7.1 Integration By Parts/65: Difference between revisions

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\int_{1}^{4}xf''(x)dx\\[2ex]
\int_{1}^{4}xf''(x)dx\\[2ex]
u=x\qquad dv=f''\cdot\;(x)\qquad \\[2ex]
u=x\qquad dv=f''(x)\cdot\;(dx)\qquad \\[2ex]
du=dx\qquad v=f'(x)\qquad \\[2ex]
du=dx\qquad v=f'(x)\qquad \\[2ex]



Latest revision as of 10:49, 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''(x)\cdot \;(dx)\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}}}