7.1 Integration By Parts/20: Difference between revisions
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u &= x^{2}+1 \quad dv= e^{-x}dx \\[2ex] | u &= x^{2}+1 \quad dv= e^{-x}dx \\[2ex] | ||
Revision as of 03:50, 29 November 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 \int_{0}^{1} \left(x^{2}+1\right)e^{-x}dx }
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 \begin{align} u &= x^{2}+1 \quad dv= e^{-x}dx \\[2ex] du &= 2xdx \qquad v= -e^{-x} \\[2ex] \end{align} }
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 \begin{align} &= (x^{2}+1)(-e^{-x})\bigg|_{0}^{1} - 2\int_{0}^{1} (-e^{x})(x)dx \\[2ex] \end{align} }
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 \begin{align} u &= x^{2}+1 \quad dv= e^{-x}dx \\[2ex] du &= 2xdx \qquad v= -e^{-x} \\[2ex] \end{align} }