Applications of Integration - 2

guest:

Dear All,

Please help to check my answer below :

Question :

MC = Q/(6Q²-7)⁵

Determine the Total Cost if the Fixed Cost is $40

Answer :

TC = ∫ MC dQ YES
= ∫ Q/(6Q²-7)⁵ dQ YES
= ∫ Q/(7776Q¹⁰-45360Q⁸+105840Q⁶-123480Q⁴+72030Q²-16807) dQ NO
= ∫ Q/(706.91Q¹¹-5040Q⁹+15120Q⁷-24696Q⁵+24010Q³-16807Q NO

I don't know the next step, please help me..

Thank you.

Sorry guest ( you really need a name )
This one is completely different from the last one because the complicated stuff is on the bottom. You can't separate them into individual terms so you have to look for hints on how else it could be done.
I'm trying to think how to explain it ummm.

TC = ∫ MC dQ
TC = ∫ Q/(6Q ²-7) ⁵ dQ
TC = ∫ Q(6Q ²-7) ^-5 dQ

Do you understand what I just did? If i swap a term from one side of the fraction line to the other I have to change the power from + to - or vise versa.

Now, I have noticed that if I differentiate the stuff inside the bracket i get 12Q. (And Q is out the front of the bracket) This is the helpful fact that I am going to use.
That is: d/dQ [ 6Q ²-7 ] = 12Q

This means that the answer will have (6Q ²-7) ^-5+1 = > (6Q ²-7) ^-4 in it.
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Now, other people may well have a more straight forward method than me but now I look at that last expression and see what happens if I differentiate it
d/dQ (6Q ²-7) ^-4
= -4(6Q ²-7) ^-4 ) ^-5 * 12Q
= -48Q(6Q ²-7) ^-4 ) ^-5

so if I divide everything by -48 the answer will be what I want.
That is

d/dQ [1/-48](6Q ²-7) ^-4
=[1/-48]* -4(6Q ²-7) ^-4 ) ^-5 * 12Q
=[1/-48]* -48Q(6Q ²-7) ^-4 ) ^-5
= Q(6Q ²-7) ^-4 ) ^-5
This is exactly what you wanted.

Therefore
TC = ∫ Q(6Q ²-7) ^-5 dQ
TC = [1/-48](6Q ²-7) ^-4 + k (where k is a constant)
TC = [1/-48](6Q ²-7) ^-4 + k
TC = - (6Q ²-7) ^-4 / 48 + k

Now it is time to look at this statement
Determine the Total Cost if the Fixed Cost is $40
This means that if the quantity is 0 the Total cost will be $40
(You have to spend $40 before you start producing anything.)
Substitute TC=40, Q=0 to find k
40 = 0 + k
so k = 40
The answer is
TC = - [(6Q ²-7) ^-4 / 48] + 40

I hope you understand

Dear Melody,

Thanks for your answer.

My name is Vee..

TC = ∫ MC dQ
TC = ∫ Q/(6Q ²-7) ⁵ dQ
TC = ∫ Q(6Q ²-7) ^-5 dQ

I understand until there.. but I'm confused with the rest of your answer..

I'm wondering if there is another easiest method to understand..

however, thank you so much Melody..

*
Hi Vee,

TC = ∫ Q(6Q ²-7) ^-5 dQ

okay, i worked out how to do it easier

If I differentiate 6Q ²-7 I get 12Q

And, just as an aside comment: ∫ X ^-5dx = X ^-4 / ( -4) (plus a constant but i don't care about that at the moment)

Now I am going to put these ideas together

TC = ∫ Q(6Q ²-7) ^-5 dQ

TC = Q(6Q ²-7) ^-4 / [ -4 * 12Q ]
The Qs cancel out and you are left with

TC = (6Q ²-7) ^-4 / [-48] + k

TC = -(6Q ²-7) ^-4 / 48 + k

Note: If the Qs did not cancel out, this method would not work.

Maybe this is easier to understand. Still, someone might add to this tomorrow.
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I have made some changes to this to make it a little clearer again but if you want someone else to look at it
(and I won't mind at all if you do, I suggest that you bump it again yourself (so that my name is not so obvious)

Dear Melody,

I understand now.. Thank You So Much..