/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Determine whether the improper i... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 22

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{-\infty}^{\infty} x^{2} e^{-x^{3}} d x $$

Short Answer

Expert verified
The improper integral \( \int_{-\infty}^{\infty} x^{2} e^{-x^{3}} d x \) converges and its value is \( \frac{2}{3} \).

Step by step solution

01

Split the integral

The integral from -鈭� to 鈭� should be split into two separate integrals at x = 0: \( \int_{-\infty}^{0} x^{2} e^{-x^{3}} d x + \int_{0}^{\infty} x^{2} e^{-x^{3}} d x \)
02

Apply substitution

For each of the two integrals, apply the substitution \( u = x^3 \). This gives you \( du = 3x^2 dx \), so \( x^2 dx = du/3 \). This transforms the integrals to \( \frac{1}{3} \) times the integral of \( e^{-u} \) from \( -\infty \) to \( 0 \) and from \( 0 \) to \( \infty \), respectively.
03

Evaluate the integrals

The integral of \( e^{-u} \) over the range from \( -\infty \) to \( 0 \) and from \( 0 \) to \( \infty \) can be evaluated as \( 1 \) for both. Hence, the overall result of both integrals is \( \frac{1}{3} + \frac{1}{3} \) = \( \frac{2}{3} \)
04

Determine Convergence or Divergence

Since we obtained a concrete and finite value for these integrals, the improper integral \( \int_{-\infty}^{\infty} x^{2} e^{-x^{3}} d x \) converges.

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