Definite Integral
Definite Integral helps to find the area of a curve in a graph. It has limits, which are the start and the end points, within which the area is calculated. The limit points can be taken as [a, b], to find the area of the curve f(x), with respect to the xaxis. The required expression for definite integrals is \(\int^b_af(x)dx\). Integration is the sum of the areas, and definite integrals are used to find the area within limits.
The study of integration started in the third century BC with the use of it to find the area of circles, parabola, ellipse. Let us learn more about definite integrals and the properties of definite integrals.
1.  What Is Definite Integral? 
2.  Properties of Definite Integral 
3.  Applications of Definite Integral 
4.  Solved Examples on Definite Integral 
5.  Practice Questions 
6.  FAQs on Definite Integral 
What is Definite Integral?
A definite integral has a defined value. The definite integral is represented as \(\int^b_af(x)dx\), where a is the lower limit and b is the upper limit, for a function f(x), defined with reference to the xaxis. The definite integrals is the antiderivative of the function f(x) to obtain the function F(x), and the upper and lower limit is applied to find the value F(b)  F(a).
\(\int^b_af(x).dx = [F(x)]^b_a = F(b)  F(a)\)
Unlike integrals, the definite integral do not have the constant C, since it is canceled on applying the upper and lower limit to the antiderivative of the function.
Properties of Definite Integral
The properties of definite integrals are helpful to integrate the given function and apply the lower and the upper limit to find the value of the integral. The definite integral formulas help for finding the integral for a function multiplied by a constant, for the sum of the functions, and for even and odd functions. Let us check the below properties of definite integrals, which are helpful to solve problems of definite integrals.
 \(\int ^b_a f(x) .dx = \int^b _a f(t).dt \)
 \(\int ^b_a f(x).dx =  \int^a _b f(x).dx \)
 \(\int ^b_a cf(x).dx = c \int^b _a f(x).dx \)
 \(\int ^b_a f(x) \pm g(x).dx = \int^b _a f(x).dx \pm \int^b_ag(x).dx\)
 \(\int ^b_a f(x) .dx = \int^c _a f(x).dx + \int^b_cf(x).dx\)
 \(\int ^b_a f(x) .dx = \int^b _a f(a + b  x).dx \)
 \(\int ^a_0 f(x) .dx = \int^a _0 f(a  x).dx \) (This is a formula derived from the above formula.)
 \(\int^{2a}_0f(x).dx = 2\int^a_0f(x).dx\) if f(2a  x) = f(x)
 \(\int^{2a}_0f(x).dx = 0\) if f(2a  x) = f(x).
 \(\int^a_{a}f(x).dx = 2\int^a_0f(x).dx\) if f(x) is an even function, and f(x) = f(x).
 \(\int^a_{a}f(x).dx = 0\) if f(x) is an odd function, and f(x) = f(x).
Applications of Definite Integral
The definite integrals are primarily used to find the areas of plane figures such as circles, parabolas, ellipse. Let us check in detail the application of definite integrals to find the areas of each of these figures.
Area of a Circle
The area of the circle is calculated by first calculating the area of the part of the circle in the first quadrant. Here the equation of the circle x^{2} + y^{2} = a^{2} is changed to an equation of a curve as y = √(a^{2}  x^{2}). Here we use the concept of definite integral to find the equation of the curve with respect to the xaxis and the limits from 0 to a.
The area of the circle is four times the area of the quadrant of the circle. The area of the quadrant is calculated by integrating the equation of the curve across the limits in the first quadrant.
A = 4\(\int^a_0 y.dx\)
= 4\(\int^a_0 \sqrt{a^2  x^2}.dx\)
= 4\([\frac{x}{2}\sqrt{a^2  x^2} + \frac{a^2}{2}Sin^{1}\frac{x}{a}]^a_0\)
= 4[((a/2)× 0 + (a^{2}/2)Sin^{1}1)  0]
= 4(a^{2}/2)(π/2)
= πa^{2}
Hence the area of the circle is πa^{2} square units.
Area of a Parabola
A parabola has an axis that divides the parabola into two symmetric parts. Here we take a parabola that is symmetric along the xaxis and has an equation y^{2} = 4ax. This can be transformed as y = √(4ax). We first find the area of the parabola in the first quadrant by using the definite integral formulas with respect to the xaxis and long the limits from 0 to a. Here the definite integral is calculated within the boundary and it is doubled to obtain the area under the whole parabola. The derivations for the area of the parabola is as follows.
\(\begin{align}A &=2 \int_0^a\sqrt{4ax}.dx\\ &=4\sqrt a \int_0^a\sqrt x.dx\\& =4\sqrt a[\frac{2}{3}.x^{\frac{3}{2}}]_0^a\\&=4\sqrt a ((\frac{2}{3}.a^{\frac{3}{2}})  0)\\&=\frac{8a^2}{3}\end{align}\)
Therefore the area under the curve enclosed by the parabola is \(\frac{8a^2}{3}\) square units.
Area of an Ellipse
The equation of the ellipse with the major axis of length 2a and a minor axis of 2b is x^{2}/a^{2} + y^{2}/b^{2} = 1. This equation can be transformed in the form as y = b/a .√(a^{2}  x^{2}). Here we use the concept of definite integral to calculate the area bounded by the ellipse in the first coordinate and with respect to the xaxis. Further, it is multiplied with 4 to obtain the area of the ellipse. The boundary limits taken on the xaxis is from 0 to a. The calculations for the area of the ellipse are as follows.
\(\begin{align}A &=4\int_0^a y.dx \\&=4\int_0^a \frac{b}{a}.\sqrt{a^2  x^2}.dx\\&=\frac{4b}{a}[\frac{x}{2}.\sqrt{a^2  x^2} + \frac{a^2}{2}Sin^{1}\frac{x}{a}]_0^a\\&=\frac{4b}{a}[(\frac{a}{2} \times 0) + \frac{a^2}{2}.Sin^{1}1)  0]\\&=\frac{4b}{a}.\frac{a^2}{2}.\frac{\pi}{2}\\&=\pi ab\end{align}\)
Therefore the area of the ellipse is πab sq units.
Related Topics
The following topic links would help for a better understanding of definite integral.
Solved Examples on Definite Integral

Example 1: Find the value of the definite integral \(\int^{\frac{\pi}{4}}_{\frac{\pi}{4}} Cos^2x.dx\).
Solution:
\(\begin{align}\int^{\frac{\pi}{4}}_{\frac{\pi}{4}} Cos^2x.dx &=2\int^{\frac{\pi}{4}}_0 Cos^2x.dx \\&=2\int^{\frac{\pi}{4}}_0 \dfrac{1 + Cos2x}{2}.dx\\&=\int^{\frac{\pi}{4}}_0(1 + cos2x).dx\\&=\left[x + \frac{Sin2x}{2}\right]^{\frac{\pi}{4}}_0\\&=\left(\frac{\pi}{4} + \frac{Sin\frac{\pi}{2}}{2}\right)  0\\&=\frac{\pi}{4} + \frac{1}{2}\end{align}\)
Answer: \(\int^{\frac{\pi}{4}}_{\frac{\pi}{4}} Cos^2x.dx=\frac{\pi}{4} + \frac{1}{2}\)

Example 2: Find the solution for the definite integral \(\int^5_{5}x^2.dx\).
Solution:
\(\begin{align}\int^5_{5}x^2.dx &= 2\int^5_0x^2.dx\\&=2\left[\frac{x^3}{3}\right]^5_0\\&=2(\frac{5^3}{3}  0)\\&=2 \times \frac{125}{3}\\&=\frac{250}{3}\end{align}\)
Note: Here the function x^{2} is an even function, and hence we have used the formula \(\int^a_{a}f(x).dx = 2\int^a_0f(x).dx\).
Answer: \(\int^5_{5}x^2.dx =\frac{250}{3}\)
FAQs on Definite Integral
What Is Definite Integral?
The definite integral is used to find the area of the curve, and it is represented as \(\int^b_af(x).dx\), where a is the lower limit and b is the upper limit., for a function f(x), defined with reference to the xaxis. The definite integrals is the antiderivative of the function f(x) to obtain the function F(x), and the upper and lower limit is applied to find the value F(b)  F(a).
\(\int^b_af(x).dx = [F(x)]^b_a = F(b)  F(a)\)
What Are the Important Formula of Definite Integrals?
The important formulas of definite integrals are as follows.
 \(\int ^b_a f(x).dx =  \int^a _b f(x).dx \)
 \(\int ^b_a f(x) .dx = \int^c _a f(x).dx + \int^b_cf(x).dx\)
 \(\int ^b_a f(x) .dx = \int^b _a f(a + b  x).dx \)
 \(\int^a_{a}f(x).dx = 2\int^a_0f(x).dx\)
How to Evaluate a Definite Integral?
The definite integrals are evaluated by first finding the integral of the given function and then applying the upper and the lower limit. Here the definite integral isis represented as \(\int^b_af(x).dx\), where a is the lower limit and b is the upper limit., for a function f(x), defined with reference to the xaxis. This antiderivative of f(x) is F(x), which is used to find definite integral.
\(\int^b_af(x).dx = [F(x)]^b_a = F(b)  F(a)\)
How to you Evaluate the Definite Integrals of Even Functions?
The definite integrals of even function also follow a similar process as any other function. Further, we have the following specific formula to find the definite integral of even functions.
 \(\int^{2a}_0f(x).dx = 2\int^a_0f(x).dx\) if f(2a  x) = f(x)
 \(\int^a_{a}f(x).dx = 2\int^a_0f(x).dx\) if (x) is an even function, and f(x) = f(x).
What Is the Definite Integral of 1?
The definite integral of 1 is simply the difference of the limits. Deriving from the formula we have \(\int^b_a 1.dx = (x)^b_a = b  a\).
What Is Definite Integral and Indefinite Integral?
The definite integrals are defined for integrals with limits. Indefinite integrals do not have any limits. The answer of a definite integral is a simple numeric value, but for an indefinite integral, the resultant answer is an antiderivative expression. All the formulas of indefinite integrals can be used with definite integrals, with the application of limits to the formula.
What Is the Practical Use of Definite Integral?
The definite integrals can be used to find the area of curves such as a circle, ellipse, parabola. Basically, integration formulas is used to find the area of irregular shapes. In definite integrals, the area of a small space is calculated by applying limits, and then it is manipulated to find the area of the entire space. The area of a circle is calculated by taking its integration with respect to the xaxis in the first quadrant with limits from origin to its radius and further it is multiplied by 4 to obtain the area of the entire circle.