Question

Step by Step help please?

Evaluate the integral from -2 to 0 of (1/2t^4+1/4t^3-t)dt

Answer 1

is the the exact problem?\[\int_{-2}^{0}\left(\frac{1}{2t^4} +\frac{1}{4t^3}-t\right)dt\]

Answer 2

yeah.. I got -1/8... I'm not sure if I did it right though

Answer 3

Hmm are you sure it was typed out correctly? D: Because I think this integral doesn't converge. After you integrate, when you try to plug in the upper boundary (zero), you'll run into a problem.

Answer 4

\[\large \int\limits_{-2}^0 \frac{1}{2}t^{-4}+\frac{1}{4}t^{-3}-t \; dt\]

\[\large = \quad -\frac{1}{6}t^{-3}-\frac{1}{8}t^{-2}-\frac{1}{2}t^2 \quad |_{-2}^0\]
\[\large =-\left(\frac{1}{6t^3}+\frac{1}{8t^2}+\frac{1}{2}t^2\right)_{-2}^0\]See how we'll have a problem if we try to plug in 0? D':

Answer 5

No, I'm sure it's right.. lol

Answer 6

I'm getting divergent when I plug it into wolfram :3 ...

But fineee, if you say so XD

Answer 7

whats that mean? divergent

Answer 8

Look at the second term with the 1/t^2.
If you were to graph this function, you would see that there is an asymptote at t=0.

So if you add up the area from -2 to 0, there is an infinite amount of area since you're approaching an asymptote.
It doesn't CONVERGE, or in other words, it doesn't add up to a nice clean number. You just keep adding more and more and more area.

It's possible that I'm missing some minor detail though :) So just do whatever.
Cause if you haven't learned about improper integrals yet, then it's probably not a problem of that type.

Either the problem was pasted incorrectly (maybe the t's weren't suppose to be in the denominator), or I'm making a silly mistake somewhere. ^^ heh

Answer 9

where did you get 1/t^2?

Answer 10

\[\int\limits \frac{1}{4t^3}dt=-\frac{1}{8t^2}\]This is from the second term :o

Answer 11

You had confirmed that ronald had wrote the problem correctly... was there in fact a mistake though?

Was it suppose to be \[\large \frac{1}{4}t^3\]instead?