Class 16
Calculate \(\displaystyle \int_{-2}^1 \frac{1}{x^2}\;dx\)
Define \(\displaystyle f(x) = \int_x^1 \frac{1}{\sqrt[3]{t}}\;dt\).
Define \(\displaystyle f(x) = \int_x^1 \frac{1}{t^2}\;dt\).
Define \(\displaystyle f(x) = \int_1^x \frac{1}{t^2}\;dt\).
Define \(\displaystyle f(x) = \int_1^x \frac{1}{\sqrt[3]{t}}\;dt\).
\[\int_0^1 \frac{1}{\sqrt{1-x}}\;dx\]
\[\int_2^\infty e^{-x}\;dx\]
\[\int_{-\infty}^0 \frac{1}{1+x^2}\;dx\]
\[\int_0^1 \frac{1}{x^p}\;dx\]
\[\int_1^\infty \frac{1}{x^p}\;dx\]
Does \(\displaystyle \int_1^\infty \frac{1}{xe^x}\;dx\) converge?
Does \(\displaystyle \int_e^\infty \frac{\ln x}{x}\;dx\) converge?