Intergral problem! !

  • Thread starter Learner123
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  • #1
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Intergral problem! plz help!

Homework Statement


[tex]\oint[/tex](x: 0 to 1)[tex]\oint[/tex](y: [tex]\sqrt{}(1 - x^2)[/tex] to e[tex]\overline{}x[/tex]) xydydx

The region bounded by y = e[tex]\overline{}x[/tex], y = [tex]\sqrt{}(1 - x^2)[/tex], and x =1
3. The Attempt at a Solution
i got stuck when i came to the part: 1/2 [tex]\oint[/tex](x: 0 to 1) (e^(2x) -1 + x^2)xdx
i appreciate any help
 

Answers and Replies

  • #2
1,015
70


The symbol you are using is the symbol for a closed line integral. You should be using a normal integral sign: [tex]\int[/tex].
Otherwise, since the integral is a linear operator, you have the following sum of integrals:
[tex]\frac{1}{2}\left(\int xe^{2x} dx - \int x dx + \int x^3 dx\right)[/tex]
Which one is giving you a problem?
 
  • #3
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the first one xe^(2x) thing
i guess it's intergral by part, but not sure
 
  • #4
16
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I tried to do part and this is how i done (for the first intergral):
u = x, du = dx, v = 1/2e^(2x), dv = e^(2x)dx
uv - [tex]\int[/tex] vdu
1/2xe^(2x) - [tex]\int[/tex] 1/2e^(2x)dx
1/2xe^(2x) - 1/4(e^2 -1 )
x runs from 0 to 1, but 1/2xe^(2x) is not in the intergral part, so how to eliminate x?
very appriciate for more help!
 
  • #5
1,015
70


1/2xe^(2x) - 1/4(e^2 -1 )
This entire expression is the indefinite integral; the entire expression must be evaluated at the endpoints of the integral if the integral is definite.
 
  • #6
16
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got it! i didn't know that after spending 3 calculus classes, what a shame of me! thank you so much for your help and your time.
 

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