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Evaluating Definite Integrals

Integration and differentiation are the two important process in Calculus. Differentiation is the process of finding the derivative of a function, whereas integration is the reverse process of differentiation. It means that the process of finding the anti-derivative of a function. In integration, the concept behind are functions, limits and integrals. The integrals are generally classified into two different types, namely:

In this article, we are going to discuss the definition of definite integrals, and the process of evaluating the definite integral using different properties.

What is a Definite Integral?

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:

\(\begin{array}{l} F(a) – F(b) = \int\limits_{a}^b f(x)dx\end{array} \)

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.

Evaluating Definite Integrals – Properties

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

Property 1:

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
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.

Property 2:

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx= -
\(\begin{array}{l}\int\limits_{b}^a \end{array} \)
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

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx= [F(x)
\(\begin{array}{l}{]}_{a}^{b}\end{array} \)
= F(b)-F(a)

Therefore

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx= F(b)-F(a) = – [F(a)-F(b)= -
\(\begin{array}{l}\int\limits_{b}^a \end{array} \)
f(x)dx

Property 3:

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
dx=
\(\begin{array}{l}\int\limits_{a}^c\end{array} \)
f(x)dx +
\(\begin{array}{l}\int\limits_{c}^b\end{array} \)
f(x)dx

From the second theorem of Calculus,

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx= [F(x)
\(\begin{array}{l}{]}_{a}^{b}\end{array} \)
= F(b)-F(a)

Therefore,

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx= F(b)-F(a)——-(1)

\(\begin{array}{l}\int\limits_{a}^c \end{array} \)
f(x)dx= F(c)-F(a)——-(2)

\(\begin{array}{l}\int\limits_{c}^b \end{array} \)
f(x)dx= F(b)-F(c)——-(3)

Adding equations 2 and 3 we have,

\(\begin{array}{l}\int\limits_{a}^c\end{array} \)
f(x)dx +
\(\begin{array}{l}\int\limits_{c}^b\end{array} \)
f(x)dx=F(c)-F(a) + F(b)-F(c)=F(b)-F(a)=
\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx

Property 4: 

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
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,

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx= -
\(\begin{array}{l}\int\limits_{b}^a \end{array} \)
f(a+b-t)dt

From the second property of definite integrals:

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx=-
\(\begin{array}{l}\int\limits_{b}^a \end{array} \)
f(x)dx

Therefore,

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx=-
\(\begin{array}{l}\int\limits_{b}^a \end{array} \)
f(a+b-t)dt=
\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(a+b-t)dt

Following the first property of definite integrals:

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{a}^b\end{array} \)
f(a+b-x)dx

Property 5:

\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
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,

\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{a}^0 \end{array} \)
f(a-t)dt

From the second property of definite integrals:

\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx=-
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx

Therefore,

\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx= -
\(\begin{array}{l}\int\limits_{a}^0 \end{array} \)
f(a-t)dt=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(a-t)dt

Following the first property of definite integrals:

\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(a-x)dx

Property 6:

\(\begin{array}{l}\int\limits_{0}^{2a}\end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)f(x)dx+
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(2a-x)dx

From the second property of definite integrals

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{a}^c\end{array} \)
f(x)dx+
\(\begin{array}{l}\int\limits_{c}^b \end{array} \)
f(x)dx

Therefore,

\(\begin{array}{l}\int\limits_{0}^{2a}\end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx+
\(\begin{array}{l}\int\limits_{a}^{2a} \end{array} \)
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:

\(\begin{array}{l}\int\limits_{a}^{2a} \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{a}^0 \end{array} \)
f(2a-t)dt=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(2a-t)dt=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(2a-x)dx

Equation 4 can, therefore, be rewritten as:

\(\begin{array}{l}\int\limits_{0}^{2a} \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx+
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(2a-x)dx

Property 7: 

\(\begin{array}{l}\int\limits_{0}^{2a} \end{array} \)
f(x)dx= 2
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx if f(2a-x)=f(x) and
\(\begin{array}{l}\int\limits_{0}^{2a} \end{array} \)
f(x)dx=0 if f(2a-x)=-f(x)

Using the sixth property of definite integrals, we have

\(\begin{array}{l}\int\limits_{0}^2a \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx+
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(2a-x)dx——-(5)

If in case f (2a-x)=f(x), therefore equation 5 can be rewritten as:

\(\begin{array}{l}\int\limits_{0}^{2a}\end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx+
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx

If f(2a – x) = -f(x), then equation 5 can be rewritten as:

\(\begin{array}{l}\int\limits_{0}^{2a} \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx -
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx

Property 8:

\(\begin{array}{l}\int\limits_{-a}^a \end{array} \)
f(x)dx=2
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx if f(-x)=f(x) and
\(\begin{array}{l}\int\limits_{-a}^a \end{array} \)
f(x)dx=0 if f(-x)= -f(x)

Using third property of definite integrals we have

\(\begin{array}{l}\int\limits_{a}^b \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{a}^c \end{array} \)
f(x)dx +
\(\begin{array}{l}\int\limits_{c}^b \end{array} \)
f(x)dx

Therefore,

\(\begin{array}{l}\int\limits_{-a}^a \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{-a}^0 \end{array} \)
f(x)dx+
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
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:

\(\begin{array}{l}\int\limits_{-a}^0 \end{array} \)
f(x)dx= -
\(\begin{array}{l}\int\limits_{a}^0 \end{array} \)
f(-t)dt

Therefore equation 6 can be written as;

\(\begin{array}{l}\int\limits_{-a}^a \end{array} \)
f(x)dx= -
\(\begin{array}{l}\int\limits_{a}^0 \end{array} \)
f(-t)dt +
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx

\(\begin{array}{l}\int\limits_{-a}^a \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(-x)dx +
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
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:

\(\begin{array}{l}\int\limits_{-a}^a \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx+
\(\begin{array}{l}\int\limits_{0}^a\end{array} \)
f(x)dx= 2
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx

In case if f is an odd function, i.e., f(-x) = – f(x), then equation 7 can be rewritten as:

\(\begin{array}{l}\int\limits_{-a}^a \end{array} \)
f(x)dx=
\(\begin{array}{l}\int\limits_{0}^a \end{array} \)
f(x)dx-
\(\begin{array}{l}\int\limits_{0}^a\end{array} \)
f(x)dx= 0

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