Question

evaluate the integral from x=1 to x=2 of (v^3 +4v^5)/v^2

Answer 1

Is this your question?
\[\int\limits_{1}^{2}\frac{v ^{3} +4v ^{5}}{v ^{2} }\]

Answer 2

yes

Answer 3

If so note common denominator: v^2.
So you can rewrite: \[\frac{ v^{3} }{ {v^{2}} } + \frac{ v^{5} }{ v^{2} }\]

Answer 4

This should give you v+v^3
So now you are integrating:
\[\int\limits_{1}^{2} v+4v^{3}\]

Answer 5

According to the properties of integrals:
\[\int\limits_{1}^{2}v dv + \int\limits_{1}^{2}v ^{3}dv\]

Answer 6

Do you understand so far?

Answer 7

Sorry I left out the 4 that should be multiplying v^3.

Answer 8

yeah it's making sense.

Answer 9

Now we integrate both and add the result:
Do you know how to integrate?

Answer 10

kind of, i'm not very good at it

Answer 11

Ok.
\[\int\limits_{1}^{2} vdv = \frac{ 1 }{ 2 }(v^{1+1})\] with the range from 1 to 2
So this evaluates to, 1/2(2^2) - 1/2(1^2)

Answer 12

First part would evaluate to 2-1/2 = 3/2

What i did was when you integrate a variable raised to a power you add 1 to the exponent and multiply the result by the reciprocal of the exponent: 1/2

Can you try the second part.

Answer 13

Okay, ya i'll try it

Answer 14

i got 30

Answer 15

So what did you get when you integrate 4v^3?
Did you get (4)*(1/4)*(v^4) ? which simplifies to v^4

Answer 16

ohh, no i did that wrong then

Answer 17

i figured it out. that would result in 15 and the total is 33/2

Answer 18

Good. Remember the first value we got was 3/2
So 3/2+15 = ?

Answer 19

Yup that is correct. Good job

Answer 20

thank you!

Answer 21

You are welcome!