Question

Please help why I am getting another answer.
I would like to know if there is an easy way to solve this or at least with th calculator......Jijiji!!
∫[1,4,(t^(7)-t^(3))/(t^(5))dx]
Thanks!!

Answer 1

Is 1 the lower limit? Try to get in the habit of writing the LOWER limit first ^^ 1,4

Cheater way using a calculator? Hmmm I'm not sure :) lol
This one is pretty straight forward though.
You just have to simplify your fraction, then you can apply the power rule for integration.

Answer 2

I just found the mistake..

Answer 3

it supposed to be this one..∫[1,4,(t^(7)-t^(3))/(t^(5)),]

Answer 4

\[\large \int\limits_1^4 \frac{t^7-t^3}{t^5}dt \quad = \quad \int\limits_1^4 \frac{t^7}{t^5}-\frac{t^3}{t^5}dt \quad = \quad \int\limits_1^4 t^{7-5}-t^{3-5}dt\]

Answer 5

Oh did you figure it out already? :D

Answer 6

Nop!! I thought so.. :(

Answer 7

I still don't get it!!

Answer 8

Ahh sorry I musta missed the blinky message up in the corner D:::

Answer 9

Understand what I did with the powers on t??

Answer 10

No, I believe you substracted them.

Answer 11

Yah when you divide terms with the same base, you SUBTRACT the exponents c:
You reposted this thread it looks like though.. so I assume you figured this out maybe? :D

Answer 12

Yes I did, but I have a concern, I have 1/3(4^3 -1) +(1/4-1) where they coming from.. How its simplify?

Answer 13

At least the fractions.. 1/3 and 1/4..

Answer 14

\[\large \int\limits\limits_1^4 t^{7-5}-t^{3-5}dt \quad = \quad \int\limits_1^4 t^2-t^{-2}dt\]Integrating gives us,\[\large \frac{1}{3}t^3+t^{-1} \quad |_1^4 \quad \rightarrow \qquad \frac{1}{3}t^3+\frac{1}{t}\quad|_1^4\]

Answer 15

\[\large \left(\frac{64}{3}+\frac{1}{4}\right)-\left(\frac{1}{3}+1\right)\]

Answer 16

I think you get something like that, hopefully I didn't mess anything up...

Answer 17

Thanks!!