7.1 Integration By Parts/24 - Revision history

Merlynm101615@students.laalliance.org at 20:31, 1 December 2022

2022-12-01T20:31:57Z

← Older revision		Revision as of 20:31, 1 December 2022
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	du&=6 \quad \quad v=sin(x)		du&=6 \quad \quad v=sin(x)
	\end{aligned}}		\end{aligned}}
	=x^3\sin(x)+3x^2\cos(x)-\bigg[6x\sin(x)-\int_{0}^{\pi} 6\sin(x)\,dx\bigg]= x^3sin(x)+3x^2\cos(x)-6x\sin(x)-6\cos(x)\|_{0}^{\pi}		=x^3\sin(x)+3x^2\cos(x)-\bigg[6x\sin(x)-\int_{0}^{\pi} 6\sin(x)\,dx\bigg]= x^3sin(x)+3x^2\cos(x)-6x\sin(x)-6\cos(x)\bigg\|_{0}^{\pi}
			\end{align}
			</math>
			<math>
			\begin{align}
			&=[(\pi^3\cdot\sin(\pi)+3(\pi)^2\cdot\cos(\pi)-6(\pi)\cdot\sin(\pi)-6\cos(\pi))-(0^3\cdot\sin(0)+3(0)^2\cdot\cos(0)-6(0)\cdot\sin(0)-6\cos(0))] \\
			&=[(\pi\cdot 0 +3\pi^2 \cdot -1 -6 \cdot -1)+6 =-3\pi^2+6+6 = -3\pi^2+12 = -17.60
	\end{align}		\end{align}
	</math>		</math>

2022-12-01T20:13:07Z

← Older revision		Revision as of 20:13, 1 December 2022
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	du&=6 \quad \quad v=sin(x)		du&=6 \quad \quad v=sin(x)
	\end{aligned}}		\end{aligned}}
	=x^3\sin(x)+3x^2\cos(x)-\bigg[6x\sin(x)-\int_{0}^{\pi} 6\sin(x)\,dx\bigg]		=x^3\sin(x)+3x^2\cos(x)-\bigg[6x\sin(x)-\int_{0}^{\pi} 6\sin(x)\,dx\bigg]= x^3sin(x)+3x^2\cos(x)-6x\sin(x)-6\cos(x)\|_{0}^{\pi}
	\end{align}		\end{align}
	</math>		</math>

2022-12-01T20:09:22Z

← Older revision		Revision as of 20:09, 1 December 2022
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	du&=6x \quad \quad v=-\cos(x)		du&=6x \quad \quad v=-\cos(x)
	\end{aligned}}		\end{aligned}}
	\,dx= x^3\sin(x)-[3x^2-\cos(x)-\int_{0}^{\pi}-6x\cos(x)\,dx]\\		\,dx= x^3\sin(x)-\bigg[3x^2-\cos(x)-\int_{0}^{\pi}-6x\cos(x)\,dx\bigg]\\
	=&x^3\sin(x)-3x^2\cos(x)-\int_{0}^{\pi}6x\cos(x)		=&x^3\sin(x)-3x^2\cos(x)-\int_{0}^{\pi}\underbrace{6x\cos(x)}_{
			\begin{aligned}
			u&=6x \quad \quad dv=cos(x) \\
			du&=6 \quad \quad v=sin(x)
			\end{aligned}}
			=x^3\sin(x)+3x^2\cos(x)-\bigg[6x\sin(x)-\int_{0}^{\pi} 6\sin(x)\,dx\bigg]
	\end{align}		\end{align}
	</math>		</math>

2022-12-01T19:45:07Z

2022-12-01T19:11:39Z

← Older revision	Revision as of 19:11, 1 December 2022
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		<math>
		\begin{align}

		&\int_{0}^{\pi}\underbrace{x^3\cos(x)}_{
		\begin{aligned}
		u&=x^3 \quad \quad dv=\cos(x) \\
		dv&=3x^2 \quad \quad v=\sin(x)
		\end{aligned}}
		\,dx =x^3\sin(x)-\int_{0}^{\pi} \underbrace{3x^2\sin(x)}_{
		\begin{aligned}
		u&=3x^2 \quad \quad dv=\sin(x) \\
		du&=6x \quad \quad v=-\cos(x)
		\end{aligned}}
		\,dx= x^3\sin(x)-[3x^2-\cos(x)-\int_{0}^{\pi}-6x\cos(x)\,dx]\\
		=&x^3\sin(x)-3x^2\cos(x)-\int_{0}^{\pi}6x\cos(x)
		\end{align}
		</math>

2022-12-01T18:00:19Z

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2022-12-01T17:38:26Z

2022-12-01T17:37:51Z

2022-12-01T17:37:43Z

Created page with "<math> \int u dv=uv-\int vd u"

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<math>
\int u dv=uv-\int vd u