7.1 Integration By Parts/65 - Revision history

Vanessav73764@students.laalliance.org at 10:49, 16 December 2022

2022-12-16T10:49:56Z

← Older revision		Revision as of 10:49, 16 December 2022
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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]

Vanessav73764@students.laalliance.org at 10:02, 16 December 2022

2022-12-16T10:02:38Z

← Older revision		Revision as of 10:02, 16 December 2022
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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]

Vanessav73764@students.laalliance.org at 09:46, 16 December 2022

2022-12-16T09:46:13Z

← Older revision		Revision as of 09:46, 16 December 2022
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	x\cdot\;f'(x)\bigg\|_{1}^{4}-\int_{1}^{4}f'(x)dx &= x\cdot\;f'(x)-f(x)\bigg\|_{1}^{4}\\[2ex]		x\cdot\;f'(x)\bigg\|_{1}^{4}-\int_{1}^{4}f'(x)dx &= x\cdot\;f'(x)-f(x)\bigg\|_{1}^{4}\\[2ex]

	&= (4\cdot\;f'(x)-f(4))-(1\cdot\;f'(1)-f(1))\\[2ex]		&= (4\cdot\;f'(4)-f(4))-(1\cdot\;f'(1)-f(1))\\[2ex]

	&= (4\cdot\;3-7)-(5-2) \\[2ex]		&= (4\cdot\;3-7)-(5-2) \\[2ex]

Vanessav73764@students.laalliance.org at 09:45, 16 December 2022

2022-12-16T09:45:13Z

← Older revision		Revision as of 09:45, 16 December 2022
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	x\cdot\;f'(x)\bigg\|_{1}^{4}-\int_{1}^{4}f'(x)dx &= x\cdot\;f'(x)-f(x)\bigg\|_{1}^{4}\\[2ex]		x\cdot\;f'(x)\bigg\|_{1}^{4}-\int_{1}^{4}f'(x)dx &= x\cdot\;f'(x)-f(x)\bigg\|_{1}^{4}\\[2ex]

	&= (4\cdot\;f'(x)-f(4))-(1\cdot\;f'(1)-r(1))\\[2ex]		&= (4\cdot\;f'(x)-f(4))-(1\cdot\;f'(1)-f(1))\\[2ex]

	&= (4\cdot\;3-7)-(5-2) \\[2ex]		&= (4\cdot\;3-7)-(5-2) \\[2ex]

Vanessav73764@students.laalliance.org: Created page with " $\begin{align} \int_{1}^{4}xf''(x)dx\\[2ex] u=x\qquad dv=f''\cdot\;(x)\qquad \\[2ex] du=dx\qquad v=f'(x)\qquad \\[2ex] x\cdot\;f'(x)\bigg|_{1}^{4}-\int_{1}^{4}f'(x)dx &= x\cdot\;f'(x)-f(x)\bigg|_{1}^{4}\\[2ex] &= (4\cdot\;f'(x)-f(4))-(1\cdot\;f'(1)-r(1))\\[2ex] &= (4\cdot\;3-7)-(5-2) \\[2ex] &= (12-7)-(3)\\[2ex] &=(5)-(3)\\[2ex] &=2 \end{align}$ "

2022-12-16T09:44:54Z

Created page with "<math> \begin{align} \int_{1}^{4}xf''(x)dx\\[2ex] u=x\qquad dv=f''\cdot\;(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] &= (4\cdot\;f'(x)-f(4))-(1\cdot\;f'(1)-r(1))\\[2ex] &= (4\cdot\;3-7)-(5-2) \\[2ex] &= (12-7)-(3)\\[2ex] &=(5)-(3)\\[2ex] &=2 \end{align} </math>"

New page

<math>
\begin{align}

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

&= (4\cdot\;f'(x)-f(4))-(1\cdot\;f'(1)-r(1))\\[2ex]

&= (4\cdot\;3-7)-(5-2) \\[2ex]

&= (12-7)-(3)\\[2ex]

&=(5)-(3)\\[2ex]

&=2

\end{align}
</math>