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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.
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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,
The main topics covered in class 12 maths chapter 8 of Application of Integrals, Miscellaneous Exercise are:
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Also Read,
Below are some useful links for subject-wise NCERT solutions for class 12.
Here are some links to subject-wise solutions for the NCERT exemplar class 12.
Frequently Asked Questions (FAQs)
There are 19 questions total in Miscellaneous exercise Chapter 8.
Simple figures like triangle, circle etc. can be tackled without integration but not the complex ones.
Yes, in the Board exam the questions are repeated every year.
Moderate level questions are asked from this Chapter.
No, as it has some good questions, miscellaneous exercise must be done.
It will take around 5-6 hours to complete for the first time.
On Question asked by student community
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Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.
From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .
If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.
The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.
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