§6.5 Improper integrals (continued)

Improper Integrals Involving Unbounded Integrands

Summary

The integral $\int_0^1\frac{1}{x^{1/3}}\,dx$ has a vertical asymptote at $x=0$, yet it converges because

The integral $\int_{-2}^2\frac{1}{x^2}\,dx$ diverges because

The integral $\int_0^{\pi/2}\tan(x)\,dx$ diverges because

Using the antiderivative $-2\sqrt{4-x}$, evaluate $\int_1^4\frac{1}{\sqrt{4-x}}\,dx$. The exact value is $2\sqrt{3}$. Report the decimal, rounded to three places.

Using the antiderivative $\arcsin(x)$, evaluate $\int_0^1\frac{1}{\sqrt{1-x^2}}\,dx$. The exact value is $\frac{\pi}{2}$. Report the decimal, rounded to three places.

The integral $\int_0^2e^{-x}\,dx$ is a proper integral with value $1-e^{-2}$, since the integrand is continuous on all of $[0,2]$.

Using the substitution $u = \ln(x)$ on integrals starting at $x = e$, the family $\int_e^\infty\frac{1}{x(\ln x)^p}\,dx$ converges exactly when

Using integration by parts and the limit $\lim_{a\to0^+}(-a\ln a) = 0$, the improper integral $\int_0^1\ln(x)\,dx$ converges. Report its exact value.

Since $\frac{1}{x^2+x+1} \lt \frac{1}{x^2}$ for $x \ge 1$ and $\int_1^\infty\frac{1}{x^2}dx$ converges, comparison tells us $\int_1^\infty\frac{1}{x^2+x+1}\,dx$