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 October 30th, 2009, 06:57 PM #1 Newbie   Joined: Oct 2009 Posts: 4 Thanks: 0 Area bounded between 2 curves Consider the area bounded between curves y=-2x and y=3-x^{2}. If 2 vertical lines that are a unit apart intersect this certain bounded area, where should they be placed so that they contain the most amount of the area between them? What's the max. area? I keep getting told that this is the rule I am supposed to use: But I am unsure of how to apply the rule here. $A = \int_a^b (f(x) - g(x)) \, dx A = \int_a^b (-2x - 3-x^{2}) \, dx$ Ok so the latex isn't working ... int a - b (-2x - 3-x^2) dx But then how to proceed?
October 30th, 2009, 08:01 PM   #2
Math Team

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Re: Area bounded between 2 curves

Hello, Civilization!

Quote:
 $\text{Consider the area bounded between curves: }\:y\:=\:-2x\,\text{ and }\,y\:=\:3-x^2$ $\text{If two vertical lines that are a unit apart intersect this bounded area,}$ $\text{where should they be placed so that they contain the most amount of the area between them?}$ $\text{What is the maximum area?}$

The graph looks like this:
Code:
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$\text{The area is: }\;A \;=\;\int^{\;\;\;a+1}_a\:\bigg[(3\,-\,x^2)\, \,- (-2x)\bigg]\,dx \;=\;\int^{\;\;\;a+1}_a\:\bigg[3 \,-\, x^2 \,+\, 2x\bigg]\,dx$

[color=beige]. . [/color]$A \;=\;3x \,-\,\frac{1}{3}x^3 \,+\, x^2\,\bigg]^{a+1}_a \;=\;\bigg[3(a+1) \,-\, \frac{1}{3}(a+1)^3 \,+\, (a+1)^2\bigg] \,-\, \bigg[3a \,-\, \frac{1}{3}a^3 \,+\, a^2\bigg]$

$\text{which simplifies to: }\;A \;=\;-a^2 \,+\, a \,+\, \frac{11}{3}$

$\text{And now you can maximize the area . . .}$

But check my work . . . please!

 October 31st, 2009, 08:34 AM #3 Newbie   Joined: Oct 2009 Posts: 4 Thanks: 0 Re: Area bounded between 2 curves Whoa, very nice diagram. I actually stopped looking at your post once you typed in (3-x^2 etc.) in the integral from a to a+1. It was easy for me to figure out what the A equals to after having the equation written. Do you have any tips on how to actually start an integration problem like this? Obviously on a test I won't have any help. After getting the right integrand, integration becomes easier. Anyway, A' = 1-2a 2a = 1 a = 1/2 So the lines should be placed at x = 1/2 and x = 3/2, right?

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