Chapter 9.4, Problem 27E

a.

To determine

Tostate: that why the integral is improper.

Answer to Problem 27E

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

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^1\frac{x+1}{\sqrt{x^2+2x}}dx}$ .

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 27E

The integralvalue is $\sqrt{3}$ .

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^1\frac{x+1}{\sqrt{x^2+2x}}dx}$ .

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

  $\begin{array}{l}{\displaystyle {\int }_0^1\frac{x+1}{\sqrt{x^2+2x}}dx}={\left[\frac{1}{2}\left(2\sqrt{x^2+2x}\right)\right]}_0^1\text{ [}{\displaystyle \int x^n=\frac{x^{n+1}}{n+1}\text{ and simplify}]}\\ {\displaystyle {\int }_0^1\frac{x+1}{\sqrt{x^2+2x}}dx}=\sqrt{{\left(1\right)}^2+2\left(1\right)}-\sqrt{{\left(0\right)}^2+2\left(0\right)}\\ {\displaystyle {\int }_0^1\frac{x+1}{\sqrt{x^2+2x}}dx}=\sqrt{1+2}-0\text{ [simplify]}\\ {\displaystyle {\int }_0^1\frac{x+1}{\sqrt{x^2+2x}}dx}=\sqrt{3}\end{array}$

Thus, the integral value is $\sqrt{3}$ .