Question

ok, so if i need to take the integral of

sqrt(t)(t-sqrt(t)) and i use substitution

let u= sqrt(t)
so, du= 1/2sqrt(t)
then why is my substitued expression

u^2(u-u^2) ?!?!?!?

Answer 1

The way to solve is almost the same as your previous post!
Simplify, then take integral one by one!

Answer 2

right, but my main question is why are my sqrt(t) terms now u^2?

because my u=sqrt(t)

Answer 3

hmm this is what i got..

let u = \(\sqrt t\)
\(u^2 = t\)
2udu = dt

therefore

\(\int \sqrt t(t - \sqrt t) dt = \int u(u^2 -u)2udu\)

factor out u...
\(2\int u(u)(u - 1)udu\)
\(2\int u^3 (u-1) du\)

Answer 4

2/5 t ^5/2 - t^2/2. You dont need substitution. Distribute first and then it will be easy to take the integral

Answer 5

how did you get 2udu?

Answer 6

implicit differentiation from u^2 = t

Answer 7

but like @shaun786 said you can distribute

Answer 8

ok

Answer 9

using what i said it would be

\(2\int (u^4 + u^3)du\)

use power rule then sub back to u = \(\sqrt t\)

Answer 10

but if you distribute first won't the square roots cancel out?

Answer 11

that's why you do algebraic sub..coz you'll get confused :)

Answer 12

ok i get it now, but after ive taken the integral and ive gotten 2((u^5)/5-(u^4)/4) i sub sqrt(t) back in. well how do i evaluate 2sqrt(t)^5?

Answer 13

i think this will be the last question the rest is really clear now thank you guys. thanks for hanging in there with me lol.

Answer 14

well uhmmm \(\sqrt {t^5} = \sqrt{(t)(t)(t)(t)(t)}\) = you have 5 t's..take out every two t's (if that makes sense) so...

\(\sqrt{\cancel{(t)(t)(t)(t)}(t)} = t^2 \sqrt t \)

Answer 15

ok, only that doesnt match up with the answer i should have which ultimately should be \[2t^(5/2)/5-t^2/2\]

Answer 16

oh lol...they didnt turn it to that...well remember how it is \(\Large \frac{2u^5}{5}?\) and how u = \(\sqrt t = t^{\frac{1}{2}}\) so

\(\Large \frac{2u^5}{5} =\frac{2(t^{\frac{1}{2}})^5}{5} = \frac{2t^{\frac{5}{2}}}{5}\)


got it?

Answer 17

ah!! brilliant lol i shoulve been able to figure that out but thank you very much!!

Answer 18

no biggie :)