Question

why does (1+2t)^3dt=1/8(1+2t)^4 and not 1/4(1+2t)^4

Answer 1

\[\int (1+2t)^3dt\]??

Answer 2

Yes the answer is \[\frac{ 1 }{ 8 } (1+2t)^{4}\] but i get\[\frac{ 1 }{ 4 }(1+2t)^{4}\]

Answer 3

Alright. Let \(u= 1+2t\)

Answer 4

\[\int (1+2t)^{3}dt\]
u = 1 + 2t => du/dt = 2
\[\int (u)^3* \frac{du}{2} => \frac{1}{2}\int(u)^3dt => \frac{1}{2}*\frac{u^4}{4} + c\]

Answer 5

\[=> \frac{u^4}{8}+c => \frac{(1+2t)^4}{3} +c\]

Answer 6

The last part should be: \[\frac{(1+2t)^4}{8}+c\]

Answer 7

so you multiply through by the derivative of U then "undo" that outside of the integral and then solve??

Answer 8

yee brah

Answer 9

It's called "u-substitution"

Answer 10

Thank you

Answer 11

np brah