NCERT Solutions for Miscellaneous Exercise Chapter 8 Class 12 - Application of Integrals
Integrals are an inseparable part of calculus, which can solve real-world problems related to areas and volumes by summing up infinitely many small pieces to make a whole. The application of integrals delves into the aspect of how integrals can be used to solve problems related to real-life scenarios. The miscellaneous exercise of the chapter, Application of Integrals, combines all the key concepts covered in the chapter, so that the students can enhance their understanding by a comprehensive review of the entire chapter and get better at problem-solving. This article on the NCERT Solutions for Miscellaneous Exercise of Class 12, Chapter 8 - Application of Integrals, offers detailed and easy-to-understand solutions for the exercise problems, so that students can strengthen their understanding of the application of integrals. For syllabus, notes, exemplar solutions and PDF, refer to this link: NCERT.
This Story also Contains
- Application of Integrals Class 12 Chapter 8 Miscellaneous: Exercise
- Topics covered in Chapter 8, Application of Integrals: Miscellaneous Exercise
- NCERT Solutions Subject Wise
- NCERT Exemplar Solutions Subject Wise
Application of Integrals Class 12 Chapter 8 Miscellaneous: Exercise
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 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)
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.
T he 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
$\\2\int_{0}^{1}x^{2}dx\\ =2\left [ \frac{x^{3}}{3} \right ]_{0}^{1}\\ =\frac{2}{3}\ units$
Also Read,
Topics covered in Chapter 8, Application of Integrals: Miscellaneous Exercise
The main topics covered in class 12 maths chapter 8 of Application of Integrals, Miscellaneous Exercise are:
- Area under curves: In this topic, we will calculate the area between a curve and the coordinate axes in a specific interval. For example, the area under the curve $y=f(x)$, between two points on the X axis, as $x=a$ and $x=b$, can be found using definite integrals as: $A=\int_a^b f(x) d x$.
- Area between two curves: This topic deals with the area between two curves. Let $f(x)$ and $g(x)$ be two curves in the interval $[a,b]$, then the area can be found using the formula, Area $=\int_a^b[f(x)-g(x)] d x$.
Also Read,
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Frequently Asked Questions (FAQs)
Q: How many questions are there in Miscellaneous exercise Chapter 8 ?
A:
There are 19 questions total in Miscellaneous exercise Chapter 8.
Q: Can we find an area without using integrals ?
A:
Simple figures like triangle, circle etc. can be tackled without integration but not the complex ones.
Q: Are questions repeated in the examination from this Chapter ?
A:
Yes, in the Board exam the questions are repeated every year.
Q: What is the level of questions asked from this Chapter ?
A:
Moderate level questions are asked from this Chapter.
Q: Can one skip Miscellaneous exercise ?
A:
No, as it has some good questions, miscellaneous exercise must be done.
Q: What is the time it will take to complete for the first time ?
A:
It will take around 5-6 hours to complete for the first time.
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