Recitation 11

Example 1: Evaluate $\int_0^3\frac{dx}{x-1}$ if possible.

Hint: Because $x=1$ is the vertical asymptote of the integrant, we have to break the integral into $\int_0^1\frac{dx}{x-1}+\int_1^3\frac{dx}{x-1}$. The first integral is divergent.

Example 2: Determine whether $\int_1^\infty\frac{\ln x}{x}dx$ is convergent or divergent.

Hint: Consider $\lim_{t\to\infty}\int_1^\infty\frac{\ln x}{x}dx$.

Problem 3: Determine whether $\int_{-2}^{14}\frac{dx}{\sqrt[4]{x+2}}$ is convergent or divergent.

Hint: Consider $\lim_{t\to -2^+}\int_t^{14}\frac{dx}{\sqrt[4]{x+4}}$.

Remark: Here is a comparison picture of graphs of $f(x)=1/x$(blue) and $g(x)=1/\sqrt[4]{x}$(red). The `vertical tail’ of $g$ is way thinner than $f$. This explains partly why $\int_0^1f(x)dx$ is divergent whereas $\int_0^1g(x)dx$ is convergent.

Example 4: Use the comparison theorem to determine whether $\int_0^\infty\frac{x}{x^3+1}dx$ is convergent or divergent.

Hint: Break the integral into $\left(\int_0^1+\int_1^\infty\right)\frac{x}{x^3+1}dx$. The second integral converges because the integral is less than $\frac{1}{x^2}$ for all $x\geq 1$.

Problem 5: Use the comparison theorem to determine whether $\int_1^\infty\frac{x+1}{\sqrt{x^4-x}}dx$ is convergent or divergent.

Hint: The integral diverges because the integrand is greater than $\frac{1}{x}$ for all $x\geq 1$.

Problem 6: Use the comparison theorem to determine whether $\int_0^\infty\frac{\arctan x}{2+e^x}dx$ is convergent or divergent.

Hint: The integral converges because the integrand is less than $\frac{\pi}{2}e^{-x}$ for all $x\geq 0$.

Problem 7: Use the comparison theorem to determine whether $\int_0^1\frac{\sec^2x}{x\sqrt{x}}dx$ is convergent or divergent.

Hint: The integral diverges because the integrand is greater than or equal to $1/x\sqrt{x}$ for all $0\leq x\leq 1$.