Question

Will someone help me with this integral problem, evaluate the integral from 1 to 3 when the equation is y^5-2y/y^3 dy

Answer 1

Just to be sure, is this what the question looks like?
\[\int\limits_{3}^{1}\frac{ y^5-2y }{ y^3 }.dy\]

Answer 2

3 is on the top but otherwise yes

Answer 3

Alright. So you can integrate it by dividing it into two separate fractions
\[\int\limits_{1}^{3}\frac{ y^5 }{ y^3 } - \int\limits_{1}^{3}\frac{ 2y }{ y^3 }\]

Answer 4

Alright

Answer 5

So you will have integral of y^2-2/y from 3 to 1?

Answer 6

Yes, just one thing. It would be 2/y^2 for the second one

Answer 7

\[\frac{ 2y }{ y^3 }=\frac{ 2 }{ y^2 }\]

Answer 8

Sorry I'm back

Answer 9

@harman.singh

Answer 10

Thats fine. I am here

Answer 11

So than antiderivative of that?

Answer 12

Yes, correct

Answer 13

So than that would be (1/3)y^3 and lets see you can rewrite the other one to -2y^-2 so is it 2/y?

Answer 14

I think I'm not sure about the second part though

Answer 15

Yes, correct so far :)

Answer 16

Awesome!! :)

Answer 17

So now, it looks like
\[\frac{ y^3 }{ 3 }+\frac{ 2 }{ y }+C\]

Answer 18

All you need to do now is compute the boundaries for 3 & 1 to get your answer

Answer 19

Ok I'll let you know what I get!

Answer 20

Sure :)

Answer 21

Is it 7.33333?

Answer 22

That's the right answer. Good job! :)

Answer 23

Thank you so much!!!

Answer 24

My pleasure!

Answer 25

:)