Definite Integral
Definite Integral helps to find the area of a curve in a graph. It has limits, which are the start and the endpoints, within which the area under a curve 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 corresponding expression of definite integral 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.  Definite Integral Formula 
3.  Properties of Definite Integral 
4.  Applications of Definite Integral 
5.  FAQs on Definite Integral 
What is Definite Integral?
A definite integral is the area under a curve between two fixed limits. 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. To find the area under a curve between two limits, we divide the area into rectangles and sum them up. The more the number of rectangles, the more accurate the area is. So we divide the area into an infinite number of rectangles each with the same (very small) size and add all the areas. This is the fundamental theory that lies behind definite integrals.
Definite Integral Formula
Definite integral formulas are used to evaluate a definite integral. We have two formulas to evaluate a definite integral as mentioned below. The first formula is called the "definite integral as a limit sum" and the second formula is called the "fundamental theorem of calculus".
 \(\int_{a}^{b} f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^{n} h f(a+r h)\), where \(h=\frac{ba}{n}\)
 \(\int_{a}^{b} f(x) d x=F(b)F(a)\), where F'(x) = f(x)
We will evaluate an example of definite integral \(\int^1_0 \, x^2 dx\) using each of these formulas and see whether we get the same answer.
Definite Integral as Limit Sum
As explained in the previous section, we can express the area under a curve between two given limits as the sum of an infinite number of rectangles. Using this concept, to evaluate a definite integral \(\int^b_af(x)dx\), we divide the area under the curve into many rectangles by dividing [a, b] into an infinite number of subintervals. Thus, the definite integral as limit sum formula is:
\(\int_{a}^{b} f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^{n} h f(a+r h)\)
Here \(h=\frac{ba}{n}\) is the length of each subinterval.
Example: We will evaluate \(\int^1_0 \, x^2 dx\) using this definite integral formula.
Comparing the integral with \(\int^b_af(x)dx\), [a, b] = [0, 1] and f(x) = x^{2}. Then h = (1  0)/n = 1/n. Applying the above formula,
\(\int_{0}^{1} \,x^2 d x=\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n} f(0+\frac{r}{n})\)
= \(\lim _{n \rightarrow \infty} \sum_{n=1}^{\infty} \frac{1}{n}\left(\frac{r}{n}\right)^{2}\)
= \(\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \sum_{r=1}^{\infty} r^{2}\)
= \(\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \cdot \frac{n(n+1)(2 n+1)}{6}\) (using summation formulas)
= \(\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \frac{n^{3}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}\)
= \(\frac{(1+0)(2+0)}{6}\)
= 1/3
Definite Integral Formula Using FTC
A definite integral \(\int^b_af(x)dx\) can be evaluated by using the fundamental theorem of calculus (FTC). This is the easiest way of evaluating a definite integral. This formula says first to find out the antiderivative (indefinite integral) of f(x) (and represent it F(x)), substitute the upper limit first and then the lower limit one by one, and subtract the results in order. i.e.,
\(\int_{a}^{b} f(x) d x=F(b)F(a)\), where F'(x) = f(x)
Example: We will evaluate \(\int^1_0 \, x^2 dx\) using this definite integral formula.
We will first evaluate \(\int x^2 dx\) using the integral formulas. Then we get \(\int x^2 dx\) = \(\frac{x^{3}}{3}\) + C.
Now, we will substitute the limits (upper and lower limits in order) and find the difference.
\(\int^1_0 \, x^2 dx\) = (1^{3}/3 + C)  (0^{3}/3 + C) = 1/3.
The integration constant C is always canceled while applying limits. So we always ignore C while evaluating a definite integral.
Properties of Definite Integral
The definite integral properties 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 (i.e., f(x) = f(x)).
 \(\int^a_{a}f(x).dx = 0\) if f(x) is an odd function (i.e., 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 Using Definite Integral
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 Using Definite Integral
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 along 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 Using Definite Integral
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 topics would help for a better understanding of definite integral.
Definite Integral Examples

Example 1: Find the value of the definite integral \(\int^{\frac{\pi}{4}}_{\frac{\pi}{4}} \cos^2x.dx\).
Solution:
\(\begin{align}\int^{\frac{\pi}{4}}_{\frac{\pi}{4}} \cos^2x.dx &=2\int^{\frac{\pi}{4}}_0 \cos^2x.dx \\&=2\int^{\frac{\pi}{4}}_0 \dfrac{1 + \cos2x}{2}.dx\\&=\int^{\frac{\pi}{4}}_0(1 + \cos 2x).dx\\&=\left[x + \frac{\sin 2x}{2}\right]^{\frac{\pi}{4}}_0\\&=\left(\frac{\pi}{4} + \frac{\sin\frac{\pi}{2}}{2}\right)  0\\&=\frac{\pi}{4} + \frac{1}{2}\end{align}\)
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: \(\int^5_{5}x^2.dx =\frac{250}{3}\)

Example 3: If \(\int_0^3 f(x) dx = 50\) and \(\int_0^1 f(x) dx = 25\) then what is \(\int_1^3 f(x) dx \)?
Solution:
Using one of the properties of definite integrals,
\(\int_{a}^{b} f(x) \cdot d x=\int_{a}^{c} f(x) \cdot d x+\int_{c}^{b} f(x) \cdot d x\)
Using this,
\(\int_{0}^{3} f(x) \cdot d x=\int_{0}^{1} f(x) \cdot d x+\int_{1}^{3} f(x) \cdot d x\)
50 = 25 + \(\int_{1}^{3} f(x) \cdot d x\)
\(\int_{1}^{3} f(x) \cdot d x\) = 75
Answer: \(\int_{1}^{3} f(x) \cdot d x\) = 75
FAQs on Definite Integral
What Is the Definition of 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.
 \(\int ^b_a f(x).dx =  \int^a _b f(x).dx \)
 \(\int ^b_a f(x) .dx = \int^c _a f(x).dx + \int^b_cf(x).dx\)
 \(\int ^b_a f(x) .dx = \int^b _a f(a + b  x).dx \)
How to Evaluate a Definite Integral?
To evaluate a definite integral:
 Evaluate the indefinite integral (i.e., without limits)
 Substitute the upper limit and then the lower limit in the answer from the above step.
 Subtract both results in order.
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?
Definite Integral  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.  For an indefinite integral, the resultant answer is mostly an expression. 
There won't be the integration constant 'C'.  We always use the integration constant 'C' in the answer. 
All the formulas of indefinite integrals can be used with definite integrals along 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.
visual curriculum