Chapter 5: Q. 38 (page 478)

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

Short Answer

Expert verified

The improper integral on $[0, 1]$ converges.

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

Step by step solution

Step 1. Given information.

The given integral is the following.

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

Step 2. Value of integral.

Solve the integral by using the definition of improper integral.

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

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

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

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.∫011xexdx

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

Recommended explanations on Math Textbooks

Calculus

Decision Maths

Discrete Mathematics

Theoretical and Mathematical Physics

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Statistics

Study anywhere. Anytime. Across all devices.

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