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Chapter 8: Application of Integrals — Online MCQ Test

MATHS · CLASS 12 SECOND PUC · Karnataka State Board
Practice Chapter 8: Application of Integrals with a free chapter-wise online MCQ test. This chapter covers: Area Under Curves: The primary objective of this chapter is to calculate the precise geometric area bounded by algebraic and trigonometric curves using definite integrals.Vertical.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter 8: Application of Integrals — Important Questions & Answers

What is the formula for finding the area under a curve y = f(x) between x = a and x = b?
  • A. Area = ∫[a to b] y dx
  • B. Area = ∫[a to b] x dy
  • C. Area = ∫[a to b] y² dx
  • D. Area = ∫[a to b] (y dx)²
Answer: A. Area = ∫[a to b] y dx
The vertical strips formula for area bounded along the x-axis is Area = ∫[a to b] y dx, where y = f(x).
For finding the area bounded by a curve between y = c and y = d, which formula should be used?
  • A. Area = ∫[c to d] y dy
  • B. Area = ∫[c to d] x dy
  • C. Area = ∫[c to d] x² dy
  • D. Area = ∫[c to d] (xy) dy
Answer: B. Area = ∫[c to d] x dy
The horizontal strips formula uses Area = ∫[c to d] x dy when integrating with respect to y-axis limits.
Find the area under the curve y = x from x = 0 to x = 2.
  • A. 1
  • B. 2
  • C. 3
  • D. 4
Answer: B. 2
Area = ∫[0 to 2] x dx = [x²/2] from 0 to 2 = 4/2 - 0 = 2.
When a curve crosses the x-axis within the interval [a, b], what strategy should be employed?
  • A. Ignore the crossing and calculate directly
  • B. Split the integral at the x-intercept and add absolute values
  • C. Use only the positive portion
  • D. Reverse the limits of integration
Answer: B. Split the integral at the x-intercept and add absolute values
The Split Region Strategy requires breaking the integral at x-intercepts to properly account for areas above and below the x-axis.
[Board-Style] A parabola y² = 8x encloses an area with the line x = 2. Draw a rough sketch and calculate the total enclosed area. Which of the following correctly represents the setup?
  • A. Area = 2∫[0 to 2] 2√(2x) dx
  • B. Area = ∫[0 to 4] √(y²/8) dy
  • C. Area = ∫[-4 to 4] 2 dy where x = y²/8
  • D. Area = 2∫[0 to 4] √(2x) dx
Answer: A. Area = 2∫[0 to 2] 2√(2x) dx
For y² = 8x at x = 2, y = ±4. Using vertical strips: Area = 2∫[0 to 2] y dx = 2∫[0 to 2] 2√(2x) dx. Evaluating: 4∫[0 to 2] √(2x) dx = 4 × 2√2 × [2x^(3/2)/3]₀² = (16√2/3) × 2^(3/2) = 32/3.