§6.5 Improper integrals

Introduction

Improper Integrals Involving Unbounded Intervals

Convergence and Divergence

The fraction of callers waiting between 5 and 10 minutes is $\int_5^{10}0.25e^{-0.25t}\,dt = e^{-1.25}-e^{-2.5}$. Report the decimal, rounded to three places.

Using the First FTC, $\int_1^b\frac{1}{x}dx = \ln(b)$. As $b \to \infty$, this shows that $\int_1^\infty\frac{1}{x}dx$

Using $\int_1^b\frac{1}{x^{3/2}}dx = 2-\frac{2}{\sqrt{b}}$, evaluate $\int_1^\infty\frac{1}{x^{3/2}}\,dx$.

The graphs of $y = \frac{1}{x}$ and $y = \frac{1}{x^{3/2}}$ both have the $x$-axis as a horizontal asymptote, but $\frac{1}{x^{3/2}}$ approaches the axis faster as $x$ grows.

Comparing $y = \frac{1}{x}$ and $y = \frac{1}{x^{3/2}}$ for $x \ge 1$, both approach the $x$-axis, yet

Evaluate $\int_1^\infty\frac{1}{x^2}\,dx$ by computing $\lim_{b\to\infty}\left(-\frac{1}{b}+1\right)$.

Using the antiderivative $-4e^{-x/4}$, evaluate the improper integral $\int_0^\infty e^{-x/4}\,dx$.

Using the antiderivative $27(x+5)^{1/3}$, the improper integral $\int_2^\infty\frac{9}{(x+5)^{2/3}}\,dx$

Using the antiderivative $-12(x+2)^{-1/4}$, the improper integral $\int_4^\infty\frac{3}{(x+2)^{5/4}}\,dx$ converges to $\frac{12}{6^{1/4}}$. Report the decimal, rounded to three places.

Using the antiderivative $-4xe^{-x/4}-16e^{-x/4}$, evaluate $\int_0^\infty xe^{-x/4}\,dx$.

For $\int_1^\infty\frac{1}{x^p}dx$ with $p \gt 0$, the integral converges exactly when