Class 12 Mathematics Chapter 8 Applications of Integrals – Extra Questions with Answers

Applications of Integrals uses definite integrals to measure area, and these questions work through finding the area under a curve, between two curves, and even a circle’s area by integration.

Last Updated: September 23, 2026

Very Short Answer Questions (1 mark)

Q1. Write the formula for area under y=f(x) from x=a to x=b.
Ans: Area=∫_a^b|f(x)|dx.

Q2. What is the area of a circle of radius a using integration?
Ans: πa².

Q3. If f(x)≥g(x) on [a,b], write the area formula between them.
Ans: ∫_a^b[f(x)−g(x)]dx.

Q4. Why do we sketch curves before setting up the integral?
Ans: To correctly identify which curve is upper/lower and the correct limits of integration.

Q5. What is the area enclosed by the parabola y²=4ax and its latus rectum x=a?
Ans: 8a²/3 (standard result).

Short Answer Questions (2–3 marks)

Q6. Find the area bounded by y=x, the x-axis, and the lines x=0, x=4.
Ans: Area=∫₀⁴x dx=[x²/2]₀⁴=8 sq units.

Q7. Find the area of the region bounded by y²=9x and x=4 in the first quadrant.
Ans: y=3√x. Area=∫₀⁴3√x dx=3[2x^3/2/3]₀⁴=2(8)=16 sq units.

Q8. Find the points of intersection of y=x and y=x².
Ans: x=x² ⇒ x²−x=0 ⇒ x(x−1)=0 ⇒ x=0,1. Points: (0,0) and (1,1).

Higher-Order Thinking / Application Questions

Q9. Find the area of the region bounded by the curves y=x² and y=x, showing the full setup with sketch reasoning and limits.
Ans: First find intersection points: x²=x ⇒ x(x−1)=0 ⇒ x=0 or x=1, giving points (0,0) and (1,1). Between x=0 and x=1, we check which curve is on top by testing a value, say x=0.5: y=x gives 0.5, y=x² gives 0.25. Since 0.5>0.25, the line y=x lies above the parabola y=x² throughout this interval. Therefore, Area=∫₀¹(x−x²)dx=[x²/2−x³/3]₀¹=(1/2−1/3)−0=3/6−2/6=1/6 sq unit. Geometrically, this represents the small lens-shaped/parabolic-segment region enclosed between the straight line and the parabola between their two intersection points, and the small value (1/6) makes sense given how close the two curves are throughout the unit interval.

Q10. Using integration, find the area of the region in the first quadrant enclosed by the circle x²+y²=4, the line x=√3y, and the x-axis, explaining how the region splits into two parts.
Ans: The circle x²+y²=4 has radius 2. The line x=√3y, i.e., y=x/√3, passes through the origin with slope 1/√3, corresponding to an angle of 30° with the x-axis. To find where the line meets the circle: substitute x=√3y into x²+y²=4: 3y²+y²=4 ⇒ 4y²=4 ⇒ y=1 (taking positive root in first quadrant), so x=√3. The intersection point is (√3,1). The region in the first quadrant bounded by the circle, the line, and the x-axis splits naturally into two parts: Part 1, from x=0 to x=√3, bounded above by the line y=x/√3 (since the line is below the circle here); Part 2, from x=√3 to x=2, bounded above by the circle y=√(4−x²). Part 1 area=∫₀^√3(x/√3)dx=(1/√3)[x²/2]₀^√3=(1/√3)(3/2)=√3/2. Part 2 area=∫_√3²√(4−x²)dx=[x√(4−x²)/2+2sin⁻¹(x/2)]_√3²=(0+2sin⁻¹(1))−(√3(1)/2+2sin⁻¹(√3/2))=(2·π/2)−(√3/2+2·π/3)=π−√3/2−2π/3=π/3−√3/2. Total area=√3/2+(π/3−√3/2)=π/3 sq units. This demonstrates the standard technique of splitting a composite region into simpler sub-regions when a single integral cannot capture the full boundary in one expression.

Frequently Asked Questions

How would you find the area under the curve y = x squared between x=0 and x=2 using integration?
Integrate x squared from 0 to 2, giving x cubed over 3 evaluated from 0 to 2, which is 8 over 3 square units.

How would you find the area enclosed between the line y=x and the curve y=x squared from x=0 to x=1?
Integrate the difference (x minus x squared) from 0 to 1, giving half minus one third = one sixth square units.

Written by Satish

NCERTBooks.org is an independent educational resource run by a small team focused on making official NCERT textbooks easy to find, read, and download for students, parents, and teachers across India. We are not affiliated with NCERT or the Ministry of Education -- we organise publicly available NCERT content by class and subject, verify links against official sources, and build tools (like our in-browser reader) that make studying more convenient. Every guide we publish is written and reviewed by our team based on the actual NCERT curriculum and syllabus.