Question

Solve: (-1to 0)∫[(3t)/(4+t^2)^4][dt]

NOTE:(By-1to 0 I mean that the -1 is below the integral and the 0 is up or above the integral)

Answer 1

\[\huge \int\limits_{-1}^0 \frac{3t}{(4+t^2)^4}dt\]We'll want to apply a u substitution.
When looking for what to substitute, it's fairly simple with polynomials.

See how we have a t^2 and a t^1 in this problem?
If we let our t^2 term be U, then we take the derivative of U, we'll get something involving t^1 for our DU.

Answer 2

wouldn't DU=2t...??

Answer 3

Yes it would! And our top is 3t, hmmm that's no good.
But it's not a problem, we can work with it.
We just need them to be of the same DEGREE (power).

Answer 4

\[\huge \text{Let} \quad u=4+t^2 \; \text{, then} \quad du=2t \cdot dt\]

Answer 5

Now the idea is, we want to try and solve for t dt.
Let me rearrange the top of the equation so you can see what I'm talking about.\[\huge 3\int\limits\limits_{-1}^0 \frac{(t\cdot dt)}{(4+t^2)^4}\]

Answer 6

So in our du, let's divide both sides by 2, so we have it written in terms of t*dt.\[\huge \frac{1}{2}du=t \cdot dt\]

Answer 7

Now we can successfully replace all of our T and DT with U DU.

Answer 8

\[\huge 3\int\limits\limits\limits_{-1}^0 \frac{(t\cdot dt)}{(4+t^2)^4} \qquad \rightarrow \qquad \huge 3\int\limits\limits\limits_{-1}^0 \frac{(\frac{1}{2}du)}{(u)^4}\]

Answer 9

Any part of that have you stumped? It's a bit tricky getting used to substitutions.

Answer 10

SO far its good. You haven't lost me yet, I was doing it out as you were explaining...

Answer 11

\[\huge \frac{3}{2}\int\limits\limits\limits\limits_{-1}^0 \frac{1}{u^4}du \qquad \rightarrow \qquad \huge \frac{3}{2}\int\limits\limits\limits\limits_{-1}^0 u^{-4}du\]
And from here we're able to apply the Power Rule for Integration.

Answer 12

ok so it'd be[ 4^-3/-3] +c ...do i have to multiply by the (3/2)??

Answer 13

why does your U look like a 4? lol
Wait this feels very familiar, -_- someone else was having that problem before also XD Is it a strange keyboard or something? :3

Answer 14

it does ?? 0_0 T.T

Answer 15

\[\huge \frac{3}{2}\cdot \left(-\frac{1}{3}u^{-3}\right)\]Yes you have the right idea.

Answer 16

ohh lolol T.T

Answer 17

i messed up T.T you are too kind

Answer 18

Now we have to be careful.... The limits of integration were for our variable T! Not U. So from here, we have 2 options.
We can either...

Find new limits of integration, plugging our limits into the U sub.

Or

Back substitute, plug our (4+t^2) back in for U and then apply the limits of integration that we started with.

They'll both take about the same amount of work to apply, so whichever method you prefer.

Answer 19

find new limits of integration, the first one pls.

Answer 20

\[\huge u=4+t^2 \]Plugging in our previous limits of integration gives us,\[\large t=-1 \qquad \rightarrow \qquad u=4+(-1)^2 \qquad \rightarrow \qquad u=5\]\[\large t=0 \qquad \rightarrow \qquad u=4+(0)^2 \qquad \rightarrow \qquad u=4\]

Answer 21

So now our integral goes from 5 to 4.

Answer 22

Cancelling the 3's out from our U solution gives us...\[\huge \frac{3}{2}\cdot \left(-\frac{1}{3}u^{-3}\right)|_4^5 \qquad \rightarrow \qquad -\frac{1}{2}\left(4^{-3}-5^{-3}\right)\]

Answer 23

Do you have an answer key or anything, so we can check our work? D':

Answer 24

ohh hahaha i was feeling dumb bc i still couldn't figure out how I came up with the 4, what happened was that when I wrote it down on my paper it my 'u" looked like a 4 T.T

Answer 25

Unfortunately, no. But I will double check with my prof tomorrow.

Answer 26

Thank you so much!! ^_^

Answer 27

Hopefully I didn't make a mistake anywhere in there c:
Yay team \:D/