NCERT Solutions for Class 12 Maths Chapter 8 Application of integrals

Application Of Integrals Class 12 Maths Chapter 8 - PDF Free Download

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NCERT Application Of Integrals Class 12 Questions And Answers (Exercise)

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

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