Chapter 5: Q. 39 (page 478)

Use limits of definite integrals to calculate each of the improper integrals in Exercises.
$\int_{0}^{\infty} \frac{1}{\sqrt{x} e^{\sqrt{x}}} d x$

Short Answer

Expert verified

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

$\int_{0}^{\infty} \frac{1}{\sqrt{x} e^{\sqrt{x}}} d x = 2 .$

Step by step solution

Step 1. Given information.

The given integral is the following.

$\int_{0}^{\infty} \frac{1}{\sqrt{x} e^{\sqrt{x}}} d x$

Step 2. Value of integral.

Solve the integral by using the definition of improper integral.

$\int_{0}^{\infty} \frac{1}{\sqrt{x} e^{\sqrt{x}}} d x = \int_{0}^{\infty} x^{\frac{- 1}{2}} e^{- \sqrt{x}} d x = \lim_{b \to \infty} \int_{0}^{b} x^{\frac{- 1}{2}} e^{- \sqrt{x}} d x = \lim_{b \to \infty} {[2 e^{- \sqrt{x}} x^{\frac{1}{2}} - (- 2 (- e^{- \sqrt{x}} \sqrt{x} - e^{- \sqrt{x}}))]}_{0}^{b} = \lim_{b \to \infty} {[2 e^{- \sqrt{x}} \sqrt{x} - 2 e^{- \sqrt{x}} \sqrt{x} - 2 e^{- \sqrt{x}}]}_{0}^{b} = \lim_{b \to \infty} {[- 2 e^{- \sqrt{x}}]}_{0}^{b} = \lim_{b \to \infty} {[- 2 e^{- \sqrt{b}} - (- 2)]}_{0}^{b} = 0 + 2 = 2$

The value of the integral is $\int_{0}^{\infty} \frac{1}{\sqrt{x} e^{\sqrt{x}}} d x = 2 .$

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Use limits of definite integrals to calculate each of the improper integrals in Exercises.
$\int_{0}^{\infty} \frac{1}{\sqrt{x} e^{\sqrt{x}}} d x$

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∞1xexdx

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.
$\int_{0}^{\infty} \frac{1}{\sqrt{x} e^{\sqrt{x}}} d x$