My Math Forum Area bounded between 2 curves

 Calculus Calculus Math Forum

 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

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 407

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:
                   |
*
*    |    *
*    *   |    |:::*
- - * - - - - - * - -|:::|- * - - - -
*             |   *::::|    *
*              |       *|     *
|           *
*               |               *
|

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

[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?

 Tags area, bounded, curves

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post PhizKid Calculus 0 December 5th, 2012 05:59 PM Trafford99 Calculus 2 June 14th, 2012 10:42 AM aaron-math Calculus 5 October 10th, 2011 02:36 PM FreaKariDunk Calculus 5 October 2nd, 2011 08:32 PM Feyenoord87 Calculus 1 December 1st, 2007 05:42 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top