Question

How do I integrate this?

Answer 1

Completely lost to be honest.

Answer 2

You can use Substitution Method here..

\[Put \; \; V_o - Q_pt = x \; \; \implies -Q_p dt = dx\]

Answer 3

You will get:

\[\frac{-KA}{Q_p}\int\limits_0^t \frac{dx}{x}\]

Answer 4

Now, I think you can proceed..

Answer 5

@waterineyes would it not be x/dx instead?

Answer 6

How??

Answer 7

I set denominator ie \(V_o - Q_pt = x\), then x must be in denominator no?

Answer 8

@waterineyes

If the limits are t and 0 then would it not be V0-Qpt-Vo

Answer 9

I have not changed the limits yet..

Answer 10

then that would be -ln( V0-Qpt/V0)

Answer 11

Yeah limits can be put after integration also.. :)

Answer 12

I'm kinda confused now..

Answer 13

Wait, dx/x integration is ln(x) right??

Answer 14

I should get this, according to my notes. Not sure what happens in between though

Answer 15

\[\implies \frac{-KA}{Q_p} [\ln(x)]_0^t\]

Answer 16

How did Qp become a demoninator???

Answer 17

Wait, one - sign is already in your question, so it will become +

Answer 18

Wait, you need explanation here..

Answer 19

I'll show you the entire equation now.

Answer 20

Just respond to my queries quickly and whereever you will have doubt, ask me there itself..

Okay??

Answer 21

Put \(V_o - Q_pt = x\) such that \(-Q_pdt = dx\) Getting till here?

Answer 22

Yep

Answer 23

Can you find dt from the second equation??

Answer 24

\[-Q_p dt = dx \implies dt = \frac{-dx}{Q_p}\] Right?

Answer 25

yep

Answer 26

That's where you have got Qp in denominator..
Let me clearly show you now..

Answer 27

\[-\int\limits_0^t \frac{KA}{x} \frac{-dx}{Q_p} \implies \color{red}{\frac{KA}{Q_p} \int\limits_0^{t}\frac{dx}{x}}\]

Answer 28

Don't worry about Limits, just concentrate on Integral..

Answer 29

Getting or not??

Answer 30

I understand the KA/x but I don't get -dx/Qp

Answer 31

i.e Qpt/Qp

Answer 32

That is the value of dt, I showed you upward..

Answer 33

See, the value of dt above...

Answer 34

Yeah I do.

Answer 35

It's just the red text

Answer 36

We put Vo - Qpt = x

then: -Qp*dt = dx ( <------- From here what is the value of dt?? )

Answer 37

Qp is a constant and you brought it on the outside of the intergrand. How can you do that if x and dx have Qp in it?

Answer 38

Figured it out at last thanks for the help @waterineyes

Answer 39

Qp is also a constant..

Suppose you have : 3 - 2t Vo = 3 and Qp = 2

Then also. 3 - 2t = dx

-2 dt = dx dt = -dx/2, this is the same I was doing there and note that 3 and 2 ie Vo and Qp are constants here.. :)

Answer 40

got ya

Answer 41

*3 - 2t = x there not dx...