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

Extra practice questions for Class 12 Maths Chapter 8 (Applications of Integrals), beyond the textbook. These Class 12 Mathematics Chapter 8 important questions are handy for last-minute exam practice.

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.

Written by Satish

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