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Integral Problem

  • Thread starter Miike012
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  • #1
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Find area of: ∫ (1 + √(9 - x^2))dx [-3,0]

Solution:

∫ (1 + √(9 - (x - 3)^2))dx [0,3]

= ∫ (1 + √(x)*√(6-x))dx [0,3]

Δx = 3/n

Ʃ√(3i/n)*Ʃ√(6 - 3i/n) +Ʃ 1 i = 1, n = n

I am stuck right here... I know that the notation for a constant is 1(n)..

but is there one for a square root? Would is be (n(n+1)/2)^(1/2) ???
 

Answers and Replies

  • #2
Dick
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No, it wouldn't. I don't think you want to try and do this one with Riemann sums. That gets to be way too hard. There's an easier way. Do you know what an antiderivative is?
 
  • #3
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No, it wouldn't. I don't think you want to try and do this one with Riemann sums. That gets to be way too hard. There's an easier way. Do you know what an antiderivative is?
Yes I know what that is how would I use the antiderivative to solve this?
 
  • #4
Dick
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Yes I know what that is how would I use the antiderivative to solve this?
Split the integral into two parts. The 1 part should be easy. The second one you can do with a trig substitution, if you've cover that. If you haven't, maybe you are expected to use geometry. What does the graph y=sqrt(9-x^2) look like?
 
  • #5
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I know all my trig derivatives...

the two parts are...

1 + (u)^(1/2)
u = 9 - x^2
Is this what your asking for?

And here is a rough sketch of the graph...
 

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  • #6
Dick
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I know all my trig derivatives...

the two parts are...

1 + (u)^(1/2)
u = 9 - x^2
Is this what your asking for?

And here is a rough sketch of the graph...
Just that substitution won't quite do it. You need something like x=3*sin(t). And the graph is a good start. The more you improve it, the more it will look like a half-circle.
 
Last edited:

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