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:
\( F(a) – F(b) = \int\limits_{a}^b f(x)dx\)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.
Evaluating Definite Integrals – Properties
Let us now discuss important properties of definite integrals and their proofs.
Property 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.
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)dxProperty 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)dxProperty 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
Property 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
Property 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)dxEquation 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
Property 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
Property 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
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 odd 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
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