7.1 Integration By Parts/11: Difference between revisions

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</math>
</math>


<math>u = x \qquad dv = e^{-x}</math> <br><br>
<math>du = dx \qquad v = -e^{-x}</math>


<math>\int (xe^{-x}) dx = x*(-e^{-x}) - \int (-e^{-x})
<math>\int (xe^{-x}) dx = x*(-e^{-x}) - \int (-e^{-x})
</math>
</math>

Revision as of 04:52, 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 \text {arctan(4t)}dt }



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 du= \frac{4}{1+(4t)^{2}} dt \qquad v=t }

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} \pi\int_1^2\left[\left(2-\frac{1}{2}x\right)^2\right]dx & = \pi\int_1^2\left[\left(4-2x+\frac{1}{4}x^2\right)\right]dx \\[2ex] &= \pi\left[4x-x^2+\frac{1}{12}x^3\right]\Bigg|_1^2 \\[2ex] &= \pi\left[\left(4(2)-(2)^2+\frac{1}{12}(2)^3\right)-\left(4(1)-(1)^2+\frac{1}{12}(1)^3\right)\right] \\[2ex] &= \pi\left[\left(8-4+\frac{8}{12}\right)-\left(4-1+\frac{1}{12}\right)\right] \\[2ex] &= \pi\left[4+\frac{8}{12}-3-\frac{1}{12}\right]= \pi\left[1+\frac{7}{12}\right] \\[2ex] &= \pi\left[\frac{12}{12}+\frac{7}{12}\right]= \pi\left[\frac{19}{12}\right] \\[2ex] &= \frac{19\pi}{12} \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 \int (xe^{-x}) dx = x*(-e^{-x}) - \int (-e^{-x}) }