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CA Foundation · Quantitative Aptitude · Differential and Integral Calculus

Evaluate ∫ x eˣ dx.

The integral equals eˣ(x − 1) + C. Using integration by parts with u = x and dv = eˣ dx gives x eˣ minus the integral of eˣ, which is x eˣ − eˣ. The second term must be subtracted, not added.

  1. Aeˣ(x − 1) + CCorrect
  2. Beˣ(x + 1) + C
  3. Cx eˣ + C
  4. Dx²eˣ/2 + C

Explanation

Take u = x and dv = eˣ dx. Then ∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C. The option eˣ(x + 1) comes from adding the second term instead of subtracting it.

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