NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1 - Application of Integrals
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Application of Integrals is like your favourite colouring brush that can fill any shape under every curve, no matter how abstract the shape is. It is a fundamental part of calculus that can help us calculate the area of a certain region, even with no straight sides. In exercise 8.1 of the chapter Application of Integrals, we go beyond the basics of integration and learn about how it's used in our real-life situations, like finding the area of a region or calculating the total distance travelled between two specific points. This article about the NCERT Solutions for Exercise 8.1 of Class 12 Maths, Chapter 8 - Application of Integrals, offers detailed and easy-to-understand solutions for the exercise problems, so that students can clear their doubts and understand the logic behind the solutions. For syllabus, notes, exemplar solutions and PDF, refer to this link: NCERT.
This Story also Contains
- Class 12 Maths Chapter 8 Exercise 8.1 Solutions: Download PDF
- Application of Integrals Class 12 Chapter 8 Exercise: 8.1
- Topics covered in Chapter 8, Application of Integrals: Exercise 8.1
- NCERT Solutions Subject Wise
- NCERT Exemplar Solutions Subject Wise
Class 12 Maths Chapter 8 Exercise 8.1 Solutions: Download PDF
Application of Integrals Class 12 Chapter 8 Exercise: 8.1
Question:1 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 = $\int_0^4 y\,dx$
$= \int^4_{0}3\sqrt{1-\frac{x^2}{16}} dx$
$= \frac{3}{4}\int^4_{0}\sqrt{16-x^2} dx$
$= \frac{3}{4}\left [ \frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}\sin^{-1}\frac{x}{4} \right ]^4_{0}$
$= \frac{3}{4}\left [ 2\sqrt{16-16} +8\sin^{-1}(1)-0-8\sin^{-1}(0)\right ]$
$= \frac{3}{4}\left [ \frac{8\pi}{2} \right ]$
$= \frac{3}{4}\left [ 4\pi \right ] =3\pi$
Therefore, the area bounded by the ellipse will be $= 4 \times 3\pi = 12\pi$ units.
Question 2: 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 $= \int_0^2 y\ dx$
$= \int^2_{0}3\sqrt{1-\frac{x^2}{4}} dx$
$= \frac{3}{2}\int^2_{0}\sqrt{4-x^2} dx$
$= \frac{3}{2}\left [ \frac{x}{2}\sqrt4-x^2 +\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]^2_{0}$
$= \frac{3}{2}\left [ \frac{2\pi}{2} \right ]$
$= \frac{3\pi}{2}$
Therefore the area bounded by the ellipse will be $= 4\times \frac{3\pi}{2} = 6\pi$ units.
Question 3: Choose the correct answer in the following
$\small (A)\hspace{1mm}\pi$
$\small (B)\hspace{1mm}\frac{\pi}{2}$
$\small (C)\hspace{1mm}\frac{\pi}{3}$
$\small (D)\hspace{1mm}\frac{\pi}{4}$
Answer:
The area bounded by circle C(0,0,4) and the line x=2 is
The required area = area of OAB
$\int_0^2 y\,dx = \int_0^2 \sqrt{4 - x^2}\,dx$
$= \left[ \frac{x}{2} \sqrt{4 - x^2} + \frac{4}{2} \sin^{-1} \frac{x}{2} \right]_0^2$
$= 2 \cdot \frac{\pi}{2}$
$= \pi$
Hence, The correct answer is $\pi$
Question 4: Choose the correct answer in the following.
(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
The required area = OAB
$= \int_0^3 x\,dy$
$= \int_0^3 \frac{y^2}{4}\,dy$
$= \frac{1}{4} \left[ \frac{y^3}{3} \right]_0^3$
$= \frac{9}{4}$
Hence, The correct answer is $\small \frac{9}{4}$
Also Read,
Topics covered in Chapter 8, Application of Integrals: Exercise 8.1
The main topic covered in class 12 maths chapter 8 of Application of Integrals, exercise 8.1 is:
Area under a curve: The exercise starts with calculating the area under the curve $y=f(x)$, between two points on the X axis, as $x=a$ and $x=b$. The area can be found using definite integrals as: $A=\int_a^b f(x) d x$.
Also Read,
NCERT Solutions Subject Wise
Below are some useful links for subject-wise NCERT solutions for class 12.
- NCERT solutions class 12 chemistry
- NCERT solutions for class 12 physics
- NCERT solutions for class 12 biology
- NCERT solutions for class 12 mathematics
JEE Main Highest Scoring Chapters & Topics
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NCERT Exemplar Solutions Subject Wise
Here are some links to subject-wise solutions for the NCERT exemplar class 12.
Frequently Asked Questions (FAQs)
Q: How many questions are there in exercise 8.1 Class 12 Maths ?
A:
13 questions are in total in exercise 8.1 Class 12 Maths.
Q: Which applications are mentioned in this exercise ?
A:
Mainly area under the curve is discussed in this exercise.
Q: Is this chapter very important for Board exams ?
A:
This is very important for boards as well as other subjects like Physics also.
Q: Is it more difficult than the previous chapter ?
A:
No, it is less difficult than Chapter 7 Integrals.
Q: What are some other topics discussed in this Chapter ?
A:
Maily area and at higher level volume etc. are discussed in this.
Q: How many exercises are there in this Chapter ?
A:
Total 3 exercises are there including miscellaneous exercise.
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