Chapter 5: Q. 36 (page 478)

Use limits of definite integrals to calculate each of the improper integrals in Exercises.
$\int_{0}^{\infty} x e^{- x} d x$

Short Answer

Expert verified

The improper integral on $[0, \infty]$ converges.

$\int_{0}^{\infty} x e^{- x} d x = 1 .$

Step by step solution

Step 1. Given information.

The given integral is the following.

$\int_{0}^{\infty} x e^{- x} d x$

Step 2. Value of integral.

Solve the integral by using the definition of improper integral.

$\int_{0}^{\infty} x e^{- x} d x = \lim_{a \to \infty} \int_{0}^{a} x e^{- x} d x = \lim_{a \to \infty} {[- x e^{- x} - e^{- x}]}_{0}^{a} = \lim_{a \to \infty} [(- a e^{- a} - e^{- a}) - (- 1)] = 0 + 1 = 1$

The value of the integral is $\int_{0}^{\infty} x e^{- x} d x = 1 .$

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Use limits of definite integrals to calculate each of the improper integrals in Exercises.
$\int_{0}^{\infty} x e^{- x} d x$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Value of integral.

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Use limits of definite integrals to calculate each of the improper integrals in Exercises.∫0∞xe-xdx

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Value of integral.

One App. One Place for Learning.

Most popular questions from this chapter

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Study anywhere. Anytime. Across all devices.

Use limits of definite integrals to calculate each of the improper integrals in Exercises.
$\int_{0}^{\infty} x e^{- x} d x$