Reduction And Standard Forms
Evaluate the definite integral from zero to \infty of e raised to negative x multiplied by x, recognizing the Gamma-function pattern embedded in this improper integral.
Select the correct option:
Solution
1
This improper integral matches the Gamma function pattern, the integral that generalizes factorials to a continuous setting and converges whenever the exponential decay dominates polynomial growth. The defining relation is \Gamma(n) = \int_0^\infty x^{n-1} e^{-x},dx, which for positive integers reduces to \Gamma(n) = (n-1)!. Here the integrand x e^{-x} corresponds to n - 1 = 1, so n = 2 and \Gamma(2) = 1! = 1. Equivalently, integrating by parts with u = x and dv = e^{-x},dx yields [-x e^{-x}]_0^\infty + \int_0^\infty e^{-x},dx. The boundary term vanishes because e^{-x} decays far faster than x grows, sending the product to zero at \infty, and the remaining integral equals [-e^{-x}]_0^\infty = 1. Option 0 ignores the surviving exponential integral after the boundary term dies. Option 2 would correspond to \Gamma(3) = 2!, which needs an extra power of x. Option \infty wrongly assumes divergence despite the dominant exponential decay guaranteeing convergence. The Gamma framework systematizes a whole family of such improper integrals. As a final plausibility check, the integrand is positive throughout and decays rapidly to zero as x grows, so a finite positive value near 1 is fully consistent with the integral converging, and the factorial identity \Gamma(2) = 1! anchors the answer to a familiar discrete value.
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About This Question
- Subject
- mathematics
- Chapter
- integral calculus
- Topic
- reduction and standard forms
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1
This improper integral matches the Gamma function pattern, the integral that generalizes factorials to a continuous setting and converges whenever the exponential decay dominates polynomial growth. The defining relation is \Gamma(n) = \int_0^\infty x^{n-1} e^{-x},dx, which for positive integers reduces to \Gamma(n) = (n-1)!. Here the integrand x e^{-x} corresponds to n - 1 = 1, so n = 2 and \Gamma(2) = 1! = 1. Equivalently, integrating by parts with u = x and dv = e^{-x},dx yields [-x e^{-x}]_0^\infty + \int_0^\infty e^{-x},dx. The boundary term vanishes because e^{-x} decays far faster than x grows, sending the product to zero at \infty, and the remaining integral equals [-e^{-x}]_0^\infty = 1. Option 0 ignores the surviving exponential integral after the boundary term dies. Option 2 would correspond to \Gamma(3) = 2!, which needs an extra power of x. Option \infty wrongly assumes divergence despite the dominant exponential decay guaranteeing convergence. The Gamma framework systematizes a whole family of such improper integrals. As a final plausibility check, the integrand is positive throughout and decays rapidly to zero as x grows, so a finite positive value near 1 is fully consistent with the integral converging, and the factorial identity \Gamma(2) = 1! anchors the answer to a familiar discrete value.
This medium difficulty mathematics question is from the chapter integral calculus, covering the topic of reduction and standard forms. It appeared in the 2025 exam.
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