7.1 Integration By Parts/53: Difference between revisions
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<math>y= xe^{-0.4x} ~~~ x=0,x=1</math> | <math>y= xe^{-0.4x} ~~~ x=0,x=1</math> | ||
<math> \underbrace{\int_{0}^{5}xe^{-0.4x}dx} | <math> \underbrace{\int_{0}^{5}xe^{-0.4x}dx} | ||
\begin{aligned} | |||
u & = x & dv &= dx \\[0.6ex] | u & = x & dv &= dx \\[0.6ex] | ||
du & = \tfrac{2\ln{(x)}}{x}dx & v &= x | du & = \tfrac{2\ln{(x)}}{x}dx & v &= x | ||
Revision as of 12:33, 13 December 2022
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle y= xe^{-0.4x} ~~~ x=0,x=1}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \underbrace{\int_{0}^{5}xe^{-0.4x}dx} \begin{aligned} u & = x & dv &= dx \\[0.6ex] du & = \tfrac{2\ln{(x)}}{x}dx & v &= x \end{aligned}} \\ [2ex] }
\begin{aligned} u & = \ln^{2}{(x)} & dv &= dx \\[0.6ex] du & = \tfrac{2\ln{(x)}}{x}dx & v &= x \end{aligned}} \\ [2ex]