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Definite Integral is a type of Integral that has a pre-existing value of limits which means that it has upper and lower limits.
- Integral is defined as a function whose derivative is another function.
- Integrals are also referred to as anti-derivatives of a function determined by Integration.
- Definite Integrals and Indefinite Integrals are the two types of Integrals.
- Definite Integral is used to calculate the area of a curve in a graph between two fixed limits.
Definite Integrals are expressed as:
| \(\int^b_af(x)dx\) |
Where a and b denote the lower and upper limit respectively for a function f(x), defined with reference to the x-axis.
Read More: NCERT Solutions For Class 12 Mathematics Integrals
Key Terms: Definite Integral, Integrals, Indefinite Integrals, Integration, Integration Formulas, Limits, Integrand, Integrating Agent
What are Definite Integrals?
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Definite Integral is defined as the area under a curve between two fixed limits.
- Definite Integrals have limits, that are considered the start and the endpoints, within which the area is calculated.
- The limit starting and ending points are [a, b], and are used to find the area of the curve f(x), with respect to the x-axis.
- Definite integrals are used to find the area within limits in Integration.
Definite Integral of a real-valued function f(x) with respect to a real variable x on an interval [a, b] is given as:
| \(\int^b_af(x)dx\) |
Where
- ∫: Integration symbol
- a: Lower Limit
- b: Upper Limit
- f(x): Integrand
- dx: Integrating Agent
Therefore, ∫ab f(x) dx expresses the definite integral of f(x) with respect to dx from a to b.
The video below explains this:
Definite Integrals Video Lecture:
Definite Integral Formula
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Definite Integral Formula is used to evaluate or simplify a definite integral. There are two important formulas to evaluate a definite integral.
The first formula is referred to as "Definite Integral as a Limit Sum" and is given as:
| \(\int_{a}^{b} f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^{n} h f(a+r h)\), Where \(h=\frac{b-a}{n}\) |
The second formula is referred to as the "Fundamental Theorem of Calculus" and is given as:
| \(\int_{a}^{b} f(x) d x=F(b)-F(a)\), Where F'(x) = f(x) |
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Definite Integral as Limit Sum
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The area under a curve between two given limits can be expressed as the sum of an infinite number of rectangles. In order to evaluate a definite integral \(\int^b_af(x)dx\), the area under the curve is divided into many rectangles by dividing [a, b] into an infinite number of subintervals.
Definite Integral as Limit Sum Formula is given as:
| \(\int_{a}^{b} f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^{n} h f(a+r h)\) |
Where, \(h=\frac{b-a}{n}\) denotes the the length of each subinterval.
Solved ExampleExample: Evaluate \(\int^1_0 \, x^2 dx\) using Definite Integral Formula. Solution: On comparing the given integral with \(\int^1_0 \, x^2 dx\), [a, b] = [0, 1] and f(x) = x2 and h = (1 - 0)/n = 1/n. Using the formula for Definite Integral as Limit Sum, \(\int_{0}^{1} \,x^2 d x=\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n} f(0+\frac{r}{n})\) = \(\lim _{n \rightarrow \infty} \sum_{n=1}^{\infty} \frac{1}{n}\left(\frac{r}{n}\right)^{2}\) = \(\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \sum_{r=1}^{\infty} r^{2}\) Now, using summation formulas, = \(\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \cdot \frac{n(n+1)(2 n+1)}{6}\) = \(\lim _{n \rightarrow \infty} \frac{1}{n^{3}} \frac{n^{3}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}\) = \(\frac{(1+0)(2+0)}{6}\) = 1/3 |
Definite Integral Using Fundamental Theorem of Calculus
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Definite Integral is also evaluated with the help of the Fundamental Theorem of Calculus. It is considered the easiest method of evaluating a definite integral.
- First, find out the antiderivative, i.e. indefinite integral of f(x), and represent it as F(x).
- Then, substitute the upper limit first and then the lower limit one by one.
- At last, subtract the results in order.
Thus, Definite Integral Using the Fundamental Theorem of Calculus is given as:
| \(\int_{a}^{b} f(x) d x=F(b)-F(a)\), Where F'(x) = f(x) |
Solved ExampleExample: Solve \(\int^1_0 \, x^2 dx\) using this definite integral formula. Solution: On evaluating ∫ x2dx using the integral formulas, we get ∫x2 dx = \(\frac{x^{3}}{3}\)+ C Substituting the upper and lower limits in order to find the difference. \(\int^1_0 \, x^2 dx\) = (13/3 + C) - (03/3 + C) = 1/3 |
Properties of Definite Integrals
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Properties of Definite Integrals help to find the integral for a function multiplied by a constant, for the sum of the functions, and even and odd functions more efficiently. Here are the important properties of Definite Integrals:
- ∫ab f(x) dx = ∫ab f(t) d(t)
- ∫ab f(x) dx = – ∫ba f(x) dx
- ∫aa f(x) dx = 0
- ∫ab f(x) dx = ∫ac f(x) dx + ∫cb f(x) dx
- ∫ab f(x) dx = ∫ab f(a + b – x) dx
- ∫0a f(x) dx = f(a – x) dx
Definite Integral by Parts
Definite integral of a function can also be evaluated by splitting it into parts.
- ∫02a f (x) dx = ∫0a f (x) dx + ∫0a f (2a – x) dx
- ∫02a f (x) dx = 2 ∫0a f (x) dx if f(2a – x) = f (x).
- ∫02a f (x) dx = 0 if f (2a – x) = – f(x).
- ∫-aa f(x) dx = 2 ∫0a f(x) dx if f(- x) = f(x) or is an even function.
- ∫-aa f(x) dx = 0 if f(- x) = – f(x) or is an odd function.
Applications of Definite Integrals
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Definite integral is used to find the area of the following:
- Area between Two Curves
- To find Volumes.
- To find Length of Plane Curve
- Area of Surface of Revolution
- It is widely used in physics, engineering statistics, machine learning, etc.
Definite Integral Solved Examples
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Here are the solved examples on definite integrals for a better understanding of the same:
Example 1: Evaluate ∫0π/4 sin 2x dx.
Solution: Assume that, I = ∫0 π/4 sin 2x dx
Now, ∫ sin 2x dx = -(1/2) cos 2x
I = ∫0 π/4 sin 2x dx
= [-(1/2) cos 2x]0π/4
= -(1/2) cos 2(π/4) – {-(½) cos 2(0)}
= -(1/2) cos π/2 + (½) cos 0
= -(1/2) (0) + (1/2)
= 1/2
Thus, ∫0 π/4 sin 2x dx = 1/2
Example 2: What is the value of ∫23 x2 dx?
Solution: Assume that, I = ∫23 x2 dx
Now, ∫x2 dx = (x3)/3
I = ∫23 x2 dx = [(x3)/3]23
= (33)/3 – (23)/3 = (27/3) – (8/3)
= (27 – 8)/3 = 19/3
Thus, ∫23 x2 dx = 19/3
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Things to Remember
- Definite Integral is a type of integral that has upper and lower limits.
- Integrals are the values of the function calculated through Integration.
- Definite Integrals help to calculate the area under a curve in a graph.
- Definite Integral Formula is given using a Limit Sum and the Fundamental Theorem of Calculus.
- Definite Integral as a Limit Sum is given as \(\int_{a}^{b} f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^{n} h f(a+r h)\), Where \(h=\frac{b-a}{n}\).
- Definite Integral using Fundamental Theorem of Calculus is \(\int_{a}^{b} f(x) d x=F(b)-F(a)\), Where F'(x) = f(x).
- They are used to find the areas of plane figures such as circles, parabolas, and ellipses.
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Previous Years’ Questions
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Sample Questions
Ques. How to evaluate a Definite Integral? (3 Marks)
Ans. Definite Integral is evaluated using the following steps:
- Find the indefinite integral, i.e., the integral without limits.
- Substitute the upper limit and then the lower limit in the obtained answer.
- At last, subtract both the results in order.
Ques. Differentiate between Definite and Indefinite Integral. (3 Marks)
Ans. The difference between Definite and Indefinite Integral is as follows:
| Definite Integral | Indefinite Integral |
|---|---|
| Definite integrals are defined as integrals with limits. | Indefinite integrals are defined as integrals that do not have any limits. |
| The solution of a definite integral is a simple numeric value. | The solution of an indefinite integral is mostly an expression. |
| There isn’t an integration constant 'C' in the answer. | Integration Constant 'C' is used in the answer. |
Ques. State the uses of Definite Integral. (3 Marks)
Ans. The uses of Definite Integral are listed as follows:
- Definite Integrals are used to find the area of curves such as a circle, ellipse, or parabola.
- The area of a small space is calculated by applying limits and is then manipulated to find the area of the entire space.
- Area of a circle is calculated by taking its integration according to the x-axis in the first quadrant with limits from origin to its radius.
- Further, it is multiplied by 4 to obtain the area of the entire circle.
Ques. Evaluate the given definite integral: \(\int^5_{-5}x^2.dx\). (3 Marks)
Ans. Given integral is \(\int^5_{-5}x^2.dx\).
= \(\begin{align}\int^5_{-5}x^2.dx &= 2\int^5_0x^2.dx\\&\end{align}\)
\(=2\left[\frac{x^3}{3}\right]^5_0\)
\(=2(\frac{5^3}{3} - 0)\)
\(=2 \times \frac{125}{3}\)
\(=\frac{250}{3}\)
Thus, \(\int^5_{-5}x^2.dx\) = 250/3.
Ques. Can Definite Integral be Negative? (1 Mark)
Ans. Yes, definite integrals can be negative, The value of a definite integral can be negative, positive, or zero.
Ques. What is the value of the definite integral \(\int^{\frac{\pi}{4}}_{\frac{-\pi}{4}} \cos^2x.dx\)? (3 Marks)
Ans. Given integral is \(\int^{\frac{\pi}{4}}_{\frac{-\pi}{4}} \cos^2x.dx\).
= \(\begin{align}\int^{\frac{\pi}{4}}_{\frac{-\pi}{4}} \cos^2x.dx &=2\int^{\frac{\pi}{4}}_0 \cos^2x.dx \end{align}\)
\(=2\int^{\frac{\pi}{4}}_0 \dfrac{1 + \cos2x}{2}.dx\)
\(=\int^{\frac{\pi}{4}}_0(1 + \cos 2x).dx\)
\(=\left[x + \frac{\sin 2x}{2}\right]^{\frac{\pi}{4}}_0\)
\(=\left(\frac{\pi}{4} + \frac{\sin\frac{\pi}{2}}{2}\right) - 0\)
\(=\frac{\pi}{4} + \frac{1}{2}\)
Thus, \(\int^{\frac{\pi}{4}}_{\frac{-\pi}{4}} \cos^2x.dx\) = \(\frac{\pi}{4} + \frac{1}{2}\).
Ques. Is C present in Definite Integrals? (1 Mark)
Ans. No, C is not present in definite integrals as it is not necessary to add an arbitrary constant, i.e. C in the case of definite integrals.
Ques. What is Definite Integral Formula? (3 Marks)
Ans. There are two formulas for Definite Integrals which are as follows:
- Definite Integral as a Limit Sum: \(\int_{a}^{b} f(x) d x=\lim _{n \rightarrow \infty} \sum_{r=1}^{n} h f(a+r h)\), Where \(h=\frac{b-a}{n}\)
- Fundamental Theorem of Calculus: \(\int_{a}^{b} f(x) d x=F(b)-F(a)\), Where F'(x) = f(x)
Ques. How is a Definite Integral expressed? (2 Marks)
Ans. A definite integral is expressed as:
\(\int^b_af(x)dx\)
Where
- ∫ is the Integration Symbol.
- a is the Lower Limit.
- b is the Upper Limit
- f(x) is the Integrand
- dx is the Integrating Agent
Ques. How is Definite Integral evaluated by Parts? (3 Marks)
Ans. Definite Integrals are evaluated using Parts with the help of the given formulas:
- ∫02a f (x) dx = ∫0a f (x) dx + ∫0a f (2a – x) dx
- ∫02a f (x) dx = 2 ∫0a f (x) dx … if f(2a – x) = f (x).
- ∫02a f (x) dx = 0 … if f (2a – x) = – f(x)
- ∫-aa f(x) dx = 2 ∫0a f(x) dx … if f(- x) = f(x) or it is an even function
- ∫-aa f(x) dx = 0 … if f(- x) = – f(x) or it is an odd function
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