Question

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

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

Split the integral

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

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.

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 \( \frac{1}{3} + \frac{1}{3} \) = \( \frac{2}{3} \)

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.