SOLUTION: Please help me solve this equation: Sketch the regions enclosed by the given curves and then find the area of the regions. a. {{{ y=e^x }}} {{{ y=(x-1)^2 }}} and {{{ x=2 }}}

Algebra ->  Test -> SOLUTION: Please help me solve this equation: Sketch the regions enclosed by the given curves and then find the area of the regions. a. {{{ y=e^x }}} {{{ y=(x-1)^2 }}} and {{{ x=2 }}}       Log On


   

Question 350018: Please help me solve this equation: Sketch the regions enclosed by the given curves and then find the area of the regions.
a. +y=e%5Ex+ +y=%28x-1%29%5E2+ and +x=2+
b. +y=-7-x%2B3x%5E2+ and +y=2x-x%5E2+
c. +y=1=cosx+ and +y=cosx+ +0%3C=x%3C=pi+
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
graph%28300%2C300%2C-2%2C4%2C-2%2C4%2Ce%5E%28x%29%2C%28x-1%29%5E2%2C2%29
Find the intersection point between y=e%5Ex and y=2
e%5Ex=2
x=ln%282%29
So then the area would be,

Expand the terms, integrate, and simplify to get the final answer.
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graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C-7-x%2B3x%5E2%2C2x-x%5E2%29
Same procedure.
Find the two intersection points x1 and x2 by solving,
-7-x%2B3x%5E2=2x-x%5E2
Proceed as above integrating from one intersection point to the other.

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graph%28300%2C300%2C-1%2C4%2C-3%2C3%2C1%2Bcos%28x%29%2C1-cos%28x%29%2Ccos%28x%29%29
I'm not sure if the first function is 1%2Bcos%28x%29 or 1-cos%28x%29
If it's 1%2Bcos%28x%29, then the integration is straightforward since
1%2Bcos%28x%29-cos%28x%29=1
Integrating that over x=0 to x=pi would give you A=%28pi-0%29=pi
If it's 1-cos%28x%29, then find the intersection point x1 again where,
1-cos%28x%29=cos%28x%29
Then you have to modify the area because at this point the functions cross.
A=int%28%281-2cos%28x%29%29%2Cdx%2C0%2Cx1%29%2Bint%28%282cos%28x%29-1%29%2Cdx%2Cx1%2Cpi%29