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1981

JEE Main 1981 Mathematics question on Definite Integrals and Areas

The value of the definite integral ∫01(1+e−x2)dx\int_0^1 (1 + e^{-x^2}) dx is
  1. A.
    −1-1
  2. B.
    22
  3. C.
    1+e−11 + e^{-1}
  4. D.
    None of the above
Show answer and solution

Correct option: D

Solution

The integral splits as ∫₀¹ 1 dx + ∫₀¹ e^(−x²) dx = 1 + ∫₀¹ e^(−x²) dx. Since e^(−x²) > 0 on [0,1], the second integral is strictly positive, making the total strictly greater than 1. Also, e^(−x²) ≥ e^(−1) on [0,1], so ∫₀¹ e^(−x²) dx > e^(−1), meaning the total exceeds 1 + e^(−1). Since the integral has no closed form in elementary functions and its value (≈ 1.747) matches none of options 1, 2, or 3 exactly, the answer is "None of the above."

[1981]

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