\[\int\limits_{0}^{4}\sqrt{3x+4}=\int\limits_{4}^{12}u^{1/2}(1/3)du\] when substituting u=sqrt{3x+4} which is all good but what i don't get is (1/3)*[(2/3)u^(3/2)]|from 4 to 16 where does the 2/3 and u^(3/2) come from?

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1/3 come from differentiating u so you have du=3dx. solve for dx=1/3du

right i got that but i don't get where the 2/3*u^(3/2) comes from

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