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What is the integral of { [(H-h)^2] / sqrt h } dh ?
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|dw:1366381904382:dw|
ok, u have\[\frac{(H-h)^2}{\sqrt{h}}=\frac{H^2-2hH+h^2}{\sqrt{h}}=H^2h^{-1/2}-2Hh^{1/2}+h^{3/2}\]do u know what is the anti-derivative of \(h^n\) because all the terms are in this form
yes \[(h ^{n+1}) / n+1\]
very good :)
thanks :)
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