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Home Q & A Board Science-Related Homework Help Mathematics
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snowbaby213 snowbaby213
wrote...
10 years ago
I can take the antiderivative of f(x) = (3 - 2x)^0.5 no problem; it equals 1/(1.5*-2) * (3 - 2x)^1.5, and with a bit of simplification it becomes -1/3 * (3 - 2x)^(3/2).

However, if I try to take the antiderivative of f(x) = (3 - 2x^2)^0.5, I get: 1/(1.5*-4x) * (3 - 2x^2)^1.5 -> -1/(6x) * (3 - 2x^2)^1.5. If I were to find the derivative of this function, I get back to the original function, but when I graph it on my calculator and take its derivative, it does NOT equal the original function for a given x value. What's wrong?
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wrote...
#1
10 years ago
There is no "chain rule for antiderivatives" that I know of. I'm not sure what technique you're using to find the antiderivative of (3 - 2x^2)^0.5. I think you may be inventing a theorem that doesn't actually exist, and that's why you aren't getting the correct answer.

In reality I think the antiderivative of this function may be an inverse trig function, but I'd have to look it up.

Edit: According to Wolfram Alpha the integral is (1/4) * [ 2x * sqrt(3 - 2x^2) + 3*sqrt(2) * arcsin((sqrt(2/3)*x)]

So like I said, there's an inverse trig function in there. Looks like integration by parts. Don't know if you've learned that.
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