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NCERT exemplar Class 12 Maths solutions chapter 8 Application Of Integrals is one of vital chapters of the NCERT Class 12 Maths Solutions. For those students who are planning to score well in their Class 12 Mathematics exam, they should not ignore integrals. Integrals and integration hold a crucial part of the question paper that one will receive in their exam. Not only integrals are important for CBSE board exams, but it is highly important for competitive examinations and higher education. Integrals are one of the base topics in calculus that is helpful in both Maths and physics. Chapter 7 of NCERT Class 12 Maths book is integrals, where students are made aware of what it is and what are its operations. In the NCERT exemplar Class 12 Maths solutions chapter 8, the topic is discussed in detail.
Question:1
Find the area of the region bounded by the curves
Answer:
It is mentioned in the question that,
belongs to a parabola and belong to a straight line which passes through origin.
Starting with a rough figure showing those equations below,
From the parabola equations it is seen that x cannot be negative, so the graph would be on the right of the X-axis. The parabola would thus be opening to the right.
In order to find the points of the two equations mentioned, you have to solve the two equations simultaneously.
We got the coordinates of x, for finding the coordinates of y.
Put x = 1 and x = 0 in the equations of y = 3x.
It is found that y coordinated are y = 3 and y = 0 respectively.
Therefore, the point of interaction of parabola and straight line is (1, 3) and (0, 0)
Now, calculate the area enclosed between the parabola and the straight line.
For calculating the area we have to minus the area under the straight line which extends from x = 0 to x = 1 from the area under the parabola.
It can be written as,
Area between parabola and straight line = area under parabola – area under straight line …. (1)
On calculating the area under parabola,
On integrating the above equation from 0 to 1
Now, for calculating the area under line y = 3x i.e. area of triangle OAB
On integrating the above equation from 0 to 1
Using equation mentioned above (1)
area between parabola and straight line
Therefore,
It is found that the area was 1/2 unit^{2} between the curves
Question:2
Find the area of the region bounded by the parabola
Answer:
The equation of the parabola is which shows no negative values for x. Therefore, the graph will be on right of Y-axis will be passing through (0, 0)
Similarly, for equations no negative values for y, therefore, the graph will be above the X axis passing through (0, 0).
For finding point of interaction, let us solve the equations simultaneously.
The interaction point was found to be (2p, 2p)
Now, we have to find out the areas between the two parabolas.
The equation can be written as
Area between the two parabola = area under parabola
For finding the area under
Integrating the equation from 0 to 2p
For finding the area under
Integrating from 0 to 2p
Using
(i)
Question:3
Find the area of the region bounded by the curve and y = x + 6 and x = 0.
Answer:
Let’s start with a rough plot of the curve along with the lines y = x + 6 and x = 0
When x = 0, it means Y axis
Questions says to find the area between curve and the line and Y axis
First solve the y = x + 6 and , in order to find the interaction point
For checking is 0, 1, 2 satisfy this cubic, shows 2 is one factor, therefore x – 2 is a factor.
Solving the equation,
Therefore, x = 2
Substituting this x = 2 in y = x + 6, y = 8
Therefore, curves intersect at (2, 8)
The area bounded will be
item Area bounded = area by on Y axis – area by y = x + 6 on Y axis
For finding area under
Integrate the equation from 0 to 8
Now, finding area under y = x + 6
For finding the area from 6 to 8 because line passes through Y axis at 6 and extends upto 8, the point where curve and line intersects.
Integrating from 6 to 8
Using the equation (1)
Area bound was found to be 12 – 2 = 10
Therefore, the area was found to be 10
Question:4
Find the area of the region bounded by the curve
Answer:
The equations is a parabola and no negative values of x are seen, therefore, this parabola lies to the right of the Y axis passing through (0, 0)
Similarly, for which is a parabola, not defined negative values of y lies above X axis and passing through (0, 0)
For finding point of interaction, solve simultaneously.
The point of interaction is therefore (4, 4)
For finding the area between two parabolas
Area between two parabolas = area under parabola
Let us find area under parabola
Integrate from 0 to 4
Let us find area under parabola
Integrate from 0 to 4
To find area under parabola
Therefore, the area is found to be
Question:5
Find the area of the region included between and y = x
Answer:
The equations has no negative values of x. Therefore it lies on right of Y axis passing through (0, 0).
And y = x depicts a straight line through origin.
For finding the area, picture shown below.
For finding the point of interaction solve the two equations simultaneously.
Therefore, point of interaction is (9, 9)
The area between the parabola and line = area under parabola – area under line
Let us find area under parabola
Now let us find area under straight line
y=x
Integrate from 0 to 9
Using (i)
area between parabola and line
Therefore, area was found to be
Question:6
Find the area of the region enclosed by the parabola and the line y = x + 2
Answer:
The equation depicts a parabola and has no negative values for y, therefore, it lies above X axis and passing through origin (0, 0).
And y = x + 2 is a straight line.
Below figure shows the area to be found out,
Let’s find the point of interaction of the two equations.
The point of interaction was found to be (-1, 1) and (2, 4)
Area between the line and parabola = area under line – area under parabola
Let us find area under line y=x+2
Now let us find area under the parabola
Therefore, the area was found to be
Question:7
Find the area of region bounded by the line x = 2 and the parabola
Answer:
The equation is a parabola not defining negative values of x, therefore it lies to the right of the Y axis passing through origin.
And x = 2 is a straight line parallel to Y axis.
Below figure shows the area to be calculated.
We have to integrate
On integrating the above equation from 0 to 2 it would give area enclosed under quadrant 1 only. Therefore, for finding the area in quadrant 2 we have to multiply it by 2 since it is symmetrical.
Therefore the area ODBC = 2 x area OBC
let us find area under parabola
Question:8
Sketch the region and x-axis. Find the area of the region using integration.
Answer:
The above equation depicts a circle with origin and radius 2
Now in
Equation of X axis is y = 0
For point of interaction,
Y = 0
Point of interaction are (-2, 0) and (2, 0)
The below figure shows the area
Now let us find the area
Integrate from -2 to 2
Question:10
Answer:
The equation of the line is 2y = 5x + 7
Now, we need point of the line, so we substitute x = 0 and then y = 0
The other two x = 2 and x = 8 are straight lines parallel to Y axis.
On plotting all the above lines.
Question:11
Answer:
Squaring both sides
The above equation has no values for x less than 1, therefore parabola will be right of x = 1
Now observe that in has to positive because of square root hence both positive hence the parabola will be drawn only in quadrant
We have to plot the curve in [1,5] so just draw the parabolic curve from x=1 to x=5 in quadrant
x=1 and x=5 are lines parallel to Y-axis
So we have to integrate from 1 to 5
let us find area under parabolic curve
Integrate from 1 to 5
Question:12
Determine the area under the curve included between the lines x = 0 and x = a
Answer:
The above equation is of a circle having centre as (0, 0) and radius a
Now in and which means x and y both positive or x negative and y positive hence the curve has to be above X -axis in and quadrant
x=0 is equation of Y -axis and x=a is a line parallel to Y-axis passing through (a, 0)
So we have to integrate from 0 to a
So we have to integrate from 0 to a
let us find area under curve
Integrate from 0 to a
Using uv rule of integration where u and v are functions of
Here and v=1
Hence
Question:13
Find the area of the region bounded by and y = x.
Answer:
This is a parabola, no negative values of x, therefore it lies on the right of Y axis passing through origin.
Now means y and x both has to be positive hence both lie in quadrant hence will be part of which is lying only in quadrant
And y=x is a straight line passing through origin
We have to find area between and y=x shown below
For finding the point of interaction, solve two equations simultaneously.
Put x=1 in y=x we get y=1
The point of interaction is (1, 1)
Area between the parabolic curve and line = area under parabolic curve – area under line
For the area under parabolic curve
Integrating from 0 to 1
For area under straight line y = x
On integrating from 0 to 1
Using (i)
Hence area bounded is
Question:14
Find the area enclosed by the curve and the straight lilne x + y + 2 = 0.
Answer:
This equation depicts a parabola defining no positive values of y therefore it lies below X axis and passes through origin \\
X + y + 2 = 0 depicts a straight line
For the point of interaction, both of the equations are solved simultaneously.
Therefore, the point of interaction is (-1, -1) and (2, -4)
Below figure shows the area to be calculated
Area between the line and parabola = Area under line – area under parabola
For finding area under line, integrate it from -1 to 2
Y = -(x + 2)
For the area under parabola, integrate it from -1 to 2
Substituting both values in (1)
area enclosed by line and parabola
The negative sign depicts the area is below the X axis
Hence the area enclosed is
Question:15
Find the area bounded by the curve , x = 2y + 3 in the first quadrant and x-axis.
Answer:
The above equation depicts a parabola which does not define negative values for x and therefore it lies to the right of the Y axis and passes through origin.
For plotting , both the values of x and y have to be positive hence this graph lies in the 1st quadrant only.
The point cannot be negative because the square root symbol itself depicts the positive root.
The equation x = 2y + 3 is a straight line.
For finding the point of interaction, solve the two equations simultaneously.
Y = -1 won’t make a difference since we are looking for the 1st quadrant only.
Therefore, the point of interaction are (9, 3)
Below is the figure shows the area to be calculated
Point of interaction of straight line on X axis can be calculated by substituting y = 0 in the equation of the line
The area enclosed = area under – area under the straight line
For the area under integrate the equation from 0 to 9
For the area under straight line
Y = (x-3) / 2
Integrating the equation from 3 to 9
Using (i)
area bounded =18-9=9
Hence area bounded by the curve and a straight line is 9
Question:16
Find the area of the region bounded by the curve and .
Answer:
depicts a parabola having no negative values for x and lying to the right of the Y axis with passing through origin.\\
The other equation of depicts a circle.
The general equation of circle is given by
Centre of circle is
In
Hence center is that is (2,0) and radius is which is 2
For the point of interaction, solve the two equations simultaneously.
Therefore, the point of interaction have been found to be (0, 0), (2, 2), and (2, -2)
Below is the diagram of the area to be calculated.
On integrating we will get area for 1st quadrant only, but since it is symmetrical we can multiple it by 2.
Area of shaded region = area under the circle – area under the parabola
For finding the area under circle
Integrate the above equation from 0 to 2
For the area under parabola,
Integrate the above equation from 0 to 2
After multiplying it by 2
Question:17
Find the area bounded by the curve y = sinx between x = 0 and x =
Answer:
Below is the graph with the shaded region whose area has to be calculated
From the plot, it is seen that the area from 0 to is positive whereas the area from to 2 is negative.
But both the areas are equal in magnitude with opposite signs.
If we integrate them, the areas will cancel each other and the answer will be 0.
Therefore, we integrate the two areas separately and split the limits in two ways –
1) Find area under sinx from 0 to and multiple by 2
2)Split the limit 0 to 2 into 0 to and to 2
Here we will solve by splitting the limits
Now the limits become negative
Hence area bounded by sinx from 0 to 2
Question:18
Answer:
The vertices of the triangle are given as (-1, 1), (0, 5) and (3, 2)
The three vertices are given alphabets as P, Q and R respectively.
Area of the region bounded by the curve , the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by
Required area
Question:19
Answer:
By solving the equations:
Through substituting for
Area of the region bounded by the curve y=f(x), the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a,b], is given by or
[By the symmetry of the image w.r.t x axis]
Question:20
Compute the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7.
Answer:
The lines given are x + 2y = 2
y – x = 1
and
Equate the values of x from 1 and 2 to get,
put the value of y in (2) to get
So, intersection point is (0,1)
On solving the equations, the point of interaction is found to be (0, 1), (2, 3) and (4, -1)
Area of the region bounded by the curve y=f(x), the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by
Required area
Question:21
Find the area bounded by the lines y = 4x + 5, y = 5 - x and 4y = x + 5
Answer:
The lines mentioned are y = 4x + 5, y = 5 - x and 4y = x + 5
Below if the figure,
By solving these equations,
From (1) and (2)
From (2) and (3)
The point of interaction have found to be (0, 5), (3, 2) and (-1, 1)
Area of the region bounded by the curve y=f(x), the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by
Required area
Question:22
Find the area bounded by the curve y = 2cos x and the x-axis from x = 0 to x = 2
Answer:
The curve given is y = 2cosx and x axis from x = 0 to x = 2
Below is the figure,
Area of the region bounded by the curve y=f(x) , the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by
From the figure;
Required area
Question:23
Answer:
Given;
Curve
Area of the region bounded by the curve y=f(x) , the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by
Required area
Question:24
The area of the region bounded by the y-axis, y = cos x and y = sin x, is.
A. sq units
B. ( + 1) sq units
C. ( – 1) sq unit
D. (2 – 1) sq unit
Answer:
The questions has given,
and
Question:25
The area of the region bounded by the curve and the straight line x = 4y – 2 is.
A. sq units
B. sq units
C. sq units
D. sq units
Answer:
The questions gives equation for curve and straight line x = 4y – 2
Below is the figure,
Question:26
The area of the region bounded by the curve and x-axis is.
A. 8π sq units
B. 20π sq units
C. 16π sq units
D. 256π sq units
Answer:
A)
Question:27
Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle is
A. 16π sq units
B. 4π sq units
C. 32π sq units
D. 24 sq units
Answer:
B)
The x-axis, the line y = x and the circle .
Question:28
Area of the region bounded by the curve y = cos x between x = 0 and x = π is
A. 2 sq units
B. 4 sq units
C. 3 sq units
D. 1 sq units
Answer:
A)
The questions mentions curve y = cos x and x = 0 and x = π
Required area
Question:29
The area of the region bounded by parabola and the straight line 2y = x is
A. sq units
B. 1 sq units
C. sq units
D. sq units
Answer:
A)
The questions gives parabola and a straight line 2y = x
By substituting for x
Required area
Question:30
The area of the region bounded by the curve y = sin x between the ordinates x = 0, x = π/2 and the x-axis is
A. 2 sq units
B. 4 sq units
C. 3 sq units
D. 1 sq units
Answer:
D)
The questions mentions y = sin x between the ordinates x = 0, x = π/2 and x axis
Question:31
The area of the region bounded by the ellipse is
A. 20π sq units
B. 20π^{2} sq units
C. 16π^{2} sq units
D. 25π sq units
Answer:
A)
The ellipse
Ellipse has a symmetry with x axis and y axis
The required area can be calculated as
Question:32
The area of the region bounded by the circle is
A. 2π sq units
B. π sq units
C. 3π sq units
D. 4π sq units
Answer:
B)
The circle
The circle is symmetrical with the x axis and y axis
Required Area
Question:33
The area of the region bounded by the curve y = x + 1 and the lines x = 2 and x = 3 is
A. sq units
B. sq units
C. sq units
D. sq units
Answer:
A)
The questions gives y = x + 1 and lines x = 2 and x = 3
Required area
Question:34
The area of the region bounded by the curve x = 2y + 3 and the y lines. y = 1 and y = –1 is
A. 4 sq units
B. sq units
C. 6 sq units
D. 8 sq units
Answer:
C)
The question mentions the curve x = 2y + 3 and the lines on y axis y = 1 and y = -1
Required area
Students can make use of NCERT exemplar Class 12 Maths solutions chapter 8 pdf download, to access it offline. We will help the students to understand the application of integrals by solving the questions given in NCERT.
NCERT exemplar Class 12 Maths solutions chapter 8 Application Of Integrals
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Our experienced teachers have covered every question and have created the NCERT exemplar Class 12 Maths chapter 8 solutions for it. The solutions given by us are simple, methodical and step oriented. We want our students to understand every step so much so that solving questions will come super easy for them.
Class 12 Maths NCERT exemplar solutions chapter 8 Application of Integrals touch upon exhaustive explanation of how the integrals can be used and to find the area under curves. There are several topics that are covered, and there are plenty of many things that one will learn.
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | Application of Integrals |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 |
This chapter contains a total 4 questions in one exercise. The questions are based on problem solving.
You need to refer to NCERT Books for solving math problems once you clarify the concept you can use formulae to solve numerous problems. Also prefer NCERT exemplar Class 12 Maths solutions chapter 8 pdf download using the webpage to pdf tools.
The NCERT exemplar solutions for Class 12 Maths chapter 8 are prepared by specialized mathematicians who refer various high-level books to find best possible answers.
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hello,
Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.
I hope this was helpful!
Good Luck
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