Question

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

Suppose f(x) and g(x) are both continuous on (a, b] but not at x = a. If ${\int }_a^{b}f\left(x\right)dx$ converges and 0 ≤ g(x) ≤ f(x)for all x ∈ (a, b], then${\int }_a^{b}g\left(x\right)dx$ also converges.

Short answer

The given statement is proved.

Step-by-step solution

Step 1. Given Information.

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

Step 2. Prove.

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

So,

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

From the inequality, we can depict that ${\int }_a^{b}f\left(x\right)dx\ge {\int }_a^{b}g\left(x\right)dx.$It is given that ${\int }_a^{b}f\left(x\right)dx$converges so if it converges then ${\int }_a^{b}g\left(x\right)dx$also converges.

Hence, it is proved.