Finding the antiderivative

[nhmllr]

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) = ax^b
F(x) = a/(b+1) * x^b+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.

 

[Mark44]

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.

$$\sqrt{2x - x^2} = \sqrt{-(x^2 - 2x)} = \sqrt{-(x^2 - 2x + 1) + 1} = \sqrt{1 - (x - 1)^2}$$

The next step is picking the right trig substitution.