Download Free PDF of Application of Integrals Exercise 8.1 for Class 12 Maths
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 CBSE Class 12 Maths Chapter 8 Application of Integrals – NCERT Solutions 2025-26
1. How can I use NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1 to score full marks in area questions?
NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1 provide accurate, stepwise answers that help you maximize marks in board exams.
- Review each question's stepwise method for solving area under curve problems.
- Practice common integration patterns and formula applications shown in solutions.
- Use summary boxes and notes to avoid calculation and step-missing errors.
- Check your answers with provided final results to build confidence for exam writing.
2. What is the best strategy to solve area between curves problems for CBSE Board exams?
The best strategy for area between curves in CBSE exams is to use a logical, structured approach:
- Draw rough sketches of the given curves and identify boundaries/intersection points.
- Set up the definite integral in the form ∫[a to b] [Upper function − Lower function] dx.
- Clearly write limits of integration based on intersection points.
- Show every calculation step, including antiderivatives and evaluations.
- Conclude with a boxed final area answer.
3. Can I download free PDF solutions for Class 12 Maths Exercise 8.1 from Vedantu?
Yes, you can download free PDF solutions for Class 12 Maths Exercise 8.1 directly from Vedantu.
- Simply use the Download PDF button on the Exercise 8.1 solutions page.
- The PDF includes stepwise answers, revision tips, and board exam pattern practice.
- It's updated for the latest CBSE 2025 syllabus.
4. Are these solutions updated according to the latest CBSE (2025) syllabus?
Yes, all NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1 are fully updated as per the CBSE 2025 syllabus.
- Solutions cover area under curve, area between curves, limits setting, and all topic requirements outlined in the official syllabus.
- Each answer follows current CBSE marking schemes for maximum exam relevance.
5. Why is Application of Integrals important for JEE/NEET aspirants?
The chapter Application of Integrals is crucial for JEE and NEET because it covers integration methods used in advanced problems.
- JEE often asks MCQs on area bounded by curves, requiring quick, precise calculations.
- NEET uses area and integration concepts in Physics and Biology numericals.
- Mastering these solutions improves your problem-solving speed and accuracy for both board and entrance exams.
6. What are frequent mistakes students make while solving Exercise 8.1?
Frequent mistakes in Exercise 8.1 include:
- Wrong limits of integration (not finding intersection points correctly).
- Confusing upper and lower functions in the area formula.
- Omitting essential calculation steps (step skipping leads to lost marks).
- Making integration errors or algebra slips.
- Not boxing the final area answer or writing correct units.
7. What is chapter 8 of maths class 12?
Chapter 8 of Class 12 Maths is Application of Integrals.
- Focuses on finding the area under curves and area between curves using definite integrals.
- Essential for CBSE Board, JEE, and NEET exams.
- Includes real-world application word problems relevant to coordinate geometry.
8. How many exercises are there in Application of Integrals class 12?
In Class 12 Maths Chapter 8 (Application of Integrals), there are typically two main exercises plus a Miscellaneous Exercise.
- Exercise 8.1 – Focuses on areas under simple curves and between lines/curves.
- Exercise 8.2 – Deals with more complex area problems.
- Miscellaneous Exercise – Contains mixed and advanced problems.
9. Is class 12 maths very tough?
Class 12 Maths can be challenging, but using structured NCERT Solutions makes each chapter, including Application of Integrals, manageable.
- Practice regularly using solved examples and revision notes.
- Follow stepwise answers to reduce mistakes in long questions.
- Use PDF solutions, summary sheets, and formula lists for revision before exams.
10. How do I set integration limits for area under curve problems?
To set integration limits for area under curve problems:
- Sketch the curves and find intersection points (these are usually your limits a and b).
- For area between two curves, limits start from the leftmost to rightmost intersection.
- Write the definite integral as ∫[a to b] [Upper function − Lower function] dx.
