Chapter 9.4, Problem 28E

a.

To determine

Tostate: that why the integral is improper.

Answer to Problem 28E

The integral has an infinite discontinuity at the interior point $x=0$ .

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^4\frac{e^{-\sqrt{x}}}{\sqrt{x}}dx}$ .

the integral has an infinite discontinuity at the interior point $x=0$ .

Thus, the integral is improper.

b.

To determine

To find: the value of integral.

Answer to Problem 28E

The integralvalue is $2\left(1-e^{-2}\right)$ .

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^4\frac{e^{-\sqrt{x}}}{\sqrt{x}}dx}$ .

Calculation:to evaluate the definite integrals first integrate then simplify,

  $\begin{array}{l}{\displaystyle {\int }_0^4\frac{e^{-\sqrt{x}}}{\sqrt{x}}dx}={\left[2\left(-e^{-\sqrt{x}}\right)\right]}_0^4\text{ [}{\displaystyle \int e^x=e^x\text{ and simplify}]}\\ {\displaystyle {\int }_0^4\frac{e^{-\sqrt{x}}}{\sqrt{x}}dx}=2\left(-e^{-\sqrt{4}}\right)-2\left(-e^{-\sqrt{0}}\right)\\ {\displaystyle {\int }_0^4\frac{e^{-\sqrt{x}}}{\sqrt{x}}dx}=-2e^{-2}+2\text{ [simplify]}\\ {\displaystyle {\int }_0^4\frac{e^{-\sqrt{x}}}{\sqrt{x}}dx}=2\left(1-e^{-2}\right)\end{array}$

Thus, the integral value is $2\left(1-e^{-2}\right)$ .