Question

I need help in evaluating the integral of the following:

t(1 + t)^(1/2)dt, from x = 0 to 3.

Can one distribute the first "t" of the integral through the parenthesis? I tried leaving it as is, and evaluating through u-substitution, with u = (1 + t), and had a "t" I couldn't get rid of :/

Any and all help is greatly appreciated! :)

Answer 1

find a substitution for "\(t\)"

Answer 2

Here is what I have worked through thus far:

u = (1 + t)
du = ((0) + t^(1-1))
= (t^0)
= (1)dt
= dt
Making the integral:
(u^(1/2)) * t * du

Right?

Answer 3

If you want a good laugh... Read the bio on the smart score page for abb0t lol.

Answer 4

._. How can you integrate t with respect to x? Wouldn't the t's just be constants then? Am I missing something?

Answer 5

Sorry, I completely missed that... I'm so used to typing dx, it slipped my mind- good catch!

Answer 6

Ah ok. Well since you have u=t+1, you will replace that last t with u-1

Answer 7

Ohh! Tell me if this works!!

du = dt
So does...
du(t) = du(t) ?

Answer 8

Oh, okay.. I see how that works. :) does my way work too?

Answer 9

Oh wait... That still leaves a t (my way). So... Combination of both thoughts:
du(t) = du(u - 1)

Answer 10

I'm not sure what you mean with du(t) = du(t). You just substitute all the t into u terms and then integrate term by term.

Answer 11

Here: You have u= 1+t so t= u-1 and du = dt. So substitute them you get integral (u-1) sqrt(u) du

Answer 12

Sweet, that's what I was putting together, but was poorly articulating. :) thanks you!

Answer 13

Ah ok. So you should be able to solve it pretty easily now.

Answer 14

Should I distribute the u^(1/2) through the (u - 1)? Can I do that?

Answer 15

(u^(3/2) - u^(1/2))du ? Can I do that, then take the anti-derivative?

Answer 16

Yup yup.

Answer 17

*working problem...

Answer 18

@nincompoop

Answer 19

Do you disagree with something abbot ?

Answer 20

I got, as a final answer, 182/15, or 12.133333. Is this what you got?

Answer 21

Hmm I got 116/15

Answer 22

After integrating and substituting I got 2/15 (t+1)^(3/2) (3 t-2)

Answer 23

Dang. Okay. Umm...
So, after that u^(1/2) was distributed the integral was...

(u^(3/2) - u^(1/2))du

Anti-deriviative taken...
= u^(5/2) - u^(3/2)
= 2/5(u^(5/2)) - 2/3(u^(3/2))
= 2/5(1 + t)^(5/2) - 2/3(1 + t)^(3/2)
= 2/5(1 + 3)^(5/2) - 2/3(1 + 0)^(3/2)
= 2/5(4)^(5/2) - 2/3(1)^(3/2)
= 2/5(32) - 2/3(1)
= (64/5) - (2/3)
= (192/15) - (10/15)
= 182/15
= 12.13333

What did I do wrong?

Answer 24

Sorry that took so long to type:/

Answer 25

It's ok. When you put in the limits, you put in the upper limit in the whole antiderivative then minus the lower limit in the whole antiderivative. I see you doing half and half here. Get what I mean?

Answer 26

So your final antiderivative is 2/5(1 + t)^(5/2) - 2/3(1 + t)^(3/2)
First you put t=3 in the whole equation. Then t=0 in the whole equation. And subtract the first with the second.

Answer 27

Yes, I totally see what you mean:) * stupid Amon... Thanks again:)

Answer 28

I don't understand how I make such stupid mistakes, on such simple problems!!

Answer 29

Nah it's fine this is what learning is. Don't beat yourself up