Formulas and Solved Examples of Application of Integrals
In mathematics, Integration and Differentiation are the most significant ways to solve complex mathematical problems. Besides mathematics, Integration and Differentiation play key roles in Science, Engineering, and various other facets of our life. We have outlined what's, ifs, and how's related to integrals and their application. But how? Let us take an example.
To determine the area of a rectangle, the formulae 'length × breadth' is used. But what if you are given to calculate the area of the shaded portion, of a shaded rectangle with an unshaded circle within?
Importance of Integrals in Maths
Integration is an important chapter in Maths that needs to be studied well in advance by the students before they attempt tests on the topic. It always comes in use even if the students take up engineering, Science in their future years. Students can read from Application of Integrals on Vedantu to know more. They need to practice sums based on Integrals too to perfect them. It is an important topic that will prove to be quite instrumental later on.
How is the Calculation done when Integrals come into the Play
Integrals
In mathematics, the application of Integrals is applied to find the area under a curve, areas bounded by any curve, and so on.
Definition of Integrals
An integral is a function, of which a given function is a derivative. It is also known as the anti-derivative or reverse of a derivative. Integrals are used to determine the area of 2D objects and the volume of 3D objects in real life.
Types of Integrals
There are two types of Integrals.
Definite Integrals
Indefinite Integrals
Definite Integrals
An integral which has a start and an end value is known as a definite integral. In simple words, the function is restricted within an interval a,b, where a and b are upper limit and lower limit, respectively. It is represented as
y = \[\int_{b}^{a}\] f(x)dx
y = \[\int_{a}^{b}\] f(x)dx
Example: y = \[\int_{2}^{4}\] 6x dx
Sol: y = \[\int_{2}^{4}\] 6x dx
y = \[\int_{4}^{2}\] 6x dx , here a = 4, b = 2
y = 6\[\left [\frac{x^{2}}{2} \right ]_{2}^{4}\]
= 6\[\left [\frac{4^{2}}{2} \right ]\]-\[\left [ \frac{2^{2}}{2} \right ]\]=6\[\left [ 8-6 \right ]\]=36
Indefinite Integrals
An integral which does not have an upper and lower limit is known as a definite integral. It is represented as
y = \[\int_{b}^{a}\] f(x)dx = F(x)+C, where ‘C’ is a constant
Example: y = \[\int_{2}^{4}\] 6x dx
Sol:
y = \[\int\] 6xdx
y = 6 \[\left [\frac{x^{2}}{2} \right ]\] = 3x + C, where C is a constant
Solved Examples
Question 1: Determine the Area Enclosed By a Circle x²+ y² =a²
Sol: It is observed that the area enclosed by the given circle is ‘4 x area of the region AOBA bounded by the curve, x-axis and the ordinates x=0 and x=a’.
As the circle is symmetrical about both x−axis and y−axis=4 \[\int_{o}^{a}\]ydx (taking vertical strips)
= 4 \[\int_{o}^{a}\] \[\sqrt{\left (a^{2}-x^{2} \right )}\]dx
Since x²+ y² = a² gives y = ± \[\sqrt{a^{2}-x^{2}}\]
The quarter AOBA lies in the first quadrant, hence 'y' is taken as positive. On integrating, we get the entire area enclosed by the given circle.
= 4 \[\left [ \left [\frac{x}{2} \right ]\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a} \right ]\]
= 4 \[\left [ \left (\left [\frac{x}{2} \right ]\times 0+\frac{a^{2}}{2}sin^{-1}1 \right )-0 \right ]\]
= 4 \[\left (\frac{a^{2}}{2} \right )\]\[\left ( \frac{\pi }{2} \right )\]
= πa²
Question 2: Determine the Area of the Region Bounded b the Curve y = x² and the line Y = 4
Sol: Since the given curve expressed by the equation y = x² is a parabola symmetric about y-axis only, therefore, the required area of the region AOBA is given bt
= 2 \[\int_{0}^{4}\] xdy
= 2 area of the region BONB bounded by the curve, y-axis and the lines y=0 and y =4
=2 x ( area of the region BONB bounced by the curve, y-axis and the lines y = 0 and y = 4)
= 2 \[\int_{0}^{4}\] ydy
= 2 \[\times \frac{2}{3}\] \[\left [ y^{\frac{3}{2}} \right ]_{0}^{4}\]
= \[\frac{4}{3}\]×8
= \[\frac{32}{3}\]
Application of Integrals
Integrals have their application in both science and maths. In maths, the application of integral is made to determine the area under a curve, the area between two curves, the center of mass of a body, and so on. Whereas in science (Physics in particular), the application of integrals is made to calculate the Centre of Gravity, Mass, Momentum, Work done, Kinetic Energy, Velocity, Trajectory, and Thrust.
Application of Integrals in Engineering Fields
There’s a vast application of integration in the fields of engineering.
In Architecture
To determine the amount of material required in a curved surface. For instance, take the construction of a dome.
In Electrical Engineering
Integrals are used in Electrical Engineering to calculate the length of a power cable required for transmission between two power stations.
Application of Integrals in Different Fields
In Medical Science
Integrals are used to determine the growth of bacterias in the laboratory by keeping variables such as a change in temperature and foodstuff.
In Medicine
To study the rate of spread of infectious disease, the field of epidemiology uses medical seine to determine how fast a disease is spreading, its origin, and how to best treat it.
In Statistics
To estimate survey data to help improve marketing plans for different companies because a survey requires many different questions with a range of possible answers.
Does Vedantu have anything on Integrals?
Vedantu has study material on Integrals that can be used by all students to study from. It has an application of Integrals on its platform which students can read and understand. The material is completely free of cost and can even be downloaded in offline mode and studied.
FAQs on Application of Integrals in Calculus
1. What are the applications of integrals in mathematics?
The applications of integrals are mainly used to find area under curves, volumes of solids, displacement, and other accumulated quantities. In calculus, integration helps measure quantities that are continuously changing.
- Area under a curve between two points.
- Area between two curves.
- Volume of solids using disk or washer method.
- Displacement and distance from velocity functions.
- Work done and mass in physics.
2. How do you find the area under a curve using integration?
The area under a curve from x = a to x = b is given by the definite integral ∫ab f(x) dx. To calculate it:
- Find the antiderivative F(x) of f(x).
- Apply the Fundamental Theorem of Calculus.
- Compute F(b) − F(a).
∫02 x² dx = [x³/3]02 = 8/3.
3. What is the formula for the area between two curves?
The area between two curves y = f(x) and y = g(x) is ∫ab [f(x) − g(x)] dx, where f(x) ≥ g(x).
- Find the points of intersection.
- Subtract the lower function from the upper function.
- Integrate over the interval.
4. How do you find the volume of a solid using integrals?
The volume of a solid using integration is commonly found using the disk method formula V = π∫ab [f(x)]² dx. Steps:
- Identify the radius function f(x).
- Square the function.
- Multiply by π and integrate.
V = π∫01 x² dx = π(1/3) = π/3.
5. What is the difference between definite and indefinite integrals?
The main difference is that an indefinite integral gives a family of functions, while a definite integral gives a numerical value.
- Indefinite integral: ∫f(x) dx = F(x) + C.
- Definite integral: ∫ab f(x) dx = F(b) − F(a).
6. How are integrals used to find displacement and distance?
The displacement is the definite integral of velocity, given by ∫ v(t) dt over a time interval.
- Displacement = ∫ab v(t) dt.
- Distance = ∫ab |v(t)| dt.
7. How do you find the area of a region bounded by a curve and the x-axis?
The area bounded by a curve and the x-axis is found using ∫ab f(x) dx, taking absolute value if the function is below the axis.
- Find points where the curve meets the x-axis.
- Integrate between those limits.
- Ensure the area is positive.
8. What is the application of integrals in finding work done?
The work done by a variable force is calculated using W = ∫ab F(x) dx.
- F(x) represents force as a function of displacement.
- Integrate over the given interval.
W = ∫02 3x² dx = [x³]02 = 8 units of work.
9. How do you find the average value of a function using integrals?
The average value of a function on [a, b] is (1/(b − a)) ∫ab f(x) dx.
- Compute the definite integral.
- Divide by (b − a).
Average value = (1/2)∫02 x dx = (1/2)(2) = 1.
10. What are some real-life applications of integrals?
In real life, integrals are used to measure accumulated quantities such as area, volume, growth, and total change.
- Finding area of irregular land plots.
- Calculating volume of tanks and containers.
- Determining population growth over time.
- Computing total profit or cost from marginal functions.
