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anti derivative of : 2/(x^.5)
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u meant intergral?
Really big hint: \[\frac{d}{dx}\sqrt{x} = \frac{1}{2\sqrt{x}}\]
anti derivative similar by integral go ahead @Otonashi, for helps
then it is \[\frac{ -1 }{ 2x ^{4} }\]
\[\frac{ 2 }{\sqrt{x} } =2x^{1/2} so F(x)=\frac{ 2x^{3/2} }{ 3/2} +C\]
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Just add 1 to your x and put it divided by your exponent, and don't forget your Constant
No. \[\frac{1}{\sqrt{x}} = x^{-1/2}\]
that is the derivitave not the Anti-derivitave, this is the part of calculus 1 that leads into Integrals
you are correct i made a mistake
\[2x^{-1/2} = \frac{ 2x^{1/2} }{ 1/2 } +C = 4\sqrt{x}+C\]
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Well, now you have an equivalence where you mean an antiderivative. Remember, notation means things. Don't write "=" unless you mean it.
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