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NCERT Solutions for Class 12 Maths Chapter 8 Application of integrals
Imagine you have a curved shape like the edge of a hill or a wave, and you want to find out how much space it takes up between the curve and the ground (x-axis). That space under the curve is called the area, and to calculate this area, we use integration. Integrals are the functions that satisfy a given differential equation for finding the area of a curvy region y = f (x), the x-axis, and the lines x = a and x = b (b>a). Therefore, finding the integral of a function with respect to x means finding the area above the x-axis from the curve. The integral is also called an anti-derivative, as it is the reverse process of differentiation. The main objective of these NCERT Solutions for class 12, the application of integrals, is to provide students with a clear understanding of how integration is used to calculate areas bounded by curves and to strengthen their problem-solving skills in real-life applications.
This Story also Contains
- Application Of Integrals Class 12 Maths Chapter 8 - PDF Free Download
- NCERT Application Of Integrals Class 12 Questions And Answers (Exercise)
- Application Of Integrals Class 12 Chapter 8: Topics
- Application Of Integrals Class 12 NCERT Solutions - Important Formulae
- What Extra Should Students Study Beyond the NCERT for JEE?
- NCERT solutions for class 12 maths - Chapter-wise
Integration teaches us how small changes add up to make a measurable difference. These NCERT Solutions for Class 12 are trustworthy and reliable, as they are created by subject matter experts at Careers360, making them an essential resource for exam preparation. NCERT Solutions are trusted by teachers for building a strong foundation in concepts. Students can additionally refer to the NCERT exemplar and notes for better practice and understanding of the topic.
Application Of Integrals Class 12 Maths Chapter 8 - PDF Free Download
The NCERT Solutions for Class 12 Maths Chapter 8 have been prepared by Careers360 experts to make learning simpler and to help you score better in exams. You can also download the solutions in PDF format.
NCERT Application Of Integrals Class 12 Questions And Answers (Exercise)
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Class 12 Maths Chapter 8 Solutions Exercise: 8.1 Page number: 296 Total questions: 4 |
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
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.
(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 =
$\\\int ^3_0xdy\\ =\int ^3_0\frac{y^2}{4}dy\\ =\frac{1}{4}.[\frac{y^3}{3}]^3_0\\ =\frac{9}{4}$
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NCERT Application of Integrals Class 12 Solutions: Exercise: Miscellaneous Exercise Page Number: 298 Total Questions: 5 |
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.
(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.
(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
$\\2\int_{0}^{1}x^{2}dx\\ =2\left [ \frac{x^{3}}{3} \right ]_{0}^{1}\\ =\frac{2}{3}\ units$
Also, read,
Application Of Integrals Class 12 Chapter 8: Topics
Here is the list of important topics that are covered in Class 12 Chapter 8 Application of Integrals:
- 8.1 Introduction
- 8.2 Area under Simpler Curves
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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 12
- 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.
Frequently Asked Questions (FAQs)
Q: What are the formulas used in Application of Integrals Class 12
A:
In Class 12 Chapter 8 - Application of Integrals, the key formulas include:
1. Area under a curve:
$
\text { Area }=\int_a^b y d x=\int_a^b f(x) d x
$
2. Area between two curves:
$ \text{Area} = \int_a^b [f(x) - g(x)] ,dx, \quad \text{where } f(x) \geq g(x) \text{ in } [a, b]. $
3. For vertical strips (in terms of $y$ ):
$\text { Area }=\int_c^d[f(y)-g(y)] d y$
Q: What are the important topics covered in NCERT Solutions for Class 12 Maths Chapter 8?
A:
NCERT Solutions for Class 12 Maths Chapter 8 – Application of Integrals – covers key topics like calculating the area under curves, the area between two curves, and the area bounded by lines and curves. It focuses on using definite integrals to find areas in both standard and complex geometrical situations. The chapter also includes a graphical representation of functions and the use of integration in real-life applications.
Q: How to find the area under curves using integrals in Class 12 Maths?
A:
To find the area under curves using integrals in Class 12 Maths, we use definite integrals. If a curve is defined by $y=f(x)$ between $x=a$ and $x=b$, the area under the curve is:
$\text { Area }=\int_a^b f(x) d x$
This formula calculates the total area between the curve and the x-axis from $x=a$ to $x=b$. If the curve lies below the $x$-axis, take the absolute value of the integral to get a positive area.
Q: How to solve area between two curves problems in NCERT Class 12 Maths?
A:
To solve area between two curves problems in NCERT Class 12 Maths, identify the two functions: y = f(x) and y = g(x), where f(x) >= g(x) in the interval [a, b]. The area between the curves from x = a to x = b is:
$\text { Area }=\int_a^b[f(x)-g(x)] d x$
This gives the vertical distance between the curves integrated over the interval. Sketching the curves helps visualize the region. Always check which function is on top within the limits.
Q: Are NCERT Solutions enough for Class 12 Maths Chapter 8 board exams?
A:
Yes, NCERT Solutions are usually enough for Class 12 Maths Chapter 8 – Application of Integrals for the board exams. The NCERT book covers all important concepts, formulas, and types of questions likely to appear in exams. It provides step-by-step solutions that help build a strong foundation. However, for better practice and confidence, solving additional problems from sample papers, previous years' questions, and reference books like RD Sharma can help master the topic thoroughly.
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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
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A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
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