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Finding the antiderivative

  • Thread starter nhmllr
  • Start date
  • #1
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1

Homework Statement


As part of a bigger problem, I am trying to find the area under the curve
f(x) = sqrt{2x - x2} between x = 0 and x = 2. To do this, I have to find the antiderivative of f(x)

Homework Equations


antiderivative
f(x) = axb
F(x) = a/(b+1) * xb+1

chain rule
f(x) = a(b(x))
f'(x) = b'(x) * a'(b(x))

The Attempt at a Solution


I took the derivative of the terms in the parenthesis, then what was outside the parenthesis, to get
2/3 * (x2 - x3/3)3/2
I don't think this is right, because you can use the chain rule to find the derivative, and it's not the original function.

NOTE- I'm not actually old enough and am not actual in a calculus course right now, but I figure that the homework forum would be the best place to put this. Also, don't give me answer, but rather gove me a direction to go in.
 
Last edited:

Answers and Replies

  • #2
33,485
5,174
The best approach, short of looking at a table of integrals, is trig substitution, which you might not have seen yet. Before doing the substitution you need to do some work first, by completing the square.

[tex]\sqrt{2x - x^2} = \sqrt{-(x^2 - 2x)} = \sqrt{-(x^2 - 2x + 1) + 1} = \sqrt{1 - (x - 1)^2}[/tex]

The next step is picking the right trig substitution.
 

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