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:
\(\begin{array}{l}\int\limits_{a}^b f(x)dx= \int\limits_{a}^b f(t)dt\end{array} \)
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:
\(\begin{array}{l}\int\limits_{a}^b f(x)dx= -\int\limits_{b}^a f(x)dx\end{array} \)
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
\(\begin{array}{l}\int\limits_{a}^b f(x)dx= [F(x){]}_{a}^{b}= F(b)-F(a)\end{array} \)
Therefore
\(\begin{array}{l}\int\limits_{a}^b f(x)dx= F(b)-F(a) = – [F(a)-F(b)= -\int\limits_{b}^a f(x)dx\end{array} \)
Property 3:
\(\begin{array}{l}\int\limits_{a}^b dx= \int\limits_{a}^cf(x)dx + \int\limits_{c}^bf(x)dx\end{array} \)
From the second theorem of Calculus,
\(\begin{array}{l}\int\limits_{a}^b f(x)dx= [F(x){]}_{a}^{b}= F(b)-F(a)\end{array} \)
Therefore,
\(\begin{array}{l}\int\limits_{a}^b f(x)dx= F(b)-F(a)…(1)\end{array} \)
\(\begin{array}{l}\int\limits_{a}^c f(x)dx= F(c)-F(a)…(2)\end{array} \)
\(\begin{array}{l}\int\limits_{c}^b f(x)dx= F(b)-F(c)…(3)\end{array} \)
Adding equations 2 and 3 we have,
\(\begin{array}{l}\int\limits_{a}^c f(x)dx + \int\limits_{c}^bf(x)dx=F(c)-F(a) + F(b)-F(c)=F(b)-F(a)=\int\limits_{a}^b f(x)dx\end{array} \)
Property 4:
\(\begin{array}{l}\int\limits_{a}^b f(x)dx= \int\limits_{a}^b f(a+b-x)dx\end{array} \)
Let us assume that t = a + b – x. Therefore dt = -dx. At x = a, t = b and at x = b, t =a.
Therefore,
\(\begin{array}{l}\int\limits_{a}^b f(x)dx= -\int\limits_{b}^a f(a+b-t)dt\end{array} \)
From the second property of definite integrals:
\(\begin{array}{l}\int\limits_{a}^b f(x)dx=-\int\limits_{b}^a f(x)dx\end{array} \)
Therefore,
\(\begin{array}{l}\int\limits_{a}^b f(x)dx=-\int\limits_{b}^a f(a+b-t)dt=\int\limits_{a}^b f(a+b-t)dt\end{array} \)
Following the first property of definite integrals:
\(\begin{array}{l}\int\limits_{a}^b f(x)dx= \int\limits_{a}^bf(a+b-x)dx\end{array} \)
Property 5:
\(\begin{array}{l}\int\limits_{0}^a f(x)dx=\int\limits_{0}^a f(a-x)dx\end{array} \)
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,
\(\begin{array}{l}\int\limits_{0}^a f(x)dx= \int\limits_{a}^0 f(a-t)dt\end{array} \)
From the second property of definite integrals:
\(\begin{array}{l}\int\limits_{0}^a f(x)dx=-\int\limits_{0}^a f(x)dx\end{array} \)
Therefore,
\(\begin{array}{l}\int\limits_{0}^a f(x)dx= -\int\limits_{a}^0 f(a-t)dt=\int\limits_{0}^a f(a-t)dt\end{array} \)
Following the first property of definite integrals:
\(\begin{array}{l}\int\limits_{0}^a f(x)dx=\int\limits_{0}^a f(a-x)dx\end{array} \)
Property 6:
\(\begin{array}{l}\int\limits_{0}^{2a}f(x)dx=\int\limits_{0}^a f(x)f(x)dx+\int\limits_{0}^a f(2a-x)dx\end{array} \)
From the second property of definite integrals
\(\begin{array}{l}\int\limits_{a}^b f(x)dx= \int\limits_{a}^c f(x)dx+\int\limits_{c}^b f(x)dx\end{array} \)
Therefore,
\(\begin{array}{l}\int\limits_{0}^{2a}f(x)dx=\int\limits_{0}^a f(x)dx+\int\limits_{a}^{2a} f(x)dx…(4)\end{array} \)
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:
\(\begin{array}{l}\int\limits_{a}^{2a} f(x)dx= \int\limits_{a}^0 f(2a-t)dt=\int\limits_{0}^a f(2a-t)dt=\int\limits_{0}^a f(2a-x)dx\end{array} \)
Equation 4 can, therefore, be rewritten as:
\(\begin{array}{l}\int\limits_{0}^{2a} f(x)dx=\int\limits_{0}^a f(x)dx+\int\limits_{0}^a f(2a-x)dx\end{array} \)
Property 7:Â
\(\begin{array}{l}\int\limits_{0}^{2a} \end{array} \)
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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