How to Solve Area Under Curves in Class 12 Maths Exercise 8.1
If you’re focusing on Class 12 Maths and want to master the Application of Integrals, this deep-dive into Exercise 8.1 offers exactly what you need—stepwise, exam-standard solutions. In board exams, questions from this chapter usually appear for about 6 marks, often as one long answer and one short answer. Valued for both its conceptual depth and direct problem-solving utility, these solutions help you confidently connect theory to real area-bounded-by-curves problems.
Table of Content
Many students specifically ask, “How do I solve exercise 8.1 class 12 questions without missing steps?” That’s why you’ll find clear walkthroughs for calculation of definite integrals and critical tips for setting up area under curve questions. Each answer reinforces pattern practice and the logic behind upper-lower boundary integration, so you solve faster and make fewer mistakes when it matters most.
Every solution here is CBSE-syllabus aligned and fact-checked for accuracy by the team at Vedantu. Use these explanations to save time in revision and to boost confidence for the “area between two curves” type questions. If you want topic-wise structure, review the Class 12 Maths syllabus for up-to-date board guidance.
Exercise 8.1
1. Find the area of the region bounded by the ellipse \[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1\].
Ans: Rewrite the given equation of ellipse
\[\frac{{{x}^{2}}}{{{4}^{2}}}+\frac{{{y}^{2}}}{{{3}^{2}}}=1\]
The intersection point of the ellipse and the x-axis are at \[x=\pm 4\].
The intersection point of the ellipse and the y-axis are at \[y=\pm 3\].
The given equation of the ellipse can be represented as
If we change \[x\] to \[-x\] or \[y\] to \[y\] the equation remains the same.
So, the ellipse is symmetrical about the x-axis and the y-axis.
Area bounded by ellipse \[=4\times\] Area of OABO
Area of OABO \[=\int\limits_{0}^{4}{ydx}\]
\[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1\]
\[\Rightarrow \frac{{{y}^{2}}}{9}=1-\frac{{{x}^{2}}}{16}\]
\[\Rightarrow {{y}^{2}}=9\left( 1-\frac{{{x}^{2}}}{16} \right)\]
\[\Rightarrow y=3\sqrt{1-\frac{{{x}^{2}}}{16}}\]
Area of OABO \[=\int\limits_{0}^{4}{3\sqrt{1-\frac{{{x}^{2}}}{16}}dx}\]
Area of OABO \[=\frac{3}{4}\int\limits_{0}^{4}{\sqrt{16-{{x}^{2}}}dx}\]
Area of OABO \[=\frac{3}{4}\int\limits_{0}^{4}{\sqrt{{{4}^{2}}-{{x}^{2}}}dx}\]
Apply the formula \[\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\frac{x}{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a}\] in above equation
Area of OABO \[=\frac{3}{4}\left[ \frac{x}{2}\sqrt{16-{{x}^{2}}}+\frac{16}{2}{{\sin }^{-1}}\frac{x}{4} \right]_{0}^{4}\]
Area of OABO \[=\frac{3}{4}\left[ 0+8{{\sin }^{-1}}(1)-0-8{{\sin }^{-1}}(0) \right]_{0}^{4}\]
Area of OABO \[=\frac{3}{4}\left( \frac{8\pi }{2} \right)\]
Area of OABO \[=3\pi \]
Therefore, the area bounded by ellipse \[=\text{4}\times \text{3}\pi\]
The area bounded by ellipse \[=\text{12}\pi \text{ sq}\text{. units}\]
2. Find the area of the region bounded by the ellipse \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1\].
Ans: Rewrite the given equation of the ellipse
\[\frac{{{x}^{2}}}{{{2}^{2}}}+\frac{{{y}^{2}}}{{{3}^{2}}}=1\]
The intersection point of the ellipse and the x-axis are at \[x=\pm 2\].
The intersection point of the ellipse and the y-axis are at \[y=\pm 3\].
The given equation of the ellipse can be represented as a vertical ellipse.
It can be observed that the ellipse is symmetrical about the x-axis and y-axis.
Since, if we change \[x\] to \[-x\] or \[y\] to \[-y\] the equation remains the same.
Area bounded by ellipse \[=4\times\]Area of OABO
Area of OABO \[=\int\limits_{0}^{2}{ydx}\]…………………….(A)
\[\Rightarrow \frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1\]
\[\Rightarrow \frac{{{y}^{2}}}{9}=1-\frac{{{x}^{2}}}{4}\]
\[\Rightarrow {{y}^{2}}=9\left( 1-\frac{{{x}^{2}}}{4} \right)\]
\[\Rightarrow y=3\sqrt{1-\frac{{{x}^{2}}}{4}}\text{ }\]
\[\Rightarrow y=\frac{3}{2}\sqrt{4-{{x}^{2}}}\text{ }\]…………………..(1)
Area of OABO \[=\int\limits_{0}^{2}{\frac{3}{2}\sqrt{4-{{x}^{2}}}\text{ }dx}\] using equation (1) in (A)
Area of OABO \[=\frac{3}{2}\int\limits_{0}^{2}{\sqrt{{{2}^{2}}-{{x}^{2}}}dx}\]
Apply the formula \[\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx=\frac{x}{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\frac{x}{a}\] in above equation
Area of OABO \[=\frac{3}{2}\left[ \frac{x}{2}\sqrt{4-{{x}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} \right]_{0}^{2}\]
Area of OABO \[=\frac{3}{2}\left[ \frac{2\pi }{2} \right]\]
Area of OABO \[=\frac{3\pi }{2}\]
Area bound by the ellipse \[=\text{4}\times \frac{\text{3}\pi }{2}\]
Area bound by the ellipse \[=\text{6}\pi \text{ sq}\text{. units}\]
Choose the correct answer in the following Exercises 3 and 4.
3. Area lying in the first quadrant and bounded by the circle \[{{\text{x}}^2} + {y^2} = 4\] and the lines \[x = 0\] and \[x = 2\] is
\[\pi \]
\[\frac{\pi }{2}\]
\[\frac{\pi }{3}\]
\[\frac{\pi }{4}\]
Ans:
Area of \[OAB=\int_{0}^{2}{ydx}\]
Since, \[{{\text{x}}^2} + {y^2} = 4\] $ [\text {Equation of circle}]$
\[{{y}^{2}}=4-{{\text{x}}^{2}}\]
\[y=\sqrt{4-{{\text{x}}^{2}}}\]
So,
Area of \[OAB=\int_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}\]
Area of \[OAB=\left[ \frac{x}{2}\sqrt{4-{{x}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} \right]_{0}^{2}\]
Area of \[OAB=\left[ \frac{2}{2}\sqrt{4-{{2}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{2}{2}-\left( \frac{0}{2}\sqrt{4-{{0}^{2}}}+\frac{4}{2}{{\sin }^{-1}}\frac{0}{2} \right) \right]\]
Area of \[OAB=\left[ \sqrt{4-4}+2{{\sin }^{-1}}1-0 \right]\]
Area of \[OAB=\left[ 2\times \frac{\pi }{2} \right]\]
Area of \[OAB=\pi \] square units
Hence, the area lying in the first quadrant and bounded by the circle \[{{\text{x}}^2} + {y^2} = 4\] and the lines \[x = 0\] and \[x = 2\] is \[\pi \] square units.
Hence, option (A) is the correct answer.
4. Area of the region bounded by the curve \[{{y}^{2}}=4x\], y-axis and the line \[y=3\] is
2
\[\frac{9}{4}\]
\[\frac{9}{3}\]
\[\frac{9}{2}\]
Ans:
The area bounded by the curve \[{{y}^{2}}=4x\], y-axis and the line \[y=3\] is the area of the region \[OAB\].
Area of \[OAB=\int_{0}^{3}{xdy}\]
Area of \[OAB=\int_{0}^{3}{\frac{{{y}^{2}}}{4}dy}\]
Area of \[OAB=\frac{1}{4}\int_{0}^{3}{{{y}^{2}}dy}\]
Area of \[OAB=\frac{1}{4}\left[ \frac{{{y}^{3}}}{3} \right]_{0}^{3}\]
Area of \[OAB=\frac{1}{4}\left[ \frac{{{3}^{3}}}{3}-0 \right]\]
Area of \[OAB=\frac{1}{4}\left[ \frac{27}{3} \right]\]
Area of \[OAB=\frac{1}{4}\left[ 9 \right]\]
Area of \[OAB=\frac{9}{4}\]square units
Hence, option (B) is the correct answer.
Conclusion
NCERT Solutions for Class 12 Chapter 8 Exercise 8.1, provided by Vedantu, offer a comprehensive understanding of the application of integrals. This exercise focuses on calculating areas under curves and between curves, which is a fundamental aspect of integral calculus. It's important to concentrate on class 12 ex 8.1 setting up integrals correctly with the appropriate limits and understanding the difference between definite integrals and the area between curves. Regular practice of these problems will enhance your problem-solving skills and solidify your grasp of integral applications, which are crucial for advanced mathematics and various real-life applications.
Class 12 Maths Chapter 8: Exercises Breakdown
CBSE Class 12 Maths Chapter 8 Other Study Materials
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.
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FAQs on Class 12 Maths Chapter 8 Exercise 8.1 NCERT Solutions (Area Under Curves)
1. What is the correct approach to solving questions from NCERT Class 12 Maths Chapter 8 using these solutions?
To effectively use the NCERT Solutions for Chapter 8, focus on the step-by-step methodology. First, attempt the problem yourself. Then, compare your method with the provided solution, paying close attention to how limits are determined, the correct integral is set up, and the final calculations are performed. This process helps you master the CBSE pattern and avoid common errors.
2. How can I correctly find the area between two curves as per the NCERT solutions for Chapter 8?
The standard method for finding the area between two curves involves these key steps:
- First, draw a rough sketch of both curves to visualise the bounded region.
- Next, find the points of intersection by solving the equations of the curves simultaneously. These points will be your limits of integration (a, b).
- Set up the definite integral as ∫ₐᵇ [f(x) - g(x)] dx, where f(x) is the upper curve and g(x) is the lower curve.
- Finally, integrate and evaluate the expression to find the required area.
3. Are the NCERT Solutions for Class 12 Maths Chapter 8 aligned with the latest CBSE 2025-26 syllabus?
Yes, all solutions provided for 'Application of Integrals' are fully updated to match the latest CBSE 2025-26 syllabus. The methods, formulas, and problem types covered are precisely what is prescribed by the NCERT curriculum, ensuring complete relevance for your board exams.
4. What are the most common mistakes to avoid when solving problems on the area under curves?
Students often make a few common errors in Chapter 8. Be careful to avoid:
- Incorrectly identifying the limits of integration. Always solve for intersection points first.
- Confusing the upper and lower functions, which can lead to a negative area if subtracted incorrectly.
- Making simple algebraic or integration errors during calculation.
- Forgetting to draw a sketch, which makes it difficult to visualise the bounded region and set up the integral correctly.
5. Why is drawing a rough sketch considered the most critical first step for finding the area under a curve?
Drawing a sketch is crucial because it provides a visual representation of the problem. It helps you to correctly:
- Identify the bounded region whose area needs to be calculated.
- Determine which function is the 'upper curve' and which is the 'lower curve'.
- Visually confirm the limits of integration (the x or y values where the region begins and ends).
- Decide whether to integrate with respect to x (vertical strips) or y (horizontal strips), which can simplify the problem.
6. How do the NCERT Solutions for the Miscellaneous Exercise of Chapter 8 differ from the regular exercises?
The NCERT Solutions for the Miscellaneous Exercise in Chapter 8 tackle more complex and application-based problems. Unlike the direct questions in Exercises 8.1 and 8.2, these often require a combination of concepts, such as finding the area of a triangle using integration or solving problems with more intricate curves. They are designed to test a deeper, more holistic understanding of the chapter.
7. When solving for a bounded area, how do I decide whether to integrate with respect to x (dx) or y (dy)?
The choice between integrating with respect to x (using vertical strips) or y (using horizontal strips) depends on the orientation of the curves.
- Use dx (vertical strips) when it is easy to express the functions in the form y = f(x) and one function is consistently above the other. This is the most common approach.
- Use dy (horizontal strips) when it is easier to express the functions in the form x = f(y) and one function is consistently to the right of the other, which is useful for curves like sideways parabolas (e.g., y² = 4ax).
8. What is the logic behind the formula ‘Area = ∫ [Upper Curve - Lower Curve] dx’?
This formula works by applying the fundamental concept of definite integrals. The integral of the upper curve, ∫ f(x) dx, calculates the entire area under it down to the x-axis. Similarly, the integral of the lower curve, ∫ g(x) dx, calculates the area under it. By subtracting the area under the lower curve from the area under the upper curve, we are left with precisely the area of the region enclosed between them.
