Definite Integrals

What is a definite integral?

A definite integral $\int_{a}^{b} f(x)\, dx$ gives the signed area under the curve $f(x)$ between $x = a$ and $x = b$ . By the fundamental theorem of calculus, if $F(x)$ is an antiderivative of $f(x)$ (that is, $F'(x) = f(x)$ ), then

\int_{a}^{b} f(x)\, dx = F(b) - F(a)

Worked example

Evaluate $\displaystyle\int_{0}^{\pi/2} \cos(2x)\, dx$ .

Solution: An antiderivative of $\cos(2x)$ is $\dfrac{\sin(2x)}{2}$ . Applying the limits:

\left[\dfrac{\sin(2x)}{2}\right]_{0}^{\pi/2} = \dfrac{\sin(\pi)}{2} - \dfrac{\sin(0)}{2} = \dfrac{0}{2} - \dfrac{0}{2} = 0

Properties of definite integrals

These properties often turn an integral with no easy antiderivative into one that is easy, or show it is zero without integrating at all.

The standard properties of definite integrals, numbered as in the NCERT textbook

	Property
P0	$\displaystyle\int_a^b f(x)\,dx = \int_a^b f(t)\,dt$ (the variable's name doesn't matter)
P1	$\displaystyle\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$ , and $\displaystyle\int_a^a f(x)\,dx = 0$
P2	$\displaystyle\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$
P3	$\displaystyle\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$
P4	$\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$
P5	$\displaystyle\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a - x)\,dx$
P6	$\displaystyle\int_0^{2a} f(x)\,dx = 2\int_0^a f(x)\,dx$ if $f(2a - x) = f(x)$ , and $0$ if $f(2a - x) = -f(x)$
P7	$\displaystyle\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$ if $f$ is even, and $0$ if $f$ is odd

Worked example: using property P4

Evaluate $I = \displaystyle\int_0^{\pi/2} \dfrac{\sin x}{\sin x + \cos x}\,dx$ .

Solution: By P4, replace $x$ with $\dfrac{\pi}{2} - x$ . Since $\sin\left(\dfrac{\pi}{2} - x\right) = \cos x$ and $\cos\left(\dfrac{\pi}{2} - x\right) = \sin x$ :

I = \int_0^{\pi/2} \dfrac{\cos x}{\cos x + \sin x}\,dx

Adding this to the original form of $I$ , the numerators combine into the denominator:

2I = \int_0^{\pi/2} \dfrac{\sin x + \cos x}{\sin x + \cos x}\,dx = \int_0^{\pi/2} 1\,dx = \dfrac{\pi}{2} \;\Rightarrow\; I = \dfrac{\pi}{4}

Worked example: an odd function

Evaluate $\displaystyle\int_{-1}^{1} x^3\cos x\,dx$ .

Solution: $x^3$ is odd and $\cos x$ is even, so their product is odd: replacing $x$ with $-x$ gives $(-x)^3\cos(-x) = -x^3\cos x$ . By P7, the integral over the symmetric interval $[-1, 1]$ is $0$ , with no antiderivative needed.

Common mistakes

Keeping the old limits after a substitution. If you substitute $u = 2x$ , the limits $0$ and $\dfrac{\pi}{2}$ become $0$ and $\pi$ . Either change them or substitute back before applying them.
Reading a definite integral as an area. It is a signed area, so parts below the axis count as negative. The worked example above is 0 for exactly that reason. For actual area, see area bounded by a curve.
Adding $+ C$ . A definite integral is a number. Any constant cancels in $F(b) - F(a)$ .
Using P7 on a function that is neither even nor odd. Check $f(-x)$ first. $x^3 + 1$ is neither, so its integral over $[-1, 1]$ is $2$ , not $0$ .