# Calculus

find the area between the two curves y=1/2x and y=x square root of 1-x^2

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1. first we have to know where they intersect, so

(1/2)x = x√(1-x^2)
times 2
x - 2x√(1-x^2) = 0
x (1 - 2√(1-x^2) = 0
x = 0 or x = ±√3/2

I made a quick sketch and found two symmetrical regions, so let's just take it from 0 to √3/2 and double that answer.

Area = ∫x(1-x^2)^.5 - x dx from 0 to √3/2
= [ -(1/3)(1-x^2)^(3/2) - (1/2)x^2 ] from 0 to √3/2
= ...

I will let you do the messy arithmetic

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2. i don't understand how to integrate x(1-x^2)^.5

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3. First, I think Reiny dropped a factor of 1/2 in his integral (-x/2, not -x), but that's not a big worry (except that the answer is wrong!) :-)

To integrate x√(1-x^2), substitute u=1-x^2

Then you have du = 2x dx

x√(1-x^2) = √u du/2 = 1/2 u^(1/2) du
integrate that to get 1/3 u^(3/2)

so, ∫x(1-x^2)^.5 = 1/3 (1-x^2)^(3/2)

After all the messy arithmetic, you should wind up with 5/48

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4. oops. dropped the minus sign, which Reiny slyly retained. . .

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5. i don't get 5/48 when i plug in root 3/2

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