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Simple Integral Question!
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\[\int\limits_{}^{} 4 \sqrt{y+1} dy\]
\[4* 2/3 (y+1)^(3/2) dy * y^2/ 2?\]
u = y + 1
Okay I know how to solve this but why do we not integrate the y+1 part?
what do you mean
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Sorta like chain rule? Why do not not integrate the expression in the parentheses?
do you see what I am saying? :) @perl
yes i do
Just take out your constant first. \[4\int\sqrt{y+1}dy\]\[u=y+1~,~ du =dy\]
\[\Large y+1 = u^2 \\ \Large dy=2udu\] Plug this in: \[\Large \int\limits 8u^2du= \frac{8}{3}u^3+C\] don't forget to resubstitute back in \[\Large \frac{8}{3} (1+y)^{3/2}+C\]
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Always one step ahead. Huge font = huge ego. =^_____^=
No, I have bad vision from staying up late too many nights doing math so I make the font large so I can read it lol
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