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Integral from 0 to 2 of: (x^3+2x^2-5x-1) / (sqrt x)
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where are you stuck?
I know to do this: \[\frac{ x^4 }{ 4 } + \frac{ 2 }{ 3 }x^3 -\frac{ 5 }{ 2 }x^2 -x\] to the top
And i know the bottom is just \[x ^{\frac{ 1 }{ 2 }}\]
NOOO!!! can't be it
do you multiply the top by x^2 first?
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The integrand only: \[\dfrac{x^3+2x^2-5x-1}{x^(1/2)}=(x^3+2x^2-5x-1)(x^{-1/2})\] distribute it, then put the integral sign in, take it as usual.
can you step up?
I know how to distribute that, but how do you know to switch the exponent to a negative 1/2?
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