NCERT Solutions for Class 12 Maths Chapter 8 Application of integrals

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NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals: Exercise Questions

NCERT Class 12 Maths Chapter 6 Application of Integrals question answers with detailed explanations are provided below.

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 : $\frac{x^2}{16}+\frac{y^2}{9}=1$

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

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

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

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

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

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

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

Question 3: Choose the correct answer in the following

The 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

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 4: 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

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

Question 1: Find the area under the given curves and given lines:

(i) $\small y=x^2,x=1,x=2$ and $\small x$ -axis

Answer:

The area bounded by the curve $\small y=x^2,x=1,x=2$ and $\small x$ -axis

The area of the required region = area of ABCD
$\\=\int_{1}^{2}ydx\\ =\int_{1}^{2}x^2dx\\ =[\frac{x^3}{3}]_1^2\\ =\frac{7}{3}$
Hence, the area of the shaded region is 7/3 units.

Question 1: Find the area under the given curves and given lines:

(ii) $\small y=x^4,x=1,x=5$ and $\small x$ -axis

Answer:

The area bounded by the curev $\small y=x^4,x=1,x=5$ and $\small x$ -axis

The area of the required region = area of ABCD
$\\=\int_{1}^{5}ydx\\ =\int_{1}^{2}x^4dx\\ =[\frac{x^5}{5}]_1^2\\ =625-\frac{1}{5}\\ =624.8$
Hence, the area of the shaded region is 624.8 units.

Question 2: Sketch the graph of $\small y=|x+3|$ and evaluate $\small \int_{-6}^{0}|x+3|dx.$

Answer:

y=|x+3|

The given modulus function can be written as

x+3>0

x>-3

for x>-3

y=|x+3|=x+3

x+3<0

x<-3

For x<-3

y=|x+3|=-(x+3)

The integral to be evaluated is

$\\\int_{-6}^{0}|x+3|dx$

$=\int_{-6}^{-3}(-x-3)dx+\int_{-3}^{0}(x+3)dx$

$ =[-\frac{x^{2}}{2}-3x]_{-6}^{-3}+[\frac{x^{2}}{2}+3x]_{-3}^{0}$

$=(-\frac{9}{2}+9)-(-18+18)+0-(\frac{9}{2}-9)\\ =9$

Question 3: Find the area bounded by the curve $\small y=\sin x$ between $\small x=0$ and $\small x=2\pi$.

Answer:

The graph of y=sinx is as follows

We need to find the area of the shaded region.

ar(OAB)+ar(BCD)

=2ar(OAB)

$\\=2\times \int_{0}^{\pi }sinxdx\\ =2\times [-cosx]_{0}^{\pi }\\ =2\times [-(-1)-(-1)]\\ =4$

The bounded area is 4 units.

Question 4: Choose the correct answer.

Area bounded by the curve $\small y=x^3$ , the $\small x$ -axis and the ordinates $\small x=-2$ and $\small x=1$ is

(A) $\small -9$ (B) $\small \frac{-15}{4}$ (C) $\small \frac{15}{4}$ (D) $\small \frac{17}{4}$

Answer:

Hence, the required area

$=\int_{-2}^1 ydx$

$=\int_{-2}^1 x^3dx = \left [ \frac{x^4}{4} \right ]_{-2}^1$

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

$= \left [ 0-\frac{(-2)^4}{4} \right ] + \left [ \frac{1}{4} - 0 \right ]$

$= -4+\frac{1}{4} = \frac{-15}{4}$

Therefore, the correct answer is B.

Question 5: Choose the correct answer.

The area bounded by the curve $\small y=x|x|$ , $\small x$ -axis and the ordinates $\small x=-1$ and $\small x=1$ is given by

(A) $\small 0$ (B) $\small \frac{1}{3}$ (C) $\small \frac{2}{3}$ (D) $\small \frac{4}{3}$

[ Hint : $y=x^2$ if $x> 0$ and $y=-x^2$ if $x<0$]

Answer:

The required area is

$\\2\int_{0}^{1}x^{2}dx\\ =2\left [ \frac{x^{3}}{3} \right ]_{0}^{1}\\ =\frac{2}{3}\ units$

Application of Integrals Class 12 NCERT Solutions: Exercise-wise

Exercise-wise NCERT Solutions of Application of Integrals Class 12 Maths Chapter 8 are provided in the links below.

• Applications Of Integrals Class 12 Exercise 8.1
• Application Of Integrals Class 12 Miscellaneous Exercise

Application Of Integrals Class 12 Chapter 8: Topics

Topics you will learn in NCERT Class 12 Maths Chapter 8 Application of Integrals include:

• 8.1 Introduction
• 8.2 Area under Simpler Curves

Application Of Integrals Class 12 NCERT Solutions - Important Formulae

1. Area Enclosed by a Curve and Lines:

The area enclosed by the curve $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$ (where $b>a$ ) is given by the formula:

Area $=\int_a^b y d x=\int_a^b f(x) d x$

2. Area Bounded by Curve and Horizontal Lines:

The area of the region bounded by the curve $x=\phi(y)$ as its $y$-axis and the lines $y=c$ and $y=d$ is given by the formula:

Area $=\int_c^d x d y=\int_c^d \phi(y) d y$

3. Area Between Two Curves and Vertical Lines:

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

Area $=\int_a^b[f(x)-g(x)] d x \quad($ Where $f(x) \geq g(x)$ in $[a, b])$

4. Area Between Curves with Different Intervals:

If $f(x) \geq g(x)$ in $[a, c]$ and $f(x) \leq g(x)$ in $[c, b]$, where $a<c<b$, then the resultant area between the curves is given as:

Area $=\int_a^c[f(x)-g(x)] d x+\int_c^b[g(x)-f(x)] d x$

Why are Class 12 Maths Chapter 8 Application of Integrals question answers important?

This Application of Integrals chapter helps us learn how to find the area under curves using integration. It connects what we learned in the previous chapter to real-life applications. These Class 12 Maths chapter 8 Application of Integrals question answers help us practise using integration to calculate areas between lines and curves. Here are some more points on why these question answers are important.

• These solutions teach you how to find areas of curved shapes that can’t be measured using normal geometry.
• Students understand how integration is applied in solving practical problems in maths and science.
• Dealing with Class 12 Maths chapter 8 Application of Integrals question answers helps us in higher studies, especially in calculus, physics, and engineering.

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

This chapter deals with the application of definite integrals, in particular, to calculate the area of the region bounded by the given curves and straight lines. Different methods of calculating the areas using integration are covered, along with the graphical interpretation of the concepts to make the problems easier to comprehend. The NCERT Solutions facilitate explaining each question in detail with stepwise explanations. Students have the answers to around 9 textbook questions across 2 exercises, which enhance conceptual knowledge and application skills. Regular practice aids in improving accuracy, reasoning ability, and conceptual clarity for examination performance. An article on advanced Calculus applications is covered.

Expert Review of NCERT Solutions for Class 12 Maths Chapter 8 - Application of Integrals.

Mathematics experts at Careers360 deem Application of Integrals as an interesting chapter because students learn how to apply the principles of integration to solve real-life problems on curves and graphs. Students who master the graphical interpretation of definite integrals can enjoy solving application-based questions. NCERT Solutions make every concept simplified with detailed and organized solutions. Experts always advise to perfectly practice the diagrams and every NCERT question to increase speed and accuracy. Adequate, methodical preparation of this chapter can help a student to excel in CBSE Board exam and Competitive exams like JEE Main And JEE Advanced.

What Extra Should Students Study Beyond the NCERT for JEE?

Here is a comparison list of the concepts in Application of Integrals that are covered in JEE and NCERT, to help students understand what extra they need to study beyond the NCERT for JEE:

NCERT Solutions for Class 12 Maths - Chapter-wise

Given below is the chapter-wise list of the NCERT Class 12 Maths solutions with their respective links:

Also read,

• NCERT Exemplar Class 12 Chemistry Solutions
• NCERT Exemplar Class 12 Mathematics Solutions
• NCERT Exemplar Class 12 Biology Solutions
• NCERT Exemplar Class 12 Physics Solutions

NCERT solutions for class 12 subject-wise

Students can check the following links for more in-depth learning.

• NCERT Solutions class 12 chemistry
• NCERT solutions for class 12 physics
• NCERT solutions for class 12 biology

NCERT Solutions Class Wise

Students can check the following links for more in-depth learning.

• NCERT solutions for class 11
• NCERT solutions for class 10
• NCERT solutions for class 9

NCERT Books and NCERT Syllabus

Students can check the following links for more in-depth learning.

• NCERT Books Class 12 Maths
• NCERT Syllabus Class 12 Maths
• NCERT Books Class 12
• NCERT Syllabus Class 12

Also, read,

• NCERT Notes Class 12 Maths Chapter 8 Application of Integrals
• NCERT Exemplar Class 12 Maths Chapter 8 Solutions Application of Integrals