Question

Prove each statement in Exercises 78–80, using the definition of improper integrals as limits of proper definite integrals.

If ${\int }_a^{\infty }f\left(x\right)dx$ diverges and 0 ≤ f(x) ≤ g(x) for all x ∈ [a,∞), then${\int }_a^{\infty }g\left(x\right)dx$also diverges.

Short answer

The given statement is proved.

Step-by-step solution

Step 1. Given Information.

The given integrals are${\int }_a^{\infty }f\left(x\right)dx\mathrm{and}{\int }_a^{\infty }\mathrm{g}\left(x\right)dx.$

Step 2. Prove.

To prove the given statement integrate each part of inequality and 0 ≤ f(x) ≤ g(x)from $a\mathrm{to}\infty \mathrm{with}\mathrm{respect}\mathrm{to}x.$

So,

$$\begin{gathered} ={\int }_a^{\infty }0dx\le {\int }_a^{\infty }f\left(x\right)dx\le {\int }_a^{\infty }g\left(x\right)dx \\ =0\le {\int }_a^{\infty }f\left(x\right)dx\le {\int }_a^{\infty }g\left(x\right)dx \end{gathered}$$

From the inequality, we can depict that ${\int }_a^{\infty }g\left(x\right)dx\ge {\int }_a^{\infty }f\left(x\right)dx.$It is given that ${\int }_a^{\infty }f\left(x\right)dx$diverges so if it is approaching torole="math" localid="1649070306841" $\infty ,$then${\int }_a^{\infty }g\left(x\right)dx$also approaching to$\infty .$

Hence, it is proved.