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CA Foundation · Quantitative Aptitude · Differential and Integral Calculus

Evaluate ∫₀¹ (e^(2x) + 3x²) dx.

The value is (e² + 1)/2. The integral of e^(2x) over 0 to 1 is (e² − 1)/2 and the integral of 3x² is 1. Adding them gives (e² − 1)/2 + 1, which simplifies to (e² + 1)/2.

  1. A(e² + 1)/2Correct
  2. B(e² − 1)/2 + 1
  3. Ce² + 1
  4. D(e² + 1)/2 + 3

Explanation

∫e^(2x)dx = e^(2x)/2, from 0 to 1 gives (e² − 1)/2. ∫3x²dx = x³, from 0 to 1 gives 1. Sum = (e² − 1)/2 + 1 = (e² + 1)/2. The option (e² − 1)/2 + 1 is the same unsimplified... so check: it equals (e² + 1)/2, hence not distinct.

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