NCERT Solutions for Exercise 8.1 Class 12 Maths Chapter 8

NCERT Solutions for Exercise 8.1 Class 12 Maths Chapter 8 - Application of Integrals

Edited By Ramraj Saini | Updated on Dec 03, 2023 10:19 PM IST | #CBSE Class 12th

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NCERT Solutions For Class 12 Maths Chapter 8 Exercise 8.1

NCERT Solutions for Exercise 8.1 Class 12 Maths Chapter 8 Application of Integrals are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for Class 12 Maths chapter 8 exercise 8.1 is the extension of last chapter i.e integrals. Concepts learnt in Class 12 Maths NCERT book chapter Integrals will be used heavily in this chapter. Exercise 8. Class 12 Maths deals with area finding of simple curves. NCERT solutions for Class 12 Maths chapter 8 exercise 8.1 requires only basic understanding of integrals which is learnt in the NCERT book previous chapter.

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NCERT Solutions For Class 12 Maths Chapter 8 Exercise 8.1
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Application of Integrals Class 12 Chapter 8 Exercise: 8.1
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Also Read| Application of Integrals Class 12 Notes
Benefits of NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1
Key Features Of NCERT Solutions for Exercise 8.1 Class 12 Maths Chapter 8
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NCERT Solutions Subject Wise
Subject Wise NCERT Exemplar Solutions

Similar questions are asked in Physics also, hence this NCERT syllabus exercise becomes more important. Upcoming exercises are also linked to this exercise, hence sincerely it should be practice. 12th class Maths exercise 8.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Access NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1

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Application of Integrals Class 12 Chapter 8 Exercise: 8.1

Question:1 Find the area of the region bounded by the curve $y^2=x$ and the lines $x=1,x=4$ and the $x$ -axis in the first quadrant.

Answer:

Area of the region bounded by the curve $y^2=x$ and the lines $x=1,x=4$ and the $x$ -axis in the first quadrant

Area = $\int_{1}^{4}ydy = \int_{1}^{4}\sqrt{x}dx$

$\left [\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]^4_1 = \frac{2}{3}\left [ (4)^\frac{3}{2}- (1)^\frac{3}{2} \right ]$

$= \frac{2}{3}\left [ 8 -1 \right ]$

= 14/3 units

Question:2 Find the area of the region bounded by $y^2=9x,x=2,x=4$ and the $x$ -axis in the first quadrant.

Answer:

Area of the region bounded by the curve $y^2=9x,x=2,x=4$ and the $x$ -axis in the first quadrant

Area = $\int_{2}^{4}ydy = \int_{2}^{4}\sqrt{9x}dx = 3\int_{2}^{4}\sqrt{x}dx$

$3\left [\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]^4_2 = 3.\frac{2}{3}\left [ (4)^\frac{3}{2}- (2)^\frac{3}{2} \right ]$

$= 2\left [ 8 -2\sqrt2 \right ]$

$= \left [ 16 -4\sqrt2 \right ]$ units

Question:3 Find the area of the region bounded by $x^2=4y,y=2,y=4$ and the $y$ -axis in the first quadrant.

Answer:

The area bounded by the curves $x^2=4y,y=2,y=4$ and the $y$ -axis in the first quadrant is ABCD.

$= \int^4_{2} x dy$

$= \int^4_{2} 2\sqrt{y} dy$

$= 2\int^4_{2} \sqrt{y} dy$

$=2\left \{ \frac{y^{\frac{3}{2}}}{\frac{3}{2}} \right \}^4_{2}$

$= \frac{4}{3}\left \{ (4)^{\frac{3}{2}}-(2)^{\frac{3}{2}} \right \}$

$= \frac{4}{3} \left \{ 8 -2\sqrt 2 \right \}$

$= \left \{ \frac{32-8\sqrt 2}{3} \right \}\ units.$

Question:4 Find the area of the region bounded by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1.$

Answer:

The area bounded by the ellipse : $\frac{x^2}{16}+\frac{y^2}{9}=1.$

Area will be 4 times the area of EAB.

Therefore, $Area\ of\ EAB= \int^4_{0} y dx$

$= \int^4_{0}3\sqrt{1-\frac{x^2}{16}} dx$

$= \frac{3}{4}\int^4_{0}\sqrt{16-x^2} dx$

$= \frac{3}{4}\left [ \frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}\sin^{-1}\frac{x}{4} \right ]^4_{0}$

$= \frac{3}{4}\left [ 2\sqrt{16-16} +8\sin^{-1}(1)-0-8\sin^{-1}(0)\right ]$

$= \frac{3}{4}\left [ \frac{8\pi}{2} \right ]$

$= \frac{3}{4}\left [ 4\pi \right ] =3\pi$

Therefore the area bounded by the ellipse will be $= 4\times {3\pi} = 12\pi\ units.$

Question: 5 Find the area of the region bounded by the ellipse $\small \frac{x^2}{4}+\frac{y^2}{9}=1$

Answer:

The area bounded by the ellipse : $\small \frac{x^2}{4}+\frac{y^2}{9}=1$

The area will be 4 times the area of EAB.

Therefore, $Area\ of\ EAB= \int^2_{0} y dx$

$= \int^2_{0}3\sqrt{1-\frac{x^2}{4}} dx$

$= \frac{3}{2}\int^2_{0}\sqrt{4-x^2} dx$

$= \frac{3}{2}\left [ \frac{x}{2}\sqrt4-x^2 +\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]^2_{0}$

$= \frac{3}{2}\left [ \frac{2\pi}{2} \right ]$

$= \frac{3\pi}{2}$

Therefore the area bounded by the ellipse will be $= 4\times \frac{3\pi}{2} = 6\pi\ units.$

Question: 6 Find the area of the region in the first quadrant enclosed by $\small x$ -axis, line $\small x=\sqrt{3}y$ and the circle $\small x^2+y^2=4$
Answer:

The area of the region bounded by $\small x=\sqrt{3}y$ and $\small x^2+y^2=4$ is ABC shown:

The point B of the intersection of the line and the circle in the first quadrant is $(\sqrt3,1)$ .

Area ABC = Area ABM + Area BMC where, M is point in x-axis perpendicular drawn from the line.

Now,area of $ABM = \frac{1}{2}\times AM\times BM = \frac{1}{2}\times \sqrt{3}\times 1 =\frac{\sqrt3}{2}$ ............(1)

and Area of $BMC = \int^2_{\sqrt{3}} ydx$

$= \int^2_{\sqrt3} \sqrt{4-x^2} dx$

$= \left [ \frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]^2_{\sqrt3}$

$= \left [ 2\times\frac{\pi}{2}-\frac{\sqrt3}{2}\sqrt{4-3}-2\sin^{-1}\left ( \frac{\sqrt3}{2} \right ) \right ]$

$= \left [ \pi - \frac{\sqrt3\pi}{2}-2\frac{\pi}{3} \right ]$

$= \left [ \pi-\frac{\sqrt3}{2}-\frac{2\pi}{3} \right ]$

$= \left [ \frac{\pi}{3}-\frac{\sqrt3}{2} \right ]$ ..................................(2)

then adding the area (1) and (2), we have then

The Area under ABC $= \frac{\sqrt3}{2} +\frac{\pi}{3}-\frac{\sqrt3}{2} = \frac{\pi}{3}\ units.$

Question: 7 Find the area of the smaller part of the circle $\small x^2+y^2=a^2$ cut off by the line
$\small x=\frac{a}{\sqrt{2}}$

Answer:

we need to find the area of smaller part of the circle
Now,
Area of ABCD = 2 X Area of ABC
Area of ABC = $\int_{\frac{a}{\sqrt2}}^{a} ydx= \int_{\frac{a}{\sqrt2}}^{a} \sqrt{a^2-x^2}dx= \left [ \frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\right ]^{a}_\frac{a}{\sqrt2}\\ \\$
$=\left [ \frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}\sin^{-1}\frac{a}{a}- \frac{a}{2\sqrt2}\sqrt{a^2-(\frac{a}{\sqrt2})^2}-\frac{a^2}{2}\sin^{-1}\frac{a}{a\sqrt2}\right ]$
$=\left [ \frac{a}{2}\sqrt{0}+\frac{a^2}{2}\sin^{-1}1- \frac{a}{2\sqrt2}\sqrt{\frac{a^2}{2}}-\frac{a^2}{2}\sin^{-1}\frac{1}{\sqrt2}\right ]$
$=\left [ 0+\frac{a^2}{2}\frac{\pi}{2}- \frac{a^2}{4}-\frac{a^2}{2}\frac{\pi}{4}\right ]$
$=\frac{a^2}{4}\left ( \frac{\pi}{2}-1 \right )$
Area of ABCD = 2 X Area of ABC
$=2\times\frac{a^2}{4}\left ( \frac{\pi}{2}-1 \right )= \frac{a^2}{2}\left ( \frac{\pi}{2}-1 \right )$
Therefore, the area of the smaller part of the circle is $\frac{a^2}{2}\left ( \frac{\pi}{2}-1 \right )$

Question:8 The area between $\small x=y^2$ and $\small x=4$ is divided into two equal parts by the line $\small x=a$ , find the value of $\small a$ .

Answer:

we can clearly see that given area is symmetrical about x - axis
It is given that
Area of OED = Area of EFCD
Area of OED = $\int_{0}^{a}ydx = \int_{0}^{a}\sqrt xdx = \left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_0^{a}= \frac{a^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2a^{\frac{3}{2}}}{3}$
Area of EFCD = $\int_{a}^{4}ydx = \int_{a}^{4}\sqrt xdx = \left [ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ]_a^{4}= \frac{4^{\frac{3}{2}}-a^\frac{3}{2}}{\frac{3}{2}} = \frac{2(8-a^\frac{3}{2})}{3}=\frac{2(8-a^\frac{3}{2})}{3}$
Area of OED = Area of EFCD
$\frac{2a^{\frac{3}{2}}}{3}= \frac{2(8-a^{\frac{3}{2}})}{3}\\ \\ 2a^\frac{3}{2} = 8\\ a^\frac{3}{2} = 4\\ a = (4)^\frac{2}{3}$
Therefore, the value of a is $a = (4)^\frac{2}{3}$

Question:9 Find the area of the region bounded by the parabola $\small y=x^2$ and $\small y=|x|$ .

Answer:

We can clearly see that given area is symmetrical about y-axis
Therefore,
Area of OCAO = Area of OBDO
Point of intersection of $y=x^2 \ and \ y = |x|$ is (1 , 1) and (-1 , 1)
Now,
Area od OCAO = Area OAM - Area of OCMO
Area of OAM = $\frac{1}{2}.OM.AM = \frac{1}{2}.1.1 = \frac{1}{2}$
Area of OCMO = $\int_{0}^{1}ydx= \int_{0}^{1}x^2dx= \left [ \frac{x^3}{3} \right ]_{0}^{1}= \frac{1}{3}$
Therefore,
Area od OCAO $=\frac{1}{2}- \frac{1}{3}= \frac{1}{6}$
Now,
Area of the region bounded by the parabola $\small y=x^2$ and $\small y=|x|$ is = 2 X Area od OCAO $=2\times \frac{1}{6} = \frac{1}{3}$ Units

Question: 10 Find the area bounded by the curve $\small x^2=4y$ and the line $\small x=4y-2$ .

Answer:

Points of intersections of $y = x^2 \ and \ x = 4y-2$ is
$A\left ( -1,\frac{1}{4} \right ) \ and \ B(2,1)$
Now,
Area of OBAO = Area of OBCO + Area of OCAO
Area of OBCO = Area of OMBCO- Area of OMBO

Area of OMBCO = $\int_{0}^{2}ydx = \int_{0}^{2}\frac{x+2}{4}dx = \left [ \frac{x^2}{8} \right ]_{0}^{2}+\left [ \frac{x}{2} \right ]_{0}^{2}= \frac{4}{8}+\frac{2}{2}=\frac{3}{2}$

Area of OMBO = $\int_{0}^{2}ydx = \int_{0}^{2}\frac{x^2}{4}dx = \left [ \frac{x^3}{12} \right ]_{0}^{2}= \frac{8}{12}= \frac{2}{3}$

Area of OBCO = Area of OMBCO- Area of OMBO
$= \frac{3}{2}-\frac{2}{3}= \frac{5}{6}$
Similarly,
Area of OCAO = Area of OCALO - Area of OALO

Area of OCALO = $\int_{-1}^{0}ydx = \int_{-1}^{0}\frac{x+2}{4}dx = \left [ \frac{x^2}{8} \right ]_{-1}^{0}+\left [ \frac{x}{2} \right ]_{-1}^{0}=- \frac{1}{8}-\frac{(-1)}{2}=-\frac{1}{8}+\frac{1}{2}=\frac{3}{8}$

Area of OALO = $\int_{-1}^{0}ydx = \int_{-1}^{0}\frac{x^2}{4}dx = \left [ \frac{x^3}{12} \right ]_{-1}^{0}= -\frac{(-1)}{12}= \frac{1}{12}$

Area of OCAO = Area of OCALO - Area of OALO
$=\frac{3}{8}- \frac{1}{12}= \frac{9-2}{24}= \frac{7}{24}$
Now,
Area of OBAO = Area of OBCO + Area of OCAO
$=\frac{5}{6}+ \frac{7}{24}= \frac{20+7}{24}= \frac{27}{24} = \frac{9}{8}$

Therefore, area bounded by the curve $\small x^2=4y$ and the line $\small x=4y-2$ is $\frac{9}{8} \ units$

Question: 11 Find the area of the region bounded by the curve $\small y^2=4x$ and the line $\small x=3$ .

Answer:
The combined figure of the curve $y^2=4x$ and $x=3$
1594726575288
The required are is OABCO, and it is symmetrical about the horizontal axis.
Therefore, Area of OABCO = 2 $\times$ Area of OAB
$\\=2[\int_{0}^{3}ydx]\\ =2\int^3_02\sqrt{x}dx\\ =4[\frac{x^{3/2}}{3/2}]^3_0\\ =8\sqrt{3}$
therefore the required area is $8\sqrt{3}$ units.

Question: 12 Choose the correct answer in the following

Area lying in the first quadrant and bounded by the circle $\small x^2+y^2=4$ and the lines $\small x=0$ and $\small x=2$ is

$\small (A)\hspace{1mm}\pi$ $\small (B)\hspace{1mm}\frac{\pi }{2}$ $\small (C)\hspace{1mm}\frac{\pi }{3}$ $\small (D)\hspace{1mm}\frac{\pi }{4}$

Answer:

The correct answer is A
The area bounded by circle C(0,0,4) and the line x=2 is
1594726886037
The required area = area of OAB
$\int^2_0ydx = \int^2_0\sqrt{4-x^2}dx$
$\\=[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2}]^2_0\\ =2(\pi/2)\\ =\pi$

Question: 13 Choose the correct answer in the following.

Area of the region bounded by the curve $\small y^2=4x$ , $\small y$ -axis and the line $\small y=3$ is

(A) $\small 2$ (B) $\small \frac{9}{4}$ (C) $\small \frac{9}{3}$ (D) $\small \frac{9}{2}$

Answer:

The area bounded by the curve $y^2=4x$ and y =3

1594727150537
the required area = OAB =
$\\\int ^3_0xdy\\ =\int ^3_0\frac{y^2}{4}dy\\ =\frac{1}{4}.[\frac{y^3}{3}]^3_0\\ =\frac{9}{4}$

Benefits of NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1

The Class 12th Maths chapter 8 exercise has 3 exercises in total maily dealing with the application of integrals.
These Class 12 Maths chapter 8 exercise 8.1 solutions are helpful in understanding the theory also.
NCERT solutions for Class 12 Maths chapter 8 exercise 8.1 provides some basic questions initially and then moderate to advanced level of questions later on.

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Key Features Of NCERT Solutions for Exercise 8.1 Class 12 Maths Chapter 8

Comprehensive Coverage: The solutions encompass all the topics covered in ex 8.1 class 12, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 maths ex 8.1, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 ex 8.1 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 8.1 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
Inclusive Approach: Solutions for ex 8.1 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
Relevance to Curriculum: The solutions for class 12 maths ex 8.1 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. How many questions are there in exercise 8.1 Class 12 Maths ?

13 questions are in total in exercise 8.1 Class 12 Maths.

2. Which applications are mentioned in this exercise ?

Mainly area under the curve is discussed in this exercise.

3. Is this chapter very important for Board exams ?

This is very important for boards as well as other subjects like Physics also.

4. Is it more difficult than the previous chapter ?

No, it is less difficult than Chapter 7 Integrals.

5. What are some other topics discussed in this Chapter ?

Maily area and at higher level volume etc. are discussed in this.

6. How many exercises are there in this Chapter ?

Total 3 exercises are there including miscellaneous exercise.

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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Exercise 8.1 Class 12 Maths Chapter 8 - Application of Integrals

NCERT Solutions For Class 12 Maths Chapter 8 Exercise 8.1

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Application of Integrals Class 12 Chapter 8 Exercise: 8.1

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Also Read| Application of Integrals Class 12 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1

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