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# NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 8 - Application of Integrals

Last updated date: 10th Aug 2024
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## NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise - Free PDF Download

NCERT Solutions for Maths Chapter 8 Application of Integrals Class 12 Miscellaneous Exercise includes solutions to all Miscellaneous Exercise problems. Application of Integrals Class 12 NCERT Solutions Miscellaneous Exercises are based on the concepts presented in Maths Chapter 8. To perform well on the board exam, download the NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise Solutions in PDF format and practice them offline. Students can download the revised Class 12 Maths NCERT Solutions from our page, which is prepared so that you can understand it easily.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise - Free PDF Download
2. Access NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals
3. Class 12 Maths Chapter 8: Exercises Breakdown
4. CBSE Class 12 Maths Chapter 8 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

Class 12 Chapter 8 Maths Miscellaneous Exercise Solutions are aligned with the updated CBSE guidelines for Class 12. Access the Class 12 Maths Syllabus here.

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals

Miscellaneous Exercise

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

(i) $\text{y=}{{\text{x}}^{\text{2}}}\text{,x=1,x=2}$ and $\text{x-axis}$

Ans:

$\text{AreaADCBA=}\int_{\text{1}}^{\text{2}}{\text{y}}\text{dx}$

substitute $y={{x}^{2}}$

$\text{=}\int_{\text{1}}^{\text{2}}{{{\text{x}}^{\text{2}}}}\text{dx}$

Substituting the limits,

$\text{=}\frac{\text{8}}{\text{3}}\text{-}\frac{\text{1}}{\text{3}}$

$\text{=}\frac{\text{7}}{\text{3}}\text{units}$

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

Ans:

$\text{AreaofADCBA=}\int_{\text{1}}^{\text{5}}{{{\text{x}}^{\text{4}}}}\text{dx}$

Integrating using the power rule,

$\text{=}\left[ \frac{{{\text{x}}^{\text{5}}}}{\text{5}} \right]_{\text{1}}^{\text{5}}$

Substituting the limits,

$\text{=}\frac{{{\text{(5)}}^{\text{5}}}}{\text{5}}\text{-}\frac{\text{1}}{\text{5}}$

Simplifying,

$\text{=(5}{{\text{)}}^{\text{4}}}\text{-}\frac{\text{1}}{\text{5}}$

$\text{=625-}\frac{\text{1}}{\text{5}}$

$\text{=624}\text{.8 units}$

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

Ans:

 $\text{x}$ $\text{-6}$ $\text{-5}$ $\text{-4}$ $\text{-3}$ $\text{-2}$ $\text{-1}$ $\text{0}$ $\text{y}$ $\text{3}$ $\text{2}$ $\text{1}$ $\text{0}$ $\text{1}$ $\text{2}$ $\text{3}$

$\left( \text{x+3} \right)\le 0$ for $\text{-6}\le \text{x}\le \text{-3}$

$\left( \text{x+3} \right)\ge \text{0}$ for $\text{-3}\le \text{x}\le 0$

Therefore,

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

Integrating using the power rule

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

Substituting the limits,

$\text{=-}\left[ \left( \frac{{{\text{(-3)}}^{\text{2}}}}{\text{2}}\text{+3(-3)} \right)\text{-}\left( \frac{{{\text{(-6)}}^{\text{2}}}}{\text{2}}\text{+3(-6)} \right) \right]\text{+}\left[ \text{0-}\left( \frac{{{\text{(-3)}}^{\text{2}}}}{\text{2}}\text{+3(-3)} \right) \right]$

Simplifying,

$\text{= -}\left[ \text{-}\frac{\text{9}}{\text{2}} \right]\text{-}\left[ \text{-}\frac{\text{9}}{\text{2}} \right]$

$\text{=9}$

3. Find the area bounded by the curve $\text{y=sinx}$ between $\text{x=0}$ and $\text{x=2 }\!\!\pi\!\!\text{ }$.

Ans:

Therefore, $\text{area = Area OABO+ Area BCDB}$

Area Bounded by Curve y=sinx

$\text{=}\int_{\text{0}}^{\text{ }\!\!\pi\!\!\text{ }}{\text{sin}}\text{xdx+}\left| \int_{\text{ }\!\!\pi\!\!\text{ }}^{\text{2 }\!\!\pi\!\!\text{ }}{\text{sin}}\text{xdx} \right|$

$\text{= }\!\![\!\!\text{ -cosx }\!\!]\!\!\text{ }_{\text{0}}^{\text{ }\!\!\pi\!\!\text{ }}\text{+}\left| \text{ }\!\![\!\!\text{ -cosx }\!\!]\!\!\text{ }_{\text{ }\!\!\pi\!\!\text{ }}^{\text{2 }\!\!\pi\!\!\text{ }} \right|$

Substituting the limits,

$\text{= }\!\![\!\!\text{ -cos }\!\!\pi\!\!\text{ +cos0 }\!\!]\!\!\text{ + }\!\!|\!\!\text{ -cos2 }\!\!\pi\!\!\text{ +cos }\!\!\pi\!\!\text{ }\!\!|\!\!\text{ }$

Simplifying,

$\text{=1+1+ }\!\!|\!\!\text{ (-1-1) }\!\!|\!\!\text{ }$

$\text{=2+ }\!\!|\!\!\text{ -2 }\!\!|\!\!\text{ }$

$\text{=2+2}$

$\text{=4 units}$

4. Area bounded by the curve $\text{y=}{{\text{x}}^{3}}$, the $\text{x-axis}$ and the ordinates $\text{x = -2}$ and $\text{x = 1}$ is

A. $\text{-9}$

B. $\text{-}\frac{\text{15}}{\text{4}}$

C. $\frac{\text{15}}{\text{4}}$

D. $\frac{\text{17}}{\text{4}}$

Ans:

As shown in the diagram, the required area is:

$\text{Required area =}\int_{\text{-2}}^{\text{1}}{\text{y}}\text{dx}$

$\text{=}\int_{\text{-2}}^{\text{1}}{{{\text{x}}^{\text{3}}}}\text{dx}$

Integrating using the power rule

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

Substituting the limits,

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

Simplifying,

$\text{=}\left( \frac{\text{1}}{\text{4}}\text{-4} \right)$

$\text{= -}\frac{\text{15}}{\text{4}}\text{ units}$

So, the correct answer is $\text{ -}\frac{\text{15}}{\text{4}}\text{ units}$ option B.

5. The area bounded by the curve $\text{y=x}\left| \text{x} \right|\text{,x-axis}$ and the ordinate $\text{x = 1}$ and $\text{x = -1}$ is given by (Hint $\text{y = }{{\text{x}}^{2}}$ if $x>0$ and $\text{y = -}{{\text{x}}^{2}}$ if $x<0$)

A. $0$

B. $\frac{\text{1}}{\text{3}}$

C. $\frac{2}{\text{3}}$

D. $\frac{4}{\text{3}}$

Ans:

$\text{Required area=}\int_{\text{-1}}^{\text{1}}{\text{y}}\text{dx}$

$\text{=}\int_{\text{-1}}^{\text{1}}{\text{x}}\text{ }\!\!|\!\!\text{ x }\!\!|\!\!\text{ dx}$

$\text{= -}\int_{\text{-1}}^{\text{0}}{{{\text{x}}^{\text{2}}}}\text{dx+}\int_{\text{0}}^{\text{1}}{{{\text{x}}^{\text{2}}}}\text{dx}$

Integrating using the power rule

$\text{= -}\left[ \frac{{{\text{x}}^{\text{3}}}}{\text{3}} \right]_{\text{-1}}^{\text{0}}\text{+}\left[ \frac{{{\text{x}}^{\text{3}}}}{\text{3}} \right]_{\text{0}}^{\text{1}}$

Substituting the limits,

$\text{= -}\left( \text{-}\frac{\text{1}}{\text{3}} \right)\text{+}\frac{\text{1}}{\text{3}}$

$\text{=}\frac{\text{2}}{\text{3}}\text{units}$

So, the correct answer is $\frac{\text{2}}{\text{3}}\text{units}$ option C.

## Conclusion

The Class 12 Maths Chapter 8 Miscellaneous Exercise Solutions is important for understanding various concepts thoroughly. Application of Integrals Class 12 Miscellaneous includes a variety of issues that call for the use of several formulas and methods. It's crucial to concentrate on comprehending the fundamental ideas behind every question, as opposed to merely learning the answers by heart. To successfully complete this task, keep in mind that you must comprehend the theory underlying each idea, practise frequently, and consult solved examples.

## Class 12 Maths Chapter 8: Exercises Breakdown

 Exercise Number of Questions Exercise 8.1 4 Questions and Solutions

## CBSE Class 12 Maths Chapter 8 Other Study Materials

 S. No Important Links for Chapter 8 Application of Integrals 1 Class 12 Application of Integrals Revision Notes 2 Class 12 Application of Integrals Important Questions 3 Class 12 Application of Integrals Formula 4 Class 12 Application of Integrals NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 8 - Application of Integrals

1. What kind of problems are covered in the Application of Integrals Class 12 NCERT Solutions Miscellaneous?

The Miscellaneous Chapter 8 Class 12 deals with various applications of definite integrals, including:

• Finding areas and volumes enclosed by curves (using definite integrals to calculate areas under curves, volumes of revolution, etc.)

• Finding lengths of curves (applying definite integrals to calculate the arc length of a curve)

• Finding moments and centres of mass (using definite integrals to calculate moments of inertia and centres of mass for various shapes)

• Solving problems involving work done by a variable force (applying definite integrals to calculate the work done by a force whose magnitude varies)

2. How do the NCERT solutions approach these problems?

The NCERT solutions typically:

• Briefly remind you of the relevant concepts from definite integrals and their applications.

• Guide you through the process of setting up the definite integral expression for the given problem scenario. This might involve identifying the region of integration, defining the integration, and setting up the limits of integration.

• Demonstrate how to evaluate the definite integral using appropriate techniques (numerical integration methods might be used in some cases).

3. Are there any tips for using the NCERT solutions effectively?

Here are some tips:

• Grasp the fundamentals: Ensure you have a clear understanding of definite integrals, their geometric interpretation, and how they relate to areas, volumes, etc.

• Practice with different applications: Solve problems involving area, volume, arc length, moments, and work done by a variable force to gain experience applying definite integrals in various contexts.

• Focus on the setup: When referring to the NCERT solutions, pay close attention to how the definite integral expression is set up for the specific problem. Understanding this setup is important for solving similar problems independently.

4. Where can I find additional resources for practising the Application of Integrals Class 12 NCERT Solutions Miscellaneous?

• The NCERT textbook itself might provide solutions to some problems within the miscellaneous exercise section.

• Vedantu offers comprehensive solutions and explanations for these problems. You can find them through a web search using terms like "NCERT Solutions Class 12 Maths Chapter 8 Miscellaneous Exercise Applications of Integrals."

5. Application of integral sample questions?

While I cannot provide specific solutions due to copyright, here are 2 sample questions to illustrate the types of problems you might encounter:

1. Find the area of the region enclosed by the curve y = x^2 and the lines x = 1 and x = 2. (This question requires setting up a definite integral to represent the area and then evaluating it)

2. Find the volume of the solid generated by revolving the region enclosed by the curve y = sin x, the x-axis, and the lines x = 0 and x = π/2 about the x-axis. (This question involves using the disc method with a definite integral to calculate the volume)

6. Are NCERT Solutions sufficient for Class 12 Maths board exams?

NCERT Solutions covers the entire CBSE Class 12 Maths syllabus comprehensively. However, it's advisable to supplement them with additional practice from reference books and previous years' question papers for thorough preparation.

7. Are NCERT Solutions useful for competitive exams like JEE and NEET?

Yes, NCERT Solutions provides a strong foundation in mathematical concepts, which is beneficial for competitive exams like JEE Main, JEE Advanced, and NEET. They cover essential topics that are part of these exams' syllabi.

8. What are the benefits of practising the Miscellaneous Exercise for exams?

Practising the Miscellaneous Chapter 8 Class 12 enhances problem-solving skills, reinforces understanding of integral applications, and prepares students for exam questions that require application-based knowledge.

9. What topics are covered in the Class 12 Maths Chapter 8 Miscellaneous Exercise Solutions?

The Miscellaneous Exercise covers various applications of definite integrals, such as finding areas between curves, volumes of solids of revolution, and calculating lengths of curves.