Properties of the Definite Integral - Practice Questions & MCQ

Definite integrals have properties that relate to the limits of integration.

Property 1

$
\int_a^a f(x) d x=0
$

If the upper and lower limits of integration are the same, the integral is just a line and contains no area, hence the value is 0
Alternatively
If $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{F}(\mathrm{x})=\mathrm{f}(\mathrm{x})$, then
$
\begin{aligned}
\int_{\mathrm{a}}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx} & =[\mathrm{F}(\mathrm{x})]_{\mathrm{a}}^{\mathrm{a}} \\
& =\mathrm{F}(\mathrm{a})-\mathrm{F}(\mathrm{a}) \\
& =0
\end{aligned}
$

Property 2

The value of the definite integral of a function over any particular interval depends on the function and the interval but not on the variable of the integration.

$
\int_a^b f(x) d x=\int_a^b f(t) d t=\int_a^b f(y) d y
$

For example,
$
\begin{aligned}
& \int_0^2 \mathrm{x}^2 \mathrm{dx}=\left[\frac{\mathrm{x}^3}{3}\right]_0^2=\frac{2^3}{3}-\frac{0^3}{3}=\frac{2^3}{3} \\
& \int_0^2 \mathrm{t}^2 \mathrm{dt}=\left[\frac{\mathrm{t}^3}{3}\right]_0^2=\frac{2^3}{3}-\frac{0^3}{3}=\frac{2^3}{3} \\
& \int_0^2 \mathrm{y}^2 \mathrm{dy}=\left[\frac{\mathrm{y}^3}{3}\right]_0^2=\frac{2^3}{3}-\frac{0^3}{3}=\frac{2^3}{3}
\end{aligned}
$

Property 3

If the limits of definite integral are interchanged, then its value changes by a minus sign only.

If $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{F}(\mathrm{x})=\mathrm{f}(\mathrm{x})$, then
$
\begin{aligned}
\int_a^b f(x) d x & =[F(x)]_a^b \\
& =F(b)-F(a)
\end{aligned}
$
and,
$
\begin{aligned}
\int_{\mathrm{b}}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} & =[\mathrm{F}(\mathrm{x})]_{\mathrm{b}}^{\mathrm{a}} \\
& =(\mathrm{F}(\mathrm{a})-\mathrm{F}(\mathrm{~b})) \\
& =-(\mathrm{F}(\mathrm{~b})-\mathrm{F}(\mathrm{a}))
\end{aligned}
$

Hence,
$
\int_a^b f(x) d x=-\int_b^a f(x) d x
$

Property 4 (King's Property)

This is one of the most important properties of definite integration.

$\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

Proof:

In R.H.S, Put $t=a+b-x$
$\Rightarrow \quad d x=-d t$
Also, when $x=a$, then $t=b$, and when $x=b, t=a$
$
\begin{aligned}
\therefore \quad \text { R.H.S } & =\int_b^a f(t)(-d t) \\
& =-\int_b^a f(t) d t \\
& =-\left(-\int_a^b f(t) d t\right) \\
& =\int_a^b f(t) d t \\
& =\int_a^b f(x) d x \\
& =\text { L.H.S }
\end{aligned}
$

Corollary:

$\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{dx}$