Question

Use limits of definite integrals to calculate each of the improper integrals in Exercises.

${\int }_0^1\frac{1}{\sqrt{x}{e}^{\sqrt{x}}}dx$

Short answer

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

${\int }_0^1\frac{1}{\sqrt{x}{e}^{\sqrt{x}}}dx=\frac{-2}{e}-2$

Step-by-step solution

Step 1. Given information. 

The given integral is the following.

${\int }_0^1\frac{1}{\sqrt{x}{e}^{\sqrt{x}}}dx$

Step 2. Value of integral. 

Solve the integral by using the definition of improper integral.

$$\begin{gathered} {\int }_0^1\frac{1}{\sqrt{x}{e}^{\sqrt{x}}}dx={\int }_0^1{x}^{\frac{-1}{2}}{e}^{-\sqrt{x}}dx \\ =\underset{b\to 0}{\lim }{\int }_b^1{x}^{\frac{-1}{2}}{e}^{-\sqrt{x}}dx \\ =\underset{b\to 0}{\lim }{\left[2e^{-\sqrt{x}}{x}^{\frac{1}{2}}-\left(-2\left(-e^{-\sqrt{x}}\sqrt{x}-e^{-\sqrt{x}}\right)\right)\right]}_b^1 \\ =\underset{b\to 0}{\lim }{\left[2e^{-\sqrt{x}}\sqrt{x}-2e^{-\sqrt{x}}\sqrt{x}-2e^{-\sqrt{x}}\right]}_b^1 \\ =\underset{b\to 0}{\lim }{\left[-2e^{-\sqrt{x}}\right]}_b^1 \\ =\underset{b\to 0}{\lim }{\left[\frac{-2}{e}-\left(-2e^{-\sqrt{b}}\right)\right]}_b^1 \\ =\frac{-2}{e}+2 \end{gathered}$$

The value of the integral is${\int }_0^1\frac{1}{\sqrt{x}{e}^{\sqrt{x}}}dx=\frac{-2}{e}-2$