Integral Test Use

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You will be shown a graph and the result of integrating from 1 to \(\infty\). Your task is to decide whether the integral test gives convergence or divergence of the corresponding sum or if the integral test cannot be used. Click on the Start button to begin.

\(\int_1^{\infty}e^{-x}dx=e^1, \displaystyle \sum^∞_{n=1}e^{-x}\)

\(\int_1^{\infty}\frac{1}{\sqrt{x}}dx=\infty, \displaystyle \sum^∞_{n=1}\frac{1}{\sqrt{n}}\)

\(\int_1^{\infty}\frac{\sin(4x)}{x}dx\sim-0.19, \displaystyle \sum^∞_{n=1}\frac{\sin(4n)}{n}\)

\(\int_1^{\infty}(x+1)^\frac{-3}{2}dx=\sqrt{2}, \displaystyle \sum^∞_{n=1}(n+1)^\frac{-3}{2}\)

\(\int_1^{\infty}\frac{1}{(x+1)\ln(x+1)}dx=\infty, \displaystyle \sum^∞_{n=1}\frac{1}{(n+1)\ln(n+1)}\)

\(\int_1^{\infty}\frac{x}{x+1}dx=\infty, \displaystyle \sum^∞_{n=1}\frac{n}{n+1}\)

\(\int_1^{\infty}\frac{x}{e^{x^2}}dx=\frac{1}{2e}, \displaystyle \sum^∞_{n=1}\frac{n}{e^{n^2}}\)

\(\int_1^{\infty}\frac{1}{\sqrt{1+x^2}}dx=\infty, \displaystyle \sum^∞_{n=1}\frac{1}{\sqrt{1+n^2}}\)

\(\int_1^{\infty}\frac{\cos(3x)}{e^x}dx\sim0.052, \displaystyle \sum^∞_{n=1}\frac{\cos(3n)}{e^n}\)

\(\int_1^{\infty}\frac{1}{1+x^2}dx=\frac{\pi}{4}, \displaystyle \sum^∞_{n=1}\frac{1}{1+n^2}\)

\(\int_1^{\infty}\frac{x}{1+x^2}dx=\infty, \displaystyle \sum^∞_{n=1}\frac{n}{1+n^2}\)

\(\int_1^{\infty}\frac{1+2\cos(2x)}{x+\sin(2x)}dx=\infty, \displaystyle \sum^∞_{n=1}\frac{1+2\cos(2n)}{n+\sin(2n)}\)

Diverges Converges Integral Test Cannot be Used