Question

Prove each statement in Exercises 74–77, using limits of definite integrals for general values of p.

If$p\ge 1,$then${\int }_0^1\frac{1}{x^p}dx$diverges.

Short answer

The given statement is proved.

Step-by-step solution

Step 1. Given Information.

The given integral is${\int }_0^1\frac{1}{x^p}dx.$

Step 2. Prove. 

To prove if $p\ge 1,$then ${\int }_0^1\frac{1}{x^p}dx$ diverges, let aand bbe any real numbers and f(x) be a function, and if fcontinuous on role="math" localid="1649068637899" $(a,b],$but not x = a, then ${\int }_a^{b}f\left(x\right)dx=\underset{A\to a^{+}}{\lim }{\int }_A^{b}f\left(x\right)dx.$

So,

$$\begin{gathered} {\int }_0^1\frac{1}{x^p}dx \\ =\underset{A\to 0^{+}}{\lim }{\int }_A^1{x}^{-p}dx \\ =\underset{A\to 0^{+}}{\lim }{\left[\frac{x^{-p+1}}{-p+1}\right]}_A^1 \\ =\underset{A\to 0^{+}}{\lim }\left[\frac{1}{1-p}-\frac{1}{1-p}{A}^{1-p}\right] \\ =\frac{1}{1-p}+\frac{1}{\left(1-p\right)0^{+}} \\ =\infty \left(p\ge 1\right) \end{gathered}$$

Thus, $p\ge 1$and integral diverges on [0, 1].

If $p\ge 1,$then $\frac{1}{x^p}\ge \frac{1}{x}\mathrm{for}x\in [0,1]\mathrm{and}{\int }_0^1\frac{1}{x^p}dx\mathrm{diverges}.$

Hence, the integral is proved.