Area Between two curves

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  • #1
just.karl
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Find the area of the region between the following curves for x in [0, 3]. Give the answer to three decimal places.
y=x
y=4e^x

I understand how to do these types of problems but I always get confused when there is an e involved. If someone could explain how to arrange it before the anti derivative or show how to do a very similar problem that would help me greatly.
 

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  • #2
HallsofIvy
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That's probably the world's simplest function! The derivative of ex is ex and the anti-derivative of ex is ex!

"Before" taking the anti-derivative, the only thing you need to know is that when x= 0, e0= 1> 0 and when x= 4, e4= 54> 4: ex is always greater than x so you integrate ex- x from 0 to 4.
 
  • #3
just.karl
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so would the final equation be e^4-.5(4^2) ?
 
  • #4
Dick
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so would the final equation be e^4-.5(4^2) ?

Just because Halls makes a mistake doesn't mean you should blindly echo it. The region of integration is [0,3]. Find the antiderivative and evaluate it between x=3 and x=0. NOT x=4. And the antiderivative isn't e^x-x^2/2 either. There a factor 4 in front of e^x. What happened to it?
 
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  • #5
just.karl
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Well my thought is that the 4 would stay where it was. So it would be 4e^x-x^2/2 but that doesn't come out to be the right answer. Could you just walk me through a problem start to finish so I can figure out how to do it? Thank you
 
  • #6
Dick
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Put in x=3 and x=0 and then subtract the two. I don't see how I can walk you through in any more detail than that. What do you get? Hint: e^0=1.
 
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  • #7
HallsofIvy
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:redface::redface::redface::redface::redface:
 
  • #8
tmclary
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Remember to use two integrals...

Your first integral evaluates 4e^x from 0 to 3, then subtract the integral of x (dx) from 0 to 3 (since it's the area between the curves, and 4e^x is always greater than y=x.) Your answer should then be (4e^3-4e^0) - ((3^2)/2 -(0^2)/2) = 4e^3-4-(9/2) = 4e^3 -(7/2).
 
  • #9
tron_2.0
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right. it helps greatly when you graph both equations first. it would just be the integral of the function that is "highter" on the interval [0,3] minus the function that is "lower" on the same interval.

for instance when you graph y=4e^x and y=x you see that y=4e^x is above y=x so you would set up the integral as:

the integral of (4e^x)-(x)dx evaluated from 0 to 3 (or whatever your integral is)
 
  • #10
BrendanH
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Well, I greatly struggled with Latex to produce the elegant looking double integral that would solve this problem, but it would be entirely unnecessary as you guys have essentially done the problem. So I won't post it, making THIS post superfluous.
 
  • #11
just.karl
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O.. I think why I couldn't get the right answer was I didn't realized that e^0 was 1. I was thinking when it was evaluated at 0 it just = 0. lol Thanks for all of your help! I knew it was just some stupid thing I was doing.

Not sure how latex has to do with this? Is it a program?
 
  • #12
BrendanH
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Latex is the language that allows users to insert math symbols. Writing out and explaining double integrals (let alone triple integrals) proves quite a prolixity.
 

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