — the letter used does not matter

Worked example (reversing). ${\int }_3^1(3x^2+2x)dx=-34$.

Worked example (splitting a modulus). Evaluate ${\int }_0^4|x-1|dx$ by splitting at 1, where the expression inside changes sign:

$${\int }_0^1(1-x)dx+{\int }_1^4(x-1)dx=\frac{1}{2}+\frac{9}{2}=5$$

Worked example (piecewise). For $f(x)=x^2$ on $[0,1]$ and $f(x)=2x-1$ on $[1,2]$, ${\int }_0^{2}f(x)dx=\frac{1}{3}+[x^2-x{]}_1^2=\frac{1}{3}+2=\frac{7}{3}$.

An everyday example. A train journey from Mumbai to Pune via Lonavala covers the same total distance as its two legs added together — additivity over adjacent intervals.

The substance. Splitting is essential for modulus and greatest integer functions — the break points must be found first.

**Replacing x by $a+b-x$ does not change a definite integral, so ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$, and in particular ${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$.

Why it works.** The substitution $t=a+b-x$ swaps the limits and turns dx into $-dt$; the two sign changes cancel.

Worked example. Evaluate $I={\int }_0^{\pi /2}\frac{\sin x}{\sin x+\cos x}dx$.

- Replacing x by $\frac{\pi }{2}-x$ gives $I={\int }_0^{\pi /2}\frac{\cos x}{\cos x+\sin x}dx$
- Adding the two forms: $2I={\int }_0^{\pi /2}1dx=\frac{\pi }{2}$
- So $I=\frac{\pi }{4}\approx 0.785$

Worked example 2. For ${\int }_2^8\frac{\sqrt{x}}{\sqrt{x}+\sqrt{10-x}}dx$, replacing x by $10-x$ and adding gives $2I={\int }_2^{8}1dx=6$, so $I=3$.

An everyday example. Reading a symmetric rangoli pattern from left to right or right to left gives the same design — the reflection property says the integral does not mind the direction either.

The substance. **The method works when $f(x)+f(a+b-x)$ is simple** — the aim of the reflection is to make that sum easy to integrate.

**Over symmetric limits, ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$ if f is even and 0 if f is odd; and ${\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x)dx$ if $f(2a-x)=f(x)$, but 0 if $f(2a-x)=-f(x)$.

Even and odd functions:

- Even** — $f(-x)=f(x)$, such as $x^2$, $\cos x$ and $|x|$
- Odd — $f(-x)=-f(x)$, such as $x^3$, $\sin x$ and $\tan x$

Worked example (odd). ${\int }_{-2}^2(x^3+x\cos x)dx=0$, since the integrand is odd.

Worked example (even). ${\int }_{-1}^1(x^4+3x^2)dx=2{\int }_0^1(x^4+3x^2)dx=2\left(\frac{1}{5}+1\right)=2.4$.

Worked example (0 to 2a). ${\int }_0^{\pi }\sin xdx=2{\int }_0^{\pi /2}\sin xdx=2$, since $\sin (\pi -x)=\sin x$; but ${\int }_0^{\pi }\cos xdx=0$, since $\cos (\pi -x)=-\cos x$.

An everyday example. Equal gains and losses on either side of a seesaw's pivot cancel exactly — the picture behind an odd function integrating to zero.

The substance. Split a mixed integrand into even and odd parts — in ${\int }_{-1}^1(x^5+x^2)dx$, the odd part vanishes and only $\frac{2}{3}$ remains.