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CA Foundation · Quantitative Aptitude · Differential and Integral Calculus

The area bounded by the curve y = x², the x-axis and the ordinates x = 1 and x = 3 is:

The area under y = x² between x = 1 and x = 3 is found by integrating x² to get x³/3. Evaluating gives 27/3 minus 1/3, which equals 26/3 square units.

  1. A26/3 square unitsCorrect
  2. B8 square units
  3. C28/3 square units
  4. D9 square units

Explanation

Area = ∫ from 1 to 3 of x² dx = [x³/3] from 1 to 3 = 27/3 − 1/3 = 26/3. Option 8 results from using x³/3 at 3 minus 1 incorrectly as (9 − 1), i.e. integrating wrongly as x² difference. So 26/3 is correct.

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