Skip to content

Reduction And Standard Forms

Mediummathematics

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 Hidden from View

Submit your answer to unlock the detailed step-by-step solution.

About This Question

Subject
mathematics
Chapter
integral calculus
Topic
reduction and standard forms
Difficulty
Medium
Year
2025
Tags
advanced-calculus-drillimproper integralgamma functionintegration by partsexponential decay

Solution

Correct Answer:

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.

Looking for more practice? Explore all mathematics questions or browse integral calculus questions on RankGuru.