NCERT Solutions for Maths Chapter 8 Application of Integrals Class 12 - FREE PDF Download
NCERT Solutions for CBSE Maths Chapter 8 Application of Integrals Class 12 are available in Vedantu. These NCERT Solutions are created as per the latest syllabus of NCERT Maths for Class 12. This PDF covers solutions for all questions that are covered in the CBSE Class 12 Maths textbook Chapter 8. All the solutions are explained in a step by step manner. Students can refer to these solutions to learn the important questions and prepare for their board exams. The NCERT Solutions for CBSE Chapter 8 Class 12 Maths are available in a PDF format and can be downloaded for free.
Table of Content
Glance on Maths Chapter 8 Class 12 - Application of Integrals
Application of Integrals Class 12 NCERT Solutions deals with finding the area between curves, such as between a curve and the x-axis, or between two curves using definite integrals. It includes step-by-step solutions for different types of curves.
Problems in this Chapter focus on calculating the area enclosed by two given curves. Solutions illustrate the method of setting up integrals with appropriate limits and finding the difference between the two areas.
Comprehensive solutions to all exercise problems in the NCERT textbook are given. This includes detailed explanations, diagrams, and the rationale behind each step to ensure thorough understanding.
There is one exercise and a miscellaneous exercise (9 fully solved questions) in Class 12th maths chapter 8 Application of Integrals.
This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 8 - Application of Integrals, which you can download as PDFs.
Access Exercise Wise NCERT Solutions for Chapter 8 Maths Class 12
Exercise 8.1: This exercise teaches students how to find the area of a region bounded by curves using integration.
Miscellaneous Exercise: This exercise contains a collection of problems that require students to apply the concepts they have learned in the previous exercises to solve real-world problems.
Access NCERT Solutions for Class 12 Maths Chapter 8 – Application of Integrals
Exercise 8.1
1. Find the area of the region bounded by the ellipse \[\frac{{{\text{x}}^{\text{2}}}}{\text{16}}\text{+}\frac{{{\text{y}}^{\text{2}}}}{\text{9}}\text{=1}\].
Ans:
Area Bounded by Ellipse
Area of ellipse \[\text{=4 }\!\!\times\!\!\text{ }\] Area of \[\text{OAB}\]
Area of \[\text{OAB=}\int_{\text{0}}^{\text{4}}{\text{y}}\text{dx}\]
Substitute $y=3\sqrt{1-\frac{{{x}^{2}}}{16}}$
\[\text{=}\int_{\text{0}}^{\text{4}}{\text{3}}\sqrt{\text{1-}\frac{{{\text{x}}^{\text{2}}}}{\text{16}}}\text{dx}\]
simplifying,
\[\text{=}\frac{\text{3}}{\text{4}}\int_{\text{0}}^{\text{4}}{\sqrt{\text{16-}{{\text{x}}^{\text{2}}}}}\text{dx}\]
\[\text{=}\frac{\text{3}}{\text{4}}\left[ \frac{\text{x}}{\text{2}}\sqrt{\text{16-}{{\text{x}}^{\text{2}}}}\text{+}\frac{\text{16}}{\text{2}}\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{x}}{\text{4}} \right]_{\text{0}}^{\text{4}}\]
Substituting the limits,
\[\text{=}\frac{\text{3}}{\text{4}}\left[ \text{2}\sqrt{\text{16-16}}\text{+8si}{{\text{n}}^{\text{-1}}}\text{(1)-0-8si}{{\text{n}}^{\text{-1}}}\text{(0)} \right]\]
\[\text{= }\frac{\text{3}}{\text{4}}\left[ \frac{\text{8 }\!\!\pi\!\!\text{ }}{\text{2}} \right]\]
\[\text{= }\frac{\text{3}}{\text{4}}\text{ }\!\![\!\!\text{ 4 }\!\!\pi\!\!\text{ }\!\!]\!\!\text{ }\]
\[\text{= 3 }\!\!\pi\!\!\text{ }\]
As a result, the ellipse's area is \[\text{= 4}\times \text{3 }\!\!\pi\!\!\text{ = 12 }\!\!\pi\!\!\text{ }\].
2. Find the area of the region bounded by the ellipse \[\frac{{{\text{x}}^{\text{2}}}}{4}\text{+}\frac{{{\text{y}}^{\text{2}}}}{\text{9}}\text{=1}\].
Ans:
Area Bounded by Ellipse
Area of ellipse \[\text{=4 }\!\!\times\!\!\text{ }\] Area of \[\text{OAB}\]
Area of \[\text{OAB=}\int_{\text{0}}^{2}{\text{y}}\text{dx}\]
Substitute $y=3\sqrt{1-\frac{{{x}^{2}}}{4}}$
\[\text{=}\int_{\text{0}}^{2}{\text{3}}\sqrt{\text{1-}\frac{{{\text{x}}^{\text{2}}}}{4}}\text{dx}\]
simplifying,
\[\text{=}\frac{\text{3}}{2}\int_{\text{0}}^{2}{\sqrt{\text{4-}{{\text{x}}^{\text{2}}}}}\text{dx}\]
\[\text{=}\frac{\text{3}}{2}\left[ \frac{\text{x}}{\text{2}}\sqrt{\text{4-}{{\text{x}}^{\text{2}}}}\text{+}\frac{4}{\text{2}}\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{x}}{2} \right]_{\text{0}}^{2}\]
Substituting the limits,
\[\text{=}\frac{\text{3}}{2}\left[ \frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{2}} \right]\]
\[\text{=}\frac{\text{3 }\!\!\pi\!\!\text{ }}{2}\]
As a result, the ellipse's area is \[\text{4}\times \frac{\text{3 }\!\!\pi\!\!\text{ }}{2}\text{=6 }\!\!\pi\!\!\text{ }\].
3. Area lying in the first quadrant and bounded by the circle \[{{\text{x}}^{\text{2}}}\text{ + }{{\text{y}}^{\text{2}}}\text{ = 4}\] and the lines \[\text{x=0}\]
A. \[\text{ }\!\!\pi\!\!\text{ }\]
B. \[\frac{\text{ }\!\!\pi\!\!\text{ }}{2}\]
C. \[\frac{\text{ }\!\!\pi\!\!\text{ }}{3}\]
D. \[\frac{\text{ }\!\!\pi\!\!\text{ }}{4}\]
Ans:
Area Lying in the First Quadrant and Bounded by the Circle
\[\text{Area OAB=}\int_{0}^{\text{2}}{\text{y}}\text{dx}\]
substitute$y=\sqrt{4-{{x}^{2}}}$
\[\text{=}\int_{\text{0}}^{\text{2}}{\sqrt{\text{4-}{{\text{x}}^{\text{2}}}}}\text{dx}\]1
\[\text{=}\left[ \frac{\text{x}}{\text{2}}\sqrt{\text{4-}{{\text{x}}^{\text{2}}}}\text{+}\frac{\text{4}}{\text{2}}\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{x}}{\text{2}} \right]_{\text{0}}^{\text{2}}\]
Substituting the limits,
\[\text{=2}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)\]
\[\text{= }\!\!\pi\!\!\text{ }\]
As a result, the required area is \[\text{ }\!\!\pi\!\!\text{ }\] option A.
4. Area of the region bounded by the curve \[{{\text{y}}^{\text{2}}}\text{ = 4x, y-axis}\] and the line \[\text{y=3}\] is
A. \[2\]
B. \[\frac{9}{4}\]
C. \[\frac{9}{3}\]
D. \[\frac{9}{2}\]
Ans:
\[\text{Area OAB=}\int_{0}^{3}{x}\text{dy}\]
substitute $x=\frac{{{y}^{2}}}{4}$
\[\text{=}\int_{\text{0}}^{3}{\frac{{{y}^{2}}}{4}}\text{dy}\]
\[\text{=}\frac{1}{4}\left[ \frac{{{y}^{3}}}{4} \right]_{\text{0}}^{3}\]
Substituting the limits,
\[\text{=}\frac{1}{\text{12}}\left( 27 \right)\]
\[\text{=}\frac{\text{9}}{\text{4}}\text{ units}\]
As a result, the correct response is \[\frac{\text{9}}{\text{4}}\text{ units}\] option B.
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:
\[\left( \text{x+3} \right)\le 0\] 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.
Overview of Deleted Syllabus for CBSE Class 12 Maths Application of Integrals
Class 12 Maths Chapter 8: Exercises Breakdown
Conclusion
Chapter 8, "Application of Integrals," is a crucial part of the Class 12 Maths curriculum. This chapter primarily focuses on using integrals to find the area under simple curves, including lines, circles, parabolas, and ellipses. Students also learn to find the area under the simple curve, which is a vital skill for various mathematical and engineering applications. Over the past few years, Chapter 8 has consistently featured in the exams with an average of 4-6 questions. These questions often include a mix of straightforward calculations and application-based problems that test the depth of understanding of integral applications.
Other Study Material for CBSE Class 12 Maths Chapter 8
NCERT Solutions for Class 12 Maths | Chapter-wise List
Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.
Related Links for NCERT Class 12 Maths in Hindi
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FAQs on NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals
1. What is the standard step-by-step method to solve problems in NCERT Solutions for Class 12 Maths Chapter 8, Application of Integrals?
To solve problems in this chapter as per the NCERT methodology, you should follow these steps:
- Sketch the Curve: Draw a rough sketch of the given curve(s) and lines to identify the required bounded region.
- Identify the Region: Clearly shade the area that needs to be calculated.
- Set Up the Integral: Determine whether to integrate with respect to the x-axis (using y dx) or the y-axis (using x dy).
- Determine Limits: Find the correct limits of integration (the starting and ending points) from the graph or the given equations.
- Integrate: Evaluate the definite integral to find the numerical value of the area.
2. How many exercises are in Chapter 8, Application of Integrals, according to the latest CBSE 2025-26 syllabus?
In the NCERT textbook for Class 12 Maths, Chapter 8 (Application of Integrals) contains one primary exercise (Exercise 8.1) and a Miscellaneous Exercise. Both focus on finding the area under simple curves as per the updated syllabus.
3. How do you correctly set up the definite integral to find the area bounded by a curve y = f(x), the x-axis, and the lines x = a and x = b?
The correct method is to set up the definite integral as Area = ∫ₐᵇ y dx. Here, 'y' is the function f(x), and 'a' and 'b' are the lower and upper limits (x-coordinates) of the region. This formula calculates the sum of the areas of infinite vertical strips within these boundaries.
4. When solving NCERT questions, when is it better to integrate with respect to the y-axis (using ∫ x dy)?
It is better to integrate with respect to the y-axis when the curve is expressed as x = f(y) and the region is bounded by the y-axis and horizontal lines (e.g., y = c and y = d). In such cases, the formula for the area becomes Area = ∫꜀ᵈ x dy, which simplifies the calculation.
5. How are the solutions for finding the area of a standard circle (x² + y² = r²) and an ellipse (x²/a² + y²/b² = 1) structured in this chapter?
The solutions for both are structured using the concept of symmetry. The standard approach involves:
- Calculating the area in the first quadrant only.
- Setting the limits of integration from 0 to the x-intercept (e.g., 0 to 'r' for a circle, 0 to 'a' for an ellipse).
- Expressing 'y' in terms of 'x' for the first quadrant.
- Evaluating the definite integral.
- Multiplying the result by 4 to get the total area of the shape.
6. Why is sketching the curve considered a mandatory first step for solving problems on Application of Integrals?
Sketching the curve is a critical first step because it provides essential visual information. It helps to:
- Accurately identify the required bounded region.
- Determine the correct upper and lower limits for integration.
- Recognise if the area lies above or below the x-axis, which is crucial for avoiding sign errors in the final area calculation.
- Identify any symmetry that can be used to simplify the integral.
7. What is a common mistake made when finding the area for a curve that lies both above and below the x-axis, like y = sin(x) from 0 to 2π?
A common mistake is to integrate the function across the entire interval in a single step. This calculates the net displacement, where the area below the axis cancels out the area above. The correct method is to split the integral at the points where the curve crosses the x-axis. You must calculate the area of the portion below the axis, take its absolute value (magnitude), and then add it to the area of the portion above the axis to find the total area.
8. How does the concept of symmetry help simplify calculations in the NCERT Solutions for this chapter?
Symmetry is a powerful tool for simplifying problems. For curves symmetric about the x-axis, y-axis, or both (like circles, ellipses, and some parabolas), you can calculate the area of one symmetric section and multiply it by the number of such sections. This often makes the limits of integration simpler (e.g., from 0 to a instead of -a to a), which reduces the complexity of calculations.
9. Geometrically, what does the element 'y dx' represent in the integral used to find the area under a curve?
The element 'y dx' represents the area of an infinitesimally thin, vertical rectangular strip. Here, 'y' is the height of the strip (the value of the function at that point) and 'dx' is its infinitesimal width. The process of integration (∫) is essentially the summation of the areas of all such an infinite number of rectangular strips across the specified interval to find the total area under the curve.
10. What is the full form of AOI in the context of Class 12 Maths?
In the context of Class 12 Maths, AOI is the common abbreviation for the chapter name Application of Integrals.
