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I want to integrate $$\int\left(\frac{y}{2}\right)^\frac{2}{3}dy$$

Doing this I get $$\cfrac{3}{5}\left(\cfrac{y}{2}\right)^\cfrac{5}{3}$$

But the answer is $$\cfrac{6}{5}\left(\cfrac{y}{2}\right)^\cfrac{5}{3}$$

What step am I missing

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    $\begingroup$ You need the chain rule to do that. You can only use the power rule on y, not y/2 $\endgroup$ – Saketh Malyala Sep 9 at 2:34
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Substitute $u=\dfrac y2$.

Then $du=\dfrac {dy}2$, so $$\int\left(\frac{y}{2}\right)^{2/3}dy=\int u^{2/3}2du=2\left(\dfrac35u^{5/3}\right)+C=\dfrac65\left(\dfrac y2\right)^{5/3}+C$$

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$$\int (\frac{y}{2})^\frac{2}{3} dy = \frac{1}{2^\frac{2}{3}} \int y^\frac{2}{3} dy = \frac{1}{2^\frac{2}{3}} \frac{3}{5} (y)^\frac{5}{3} = \frac{2 \cdot 3}{2^\frac{5}{3} \cdot 5} (y)^\frac{5}{3} = \frac{6}{5}(\frac{y}{2})^\frac{5}{3}$$

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You are exponentiating the 1/2 as well. That's why you have an additional factor of two in the numerator.

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