Question

Evaluate the improper integral. int_0^oo 6e^(-6x)dx

Integrals
ANSWERED
asked 2021-09-13
Evaluate the improper integral.
\(\displaystyle{\int_{{0}}^{\infty}}{6}{e}^{{-{6}{x}}}{\left.{d}{x}\right.}\)

Expert Answers (1)

2021-09-14
\(\displaystyle{\int_{{0}}^{\infty}}{6}{e}^{{-{6}{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle={6}{\int_{{0}}^{\infty}}{e}^{{-{6}{x}}}{\left.{d}{x}\right.}=\)
\(\displaystyle={6}{x}{{\left[\frac{{{e}^{{-{6}{x}}}}}{{-{6}}}\right]}_{{0}}^{\infty}}=\)
\(\displaystyle=\frac{{6}}{{-{6}}}{x}{{\left[\frac{{1}}{{e}^{{{6}{x}}}}\right]}_{{0}}^{\infty}}=\)
\(\displaystyle={\left(-{1}\right)}{x}{\left[\frac{{1}}{{e}^{{{6}{\left(\infty\right)}}}}–\frac{{1}}{{e}^{{{6}\times{0}}}}\right]}=\)
\(\displaystyle={\left(-{1}\right)}{x}{\left[\frac{{1}}{{e}^{\infty}}–\frac{{1}}{{e}^{{0}}}\right]}=\)
\(\displaystyle={\left(-{1}\right)}{x}{\left[\frac{{1}}{\infty}–\frac{{1}}{{1}}\right]}=\)
\(\displaystyle={\left(-{1}\right)}{x}{\left[{0}-{1}\right]}=\)
\(\displaystyle={\left(-{1}\right)}{x}\times{\left(-{1}\right)}={1}\)
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