7.1 Integration By Parts/65: Difference between revisions
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du=dx\qquad v=f'(x)\qquad \\[2ex] | du=dx\qquad v=f'(x)\qquad \\[2ex] | ||
x\cdot\;f'(x)\bigg|_{1}^{4}-\int_{1}^{4}f'(x)dx &= x\cdot\;f'(x)-f(x)\bigg|_{1}^{4}\\[2ex] | \int_{1}^{4}xf''(x)dx&=x\cdot\;f'(x)\bigg|_{1}^{4}-\int_{1}^{4}f'(x)dx \\[2ex] | ||
&= x\cdot\;f'(x)-f(x)\bigg|_{1}^{4}\\[2ex] | |||
&= (4\cdot\;f'(4)-f(4))-(1\cdot\;f'(1)-f(1))\\[2ex] | &= (4\cdot\;f'(4)-f(4))-(1\cdot\;f'(1)-f(1))\\[2ex] | ||
Revision as of 10:02, 16 December 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{1}^{4}xf''(x)dx\\[2ex]u=x\qquad dv=f''\cdot \;(x)\qquad \\[2ex]du=dx\qquad v=f'(x)\qquad \\[2ex]\int _{1}^{4}xf''(x)dx&=x\cdot \;f'(x){\bigg |}_{1}^{4}-\int _{1}^{4}f'(x)dx\\[2ex]&=x\cdot \;f'(x)-f(x){\bigg |}_{1}^{4}\\[2ex]&=(4\cdot \;f'(4)-f(4))-(1\cdot \;f'(1)-f(1))\\[2ex]&=(4\cdot \;3-7)-(5-2)\\[2ex]&=(12-7)-(3)\\[2ex]&=(5)-(3)\\[2ex]&=2\end{aligned}}}