CBSE Class 12 Mathematics Notes Chapter 8 Application of Integrals

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The application of integrals is used to calculate the area and volume of different 2-D and 3-D curves, and they have various applications in mathematics.

They help us calculate the area of an arc of a circle, irregular boundaries, the volume of various curves, and the area between the two curves.
Integrals include the summation of discrete data, and their applications cover basic integral concepts such as the fundamental theorem of calculus.
It is used in the fields of architecture, electrical engineering, medical science and statistics.
In mathematics, integrals are used to calculate the area under a curve and the area of a region bounded by a curve and a line.

Application of integrals is important for class 12 students, as well as engineering mathematics. CBSE Class 12 Mathematics Notes for Chapter 8 Application of Integrals are given in the article below for easy preparation and understanding of the concepts involved.

Read More:

Preparation Resources
Application of Integrals	NCERT Solutions For Class 12 Mathematics Chapter 8: Application of Integrals
NCERT Solutions For Class 12 Mathematics Chapter 8 Exercise 8.1	NCERT Solutions for Class 12 Maths Chapter 8 Applications of Integrals Miscellaneous Exercise Solutions

Application of Integrals

There are various applications of integrals. Some of them are as follows:

It is used to calculate the area under simple curves.
The concept is used to find areas enclosed by lines, arcs of circles, parabolas, and ellipses.
Integrals are used to find out the area between two curves.
It is used to find out the centroids of areas of the triangle with curved boundaries.
In the field of statistics, it is used to determine survey data to improve marketing plans for different companies.

Area Between the Curve and the Axis

The area of the region bounded by the curve y = f(x), x-axis, and the lines x = a and x = b (b > a) is given by the formula

Area = ∫_a^b y dx = ∫_a^b f(x) dx

The area between the curve and the axis

The area between the curve and the axis

The area of the region bounded by the curve x = Φ(y), y-axis, and the lines y = c and y = d (b > a) is given by the formula

Area = ∫_c^dx dy = ∫_c^dΦ(y) dy

Area Between Two Curve

The area of the region enclosed between two curves y = f(x), y = g(x), and the lines x = a, x = b is given by the formula

Area = ∫_a^b [f(x) - g(x)] dx

Where f(x) ≥ g (x) in [a, b]

Area Between Two Curve

Area Between Two Curve

If f(x) ≥ g (x) in [a, c] and f(x) ≤ g (x) in [c, b], a < c < b, then

Area = ∫_a^c [f(x) - g(x)] dx + ∫_c^b [g(x) - f(x)] dx

Method to Find Area Under Curve

To calculate area, first, find the equation of the curve, y = f(x), as well as its limits and axis.
The integration i.e. antiderivative of the curve is found.
The upper and lower limits are applied to the integral result, and the difference provides the area under the curve.

Area in Polar Coordinates

Consider the region OKM bounded by a polar curve r = f(θ) and two semi-straight lines θ = ⍺ and θ = ꞵ.
The area of the polar region is given by

Area = 1/2 ∫_⍺^ꞵ r²dθ = 1/2 ∫_⍺^ꞵ f²(θ) dθ

Area in Polar Coordinates

Area in Polar Coordinates

Area Between Two Polar Curves

The area of a region between two polar curves r = f(θ) and r = g(θ) in the sector [⍺, ꞵ] is expressed by the integral

Area = 1/2 ∫_⍺^ꞵ [f²(θ) – g²(θ)] dθ

Area Between Two Polar Curves

Area Between Two Polar Curves

Some Standard Curves and their Equation

Straight Line:

x = a and x = – a, where a > 0

Straight Line

Straight Line

Circle:

x² + y² = a²

Circle

Circle

Parabola:

y² = 4ax or y² = – 4ax

Parabola

Parabola

Ellipse

x²/a² + y²/b² = 1

Ellipse

Ellipse

There are Some important List Of Top Mathematics Questions On Applications Of Integrals Asked In CBSE CLASS XII

CBSE CLASS XII Related Questions

1.
Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$

View Solution

2.
The integrating factor of the differential equation $ (x + 2y^3) \frac{dy}{dx} = 2y $ is:

$ e^{y^2} $
$ \frac{1}{\sqrt{y}} $
$ e^{-\frac{1}{y^2}} $
$ e^{y^2} $

View Solution

3.
Evaluate : \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} \]

View Solution

4.
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.

View Solution

5.
Let $ 2x + 5y - 1 = 0 $ and $ 3x + 2y - 7 = 0 $ represent the equations of two lines on which the ants are moving on the ground. Using matrix method, find a point common to the paths of the ants.

View Solution

6.
If $ \mathbf{a} $ and $ \mathbf{b} $ are position vectors of two points $ P $ and $ Q $ respectively, then find the position vector of a point $ R $ in $ QP $ produced such that \[ QR = \frac{3}{2} QP. \]

View Solution

CBSE Class 12 Mathematics Notes Chapter 8 Application of Integrals

Application of Integrals

Area Between the Curve and the Axis

Area Between Two Curve

Method to Find Area Under Curve

Area in Polar Coordinates

Area Between Two Polar Curves

Some Standard Curves and their Equation

CBSE CLASS XII Related Questions

1.
Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$

2.
The integrating factor of the differential equation \( (x + 2y^3) \frac{dy}{dx} = 2y \) is:

3.
Evaluate : \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} \]

4.
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.

5.
Let \( 2x + 5y - 1 = 0 \) and \( 3x + 2y - 7 = 0 \) represent the equations of two lines on which the ants are moving on the ground. Using matrix method, find a point common to the paths of the ants.

6.
If \( \mathbf{a} \) and \( \mathbf{b} \) are position vectors of two points \( P \) and \( Q \) respectively, then find the position vector of a point \( R \) in \( QP \) produced such that \[ QR = \frac{3}{2} QP. \]

Similar Mathematics Concepts

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CBSE Class 12 Mathematics Notes Chapter 8 Application of Integrals

Application of Integrals

Area Between the Curve and the Axis

Area Between Two Curve

Method to Find Area Under Curve

Area in Polar Coordinates

Area Between Two Polar Curves

Some Standard Curves and their Equation

CBSE CLASS XII Related Questions

1.Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$

2.The integrating factor of the differential equation \( (x + 2y^3) \frac{dy}{dx} = 2y \) is:

3. Evaluate : \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} \]

4.Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.

5.Let \( 2x + 5y - 1 = 0 \) and \( 3x + 2y - 7 = 0 \) represent the equations of two lines on which the ants are moving on the ground. Using matrix method, find a point common to the paths of the ants.

6.If \( \mathbf{a} \) and \( \mathbf{b} \) are position vectors of two points \( P \) and \( Q \) respectively, then find the position vector of a point \( R \) in \( QP \) produced such that \[ QR = \frac{3}{2} QP. \]

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Evaluate: $ \tan^{-1} \left[ 2 \sin \left( 2 \cos^{-1} \frac{\sqrt{3}}{2} \right) \right]$

2.
The integrating factor of the differential equation \( (x + 2y^3) \frac{dy}{dx} = 2y \) is:

3.
Evaluate : \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} \]

4.
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.

5.
Let \( 2x + 5y - 1 = 0 \) and \( 3x + 2y - 7 = 0 \) represent the equations of two lines on which the ants are moving on the ground. Using matrix method, find a point common to the paths of the ants.

6.
If \( \mathbf{a} \) and \( \mathbf{b} \) are position vectors of two points \( P \) and \( Q \) respectively, then find the position vector of a point \( R \) in \( QP \) produced such that \[ QR = \frac{3}{2} QP. \]