# Evaluating Definite Integrals

If the upper limit and the lower limit of the independent variable of the given function or integrand is specified, its integration is expressed using definite integrals. A definite integral is denoted as:

$F(a) – F(b) = \int\limits_{a}^b f(x)dx$

Here R.H.S. of the equation means integral of f(x) with respect to x.

f(x)is called the integrand.

dx is called the integrating agent.

a is the upper limit of the integral and b is the lower limit of the integral.

Let us now discuss important properties of definite integrals and their proofs.

Let us now discuss important properties of definite integrals and their proofs.

1. $\int\limits_{a}^b$f(x)dx= $\int\limits_{a}^b$f(t)dt

Let us consider x = t. Therefore dx = dt. Substituting these values in the LHS of the above equation we can prove this property.

2. $\int\limits_{a}^b$f(x)dx= -$\int\limits_{b}^a$f(x)dx

According to second fundamental theorem of Calculus, if f(x) is a continuous function defined on the closed interval [a, b] and F(x) denotes the anti-derivative of f(x), then

$\int\limits_{a}^b$f(x)dx= [F(x)${]}_{a}^{b}$= F(b)-F(a)

Therefore

$\int\limits_{a}^b$f(x)dx= F(b)-F(a) = – [F(a)-F(b)= -$\int\limits_{b}^a$f(x)dx

3. $\int\limits_{a}^b$dx= $\int\limits_{a}^c$f(x)dx + $\int\limits_{c}^b$f(x)dx

From the second theorem of Calculus,

$\int\limits_{a}^b$f(x)dx= [F(x)${]}_{a}^{b}$= F(b)-F(a)

Therefore,

$\int\limits_{a}^b$f(x)dx= F(b)-F(a)——-(1)

$\int\limits_{a}^c$f(x)dx= F(c)-F(a)——-(2)

$\int\limits_{c}^b$f(x)dx= F(b)-F(c)——-(3)

Adding equations 2 and 3 we have,

$\int\limits_{a}^c$f(x)dx + $\int\limits_{c}^b$f(x)dx=F(c)-F(a) + F(b)-F(c)=F(b)-F(a)=$\int\limits_{a}^b$f(x)dx

4.  $\int\limits_{a}^b$f(x)dx= $\int\limits_{a}^b$f(a+b-x)dx

Let us assume that t = a + b – x. Therefore dt = -dx. At x = a, t = b and at x = b, t =a.

Therefore,
$\int\limits_{a}^b$f(x)dx= -$\int\limits_{b}^a$f(a+b-t)dt

From the second property of definite integrals:
$\int\limits_{a}^b$f(x)dx=-$\int\limits_{b}^a$f(x)dx

Therefore,
$\int\limits_{a}^b$f(x)dx=-$\int\limits_{b}^a$f(a+b-t)dt=$\int\limits_{a}^b$f(a+b-t)dt

Following the first property of definite integrals:
$\int\limits_{a}^b$f(x)dx= $\int\limits_{a}^b$f(a+b-x)dx

5. $\int\limits_{0}^a$f(x)dx=$\int\limits_{0}^a$f(a-x)dx

This property is a special case of fourth property of integrals as discussed above.
Let us assume that t = a – x. Therefore dt = -dx. At x = 0, t = a and at x = a, t =0.

Therefore,
$\int\limits_{0}^a$f(x)dx= $\int\limits_{a}^0$f(a-t)dt

From the second property of definite integrals:
$\int\limits_{0}^a$f(x)dx=-$\int\limits_{0}^a$f(x)dx

Therefore,
$\int\limits_{0}^a$f(x)dx= -$\int\limits_{a}^0$f(a-t)dt=$\int\limits_{0}^a$f(a-t)dt

Following the first property of definite integrals:
$\int\limits_{0}^a$f(x)dx=$\int\limits_{0}^a$f(a-x)dx

6. $\int\limits_{0}^{2a}$f(x)dx=$\int\limits_{0}^a$f(x)f(x)dx+$\int\limits_{0}^a$f(2a-x)dx

From the second property of definite integrals
$\int\limits_{a}^b$f(x)dx= $\int\limits_{a}^c$f(x)dx+$\int\limits_{c}^b$f(x)dx

Therefore,
$\int\limits_{0}^{2a}$f(x)dx=$\int\limits_{0}^a$f(x)dx+$\int\limits_{a}^{2a}$f(x)dx——–(4)

Let us assume that t = 2a – x. Then dt = – dx. In such a case, when x = 2a, t = 0. Therefore, the second integral can be expressed as:

$\int\limits_{a}^{2a}$f(x)dx= $\int\limits_{a}^0$f(2a-t)dt=$\int\limits_{0}^a$f(2a-t)dt=$\int\limits_{0}^a$f(2a-x)dx

Equation 4 can, therefore, be rewritten as:
$\int\limits_{0}^{2a}$f(x)dx=$\int\limits_{0}^a$f(x)dx+$\int\limits_{0}^a$f(2a-x)dx

7.  $\int\limits_{0}^{2a}$f(x)dx= 2$\int\limits_{0}^a$f(x)dx if f(2a-x)=f(x)

and

$\int\limits_{0}^{2a}$f(x)dx=0 if f(2a-x)=-f(x)

Using the sixth property of definite integrals, we have
$\int\limits_{0}^2a$f(x)dx=$\int\limits_{0}^a$f(x)dx+$\int\limits_{0}^a$f(2a-x)dx——-(5)

If in case f (2a-x)=f(x), therefore equation 5 can be rewritten as:
$\int\limits_{0}^{2a}$f(x)dx=$\int\limits_{0}^a$f(x)dx+$\int\limits_{0}^a$f(x)dx

If f(2a – x) = -f(x), then equation 5 can be rewritten as:
$\int\limits_{0}^{2a}$f(x)dx=$\int\limits_{0}^a$f(x)dx -$\int\limits_{0}^a$f(x)dx

8.  $\int\limits_{-a}^a$f(x)dx=2$\int\limits_{0}^a$f(x)dx if f(-x)=f(x)

and

$\int\limits_{-a}^a$f(x)dx=0 if f(-x)= -f(x)

Using third property of definite integrals we have
$\int\limits_{a}^b$f(x)dx=$\int\limits_{a}^c$f(x)dx +$\int\limits_{c}^b$f(x)dx

Therefore,
$\int\limits_{-a}^a$f(x)dx=$\int\limits_{-a}^0$f(x)dx+$\int\limits_{0}^a$f(x)dx——–(6)

Let us assume that t = -x. Then dt = -dx. Thus when x = -a, t = a and x =0, t = 0. Then the first integral in right hand side can be written as:
$\int\limits_{-a}^0$f(x)dx= -$\int\limits_{a}^0$f(-t)dt

Therefore equation 6 can be written as;
$\int\limits_{-a}^a$f(x)dx= -$\int\limits_{a}^0$f(-t)dt + $\int\limits_{0}^a$f(x)dx

$\int\limits_{-a}^a$f(x)dx= $\int\limits_{0}^a$f(-x)dx + $\int\limits_{0}^a$f(x)dx———(7)

In case if f is an even function, i.e., f(-x) = f(x), then equation 7 can be rewritten as:
$\int\limits_{-a}^a$f(x)dx= $\int\limits_{0}^a$f(x)dx+ $\int\limits_{0}^a$f(x)dx= 2$\int\limits_{0}^a$f(x)dx

In case if f is an even function, i.e., f(-x) = – f(x), then equation 7 can be rewritten as:
$\int\limits_{-a}^a$f(x)dx= $\int\limits_{0}^a$f(x)dx- $\int\limits_{0}^a$f(x)dx= 0

#### Practise This Question

Raghu wants to make a closed aluminium cone for his science project. The slant height and radius of the cone are 10 cm and 7 cm respectively. The area of aluminium sheet required to make the cone is
(Use π=3.14)