6.5 Average Value of a Function/17 - Revision history

Merlynm101615@students.laalliance.org at 17:25, 1 December 2022

2022-12-01T17:25:42Z

← Older revision		Revision as of 17:25, 1 December 2022
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	In a certain city the temperature (in ~~\text{F}^{\circ}~~) t hours after a 9 AM was modeled by the function		In a certain city the temperature (in Fahrenheit) t hours after a 9 AM was modeled by the function

	<math>		<math>
	T(t)=50+14\sin(\frac{\pi}{12}t)		T(t)=50+14\sin(\frac{\pi}{12}t)
	</math>		</math>
	Find the average temperature during the period 9 AM to 9 PM \\
			Find the average temperature during the period 9 AM to 9 PM

	1. Use the Average Value from a to b:		1. Use the Average Value from a to b:

Merlynm101615@students.laalliance.org at 17:22, 1 December 2022

2022-12-01T17:22:18Z

← Older revision		Revision as of 17:22, 1 December 2022
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	<math>		<math>
	\frac{1}{12-0}\int_{0}^{12} 50+14\sin\left(\frac{\pi}{12}t\right)=\frac{1}{12}\int_{0}^{12} 50 + \underbrace{14\sin\left(\frac{\pi}{12}t\right)}_{		\frac{1}{12-0}\int_{0}^{12} 50+14\sin\left(\frac{\pi}{12}t\right)= \frac{1}{12}\int_{0}^{12} 50 + \underbrace{14\sin\left(\frac{\pi}{12}t\right)}_{
	\begin{aligned}		\begin{aligned}
	u &= \frac{\pi}{12}t\\dt\cdot\frac{du}{dt} &= dt\\ \frac{12}{\pi}du &= dt \\ integrate for\, 14\sin(u)\frac{12}{\pi}\\ \int14\sin(u)\frac{12}{\pi}\,du		u &= \frac{\pi}{12}t\\dt\cdot\frac{du}{dt} &= dt\\ \frac{12}{\pi}du &= dt \\ integrate for\, 14\sin(u)\frac{12}{\pi}\\ \int14\sin(u)\frac{12}{\pi}\,du

Merlynm101615@students.laalliance.org at 17:21, 1 December 2022

2022-12-01T17:21:51Z

← Older revision		Revision as of 17:21, 1 December 2022
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	<math>		<math>
	f_{\text{avg}} = \frac{1}{b-a}\int_{a}^{b}f(x)\,dx		f_{\text{avg}} = \frac{1}{b-a}\int_{a}^{b}f(x)\,dx
	</math>		</math>
	a= 0 (start at 9 AM) b= 12 (From 9 AM to 9 PM)
			a=0 (start at 9 AM) b=12 (From 9 AM to 9 PM)

	<math>		<math>
	\frac{1}{12}\int_{0}^{12} 50 + \underbrace{14\sin\left(\frac{\pi}{12}t\right)}_{		\frac{1}{12-0}\int_{0}^{12} 50+14\sin\left(\frac{\pi}{12}t\right)=\frac{1}{12}\int_{0}^{12} 50 + \underbrace{14\sin\left(\frac{\pi}{12}t\right)}_{
	\begin{aligned}		\begin{aligned}
	u &= \frac{\pi}{12}t\\dt\cdot\frac{du}{dt} &= dt\\ \frac{12}{\pi}du &= dt \\ integrate for\, 14\sin(u)\frac{12}{\pi}\\ \int14\sin(u)\frac{12}{\pi}\,du		u &= \frac{\pi}{12}t\\dt\cdot\frac{du}{dt} &= dt\\ \frac{12}{\pi}du &= dt \\ integrate for\, 14\sin(u)\frac{12}{\pi}\\ \int14\sin(u)\frac{12}{\pi}\,du

Merlynm101615@students.laalliance.org at 17:19, 1 December 2022

2022-12-01T17:19:40Z

← Older revision		Revision as of 17:19, 1 December 2022
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	In a certain city the temperature (in \text{F}^{\circ}) t hours after a ~~9AM~~ was modeled by the function		In a certain city the temperature (in \text{F}^{\circ}) t hours after a 9 AM was modeled by the function

	<math>		<math>
	T(t)=50+14\sin(\frac{\pi}{12}t)		T(t)=50+14\sin(\frac{\pi}{12}t)
	</math>		</math>
			Find the average temperature during the period 9 AM to 9 PM \\
	1. Use the Average Value from a to b:		1. Use the Average Value from a to b:

Line 9:		Line 9:
	f_{\text{avg}} = \frac{1}{b-a}\int_{a}^{b}f(x)\,dx		f_{\text{avg}} = \frac{1}{b-a}\int_{a}^{b}f(x)\,dx
	</math>		</math>
			a= 0 (start at 9 AM) b= 12 (From 9 AM to 9 PM)

	<math>		<math>

Merlynm101615@students.laalliance.org at 17:16, 1 December 2022

2022-12-01T17:16:37Z

← Older revision		Revision as of 17:16, 1 December 2022
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	In a certain city the temperature (in \text{F}^{\circ}) t hours after a 9AM was modeled by the function		In a certain city the temperature (in \text{F}^{\circ}) t hours after a 9AM was modeled by the function

	<math>		<math>
	T(t)=50+14\sin(\frac{\pt}{12}t)		T(t)=50+14\sin(\frac{\pi}{12}t)
	</math>		</math>
	1. Use the Average Value from a to b:		1. Use the Average Value from a to b:

Merlynm101615@students.laalliance.org at 17:16, 1 December 2022

2022-12-01T17:16:16Z

← Older revision		Revision as of 17:16, 1 December 2022
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			In a certain city the temperature (in \text{F}^{\circ}) t hours after a 9AM was modeled by the function
			<math>
			T(t)=50+14\sin(\frac{\pt}{12}t)
			</math>
	1. Use the Average Value from a to b:		1. Use the Average Value from a to b:

Merlynm101615@students.laalliance.org at 17:10, 1 December 2022

2022-12-01T17:10:20Z

← Older revision		Revision as of 17:10, 1 December 2022
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	<math>		<math>
	=\frac{1}{12}[600-\frac{168}{\pi}(-1)+\frac{168}{\pi}(1)] =\frac{1}{12}[600+\frac{168}{\pi}+\frac{168}{\pi}]=\frac{1}{12}[600+\frac{336}{\pi}]= 50+\frac{336}{12\pi}=50+\frac{28}{\pi}= 59		=\frac{1}{12}[600-\frac{168}{\pi}(-1)+\frac{168}{\pi}(1)] =\frac{1}{12}[600+\frac{168}{\pi}+\frac{168}{\pi}]=\frac{1}{12}[600+\frac{336}{\pi}]= 50+\frac{336}{12\pi}=50+\frac{28}{\pi}= 59\text{F}^{\circ}
	</math>		</math>

Merlynm101615@students.laalliance.org at 21:09, 29 November 2022

2022-11-29T21:09:10Z

← Older revision		Revision as of 21:09, 29 November 2022
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	<math>		<math>
	\frac{1}{12}\int_{0}^{12} 50 + \underbrace{14\sin(\frac{\pi}{12}t)}\~~,dt~~ =\frac{1}{12}~~[50t-~~\frac{~~168~~}{\pi}\~~cos~~(\frac{\pi}~~{12}t~~)]\~~bigg\|_~~{12}^{0}=\frac{1}{12}[(~~50)(12~~)-\frac{168}{\pi}\cos(\~~pi))(0~~-\frac{168}{\pi}\cos(0)]=\frac{1}{12}[~~600~~-\frac{168}{\pi}(~~-1)+~~\frac{~~168~~}{~~\pi~~}(1)] =\frac{1}{12}[~~600+~~\frac{168}{\pi}+\frac{168}{\pi}]		\frac{1}{12}\int_{0}^{12} 50 + \underbrace{14\sin\left(\frac{\pi}{12}t\right)}_{
			\begin{aligned}
			u &= \frac{\pi}{12}t\\dt\cdot\frac{du}{dt} &= dt\\ \frac{12}{\pi}du &= dt \\ integrate for\, 14\sin(u)\frac{12}{\pi}\\ \int14\sin(u)\frac{12}{\pi}\,du
			14\cdot\frac{12}{\pi}\int\sin(u)\,du \\ -\frac{168}{\pi}\cos(u) \\ -\frac{168}{\pi}\cos(\frac{\pi}{12}t)
			\end{aligned}}

			\,dt =\frac{1}{12}[50t-\frac{168}{\pi}\cos(\frac{\pi}{12}t)]\bigg\|_{12}^{0}=\frac{1}{12}[(50)(12)-\frac{168}{\pi}\cos(\pi))(0-\frac{168}{\pi}\cos(0)]
	</math>		</math>


	~~<math>~~
	~~\begin{aligned}~~

	~~u &= \frac{\pi}{12}t \\~~
	~~dt\cdot\frac{du}{dt} &= dt \\~~
	~~\frac{12}{\pi}du &= dt \\~~
	~~integrate for\, 14\sin(u)\frac{12}{\pi} \\~~
	~~\int14\sin(u)\frac{12}{\pi}\,du \\~~
	~~14\cdot\frac{12}{\pi}\int\sin(u)\,du \\~~
	~~-\frac{168}{\pi}\cos(u) \\~~
	~~-\frac{168}{\pi}\cos(\frac{\pi}{12}t) \\~~
	~~\end{aligned}~~
	~~</math>~~


	<math>		<math>
	=\frac{1}{12}[600+\frac{336}{\pi}]= 50+\frac{336}{12\pi}=50+\frac{28}{\pi}= 59		=\frac{1}{12}[600-\frac{168}{\pi}(-1)+\frac{168}{\pi}(1)] =\frac{1}{12}[600+\frac{168}{\pi}+\frac{168}{\pi}]=\frac{1}{12}[600+\frac{336}{\pi}]= 50+\frac{336}{12\pi}=50+\frac{28}{\pi}= 59
	</math>		</math>

Merlynm101615@students.laalliance.org at 20:42, 29 November 2022

2022-11-29T20:42:29Z

← Older revision		Revision as of 20:42, 29 November 2022
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	<math>		<math>
	=\frac{1}{12}[600+\frac{336}{\pi}]=		=\frac{1}{12}[600+\frac{336}{\pi}]= 50+\frac{336}{12\pi}=50+\frac{28}{\pi}= 59
	</math>		</math>

Merlynm101615@students.laalliance.org at 20:37, 29 November 2022

2022-11-29T20:37:12Z

← Older revision		Revision as of 20:37, 29 November 2022
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	\begin{aligned}		\begin{aligned}

	u &= \frac{\pi}{12}t \\		u &= \frac{\pi}{12}t \\
	dt\cdot\frac{du}{dt} &= dt \\		dt\cdot\frac{du}{dt} &= dt \\
	\frac{12}{\pi}du &= dt \\		\frac{12}{\pi}du &= dt \\
	integrate for 14\sin(u)\frac{12}{\pi} \\		integrate for\, 14\sin(u)\frac{12}{\pi} \\
	\int14\sin(u)\frac{12}{\pi}\,du \\		\int14\sin(u)\frac{12}{\pi}\,du \\
	14\cdot\frac{12}{\pi}\int\sin(u)\,du \\		14\cdot\frac{12}{\pi}\int\sin(u)\,du \\
	-\frac{168}{\pi}\cos(u) \\		-\frac{168}{\pi}\cos(u) \\
	-\frac{168}{\pi}\cos(\frac{\pi}{12}t) \\		-\frac{168}{\pi}\cos(\frac{\pi}{12}t) \\
	\end{aligned}		\end{aligned}
			</math>


			<math>
			=\frac{1}{12}[600+\frac{336}{\pi}]=
	</math>		</math>