Chapter 9.4, Problem 29E

a.

To determine

Tostate: that why the integral is improper.

Answer to Problem 29E

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

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^{1}x\ln xdx}$ .

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 29E

The integralvalue is $-\frac{1}{4}$ .

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^{1}x\ln xdx}$ .

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

  $\begin{array}{l}{\displaystyle {\int }_0^{1}x\ln xdx}={\left[\ln x\left(\frac{x^2}{2}\right)-\frac{x^2}{4}\right]}_0^1\text{ [}{\displaystyle \int uvdx=u{\displaystyle \int vdx-{\displaystyle \int \frac{du}{dx}}}{\displaystyle \int vdxdx}\text{ and simplify}]}\\ {\displaystyle {\int }_0^{1}x\ln xdx}=\ln \left(1\right)\left(\frac{{\left(1\right)}^2}{2}\right)-\frac{{\left(1\right)}^2}{4}-\ln \left(0\right)\left(\frac{{\left(0\right)}^2}{2}\right)+\frac{{\left(0\right)}^2}{4}\\ {\displaystyle {\int }_0^{1}x\ln xdx}=0-\frac{1}{4}-0+0\text{ [simplify]}\\ {\displaystyle {\int }_0^{1}x\ln xdx}=-\frac{1}{4}\end{array}$

Thus, the integral value is $-\frac{1}{4}$ .