Applications of Integration

[quote="guest"]Dear All,

Please help to check my answer of below question :

Question :

Marginal Revenue function of a product is = MR = 12Q (2Q ²+5) ³

Determine the Revenue function.


Answer :

TR = ∫ MR dQ
= ∫ 12Q (2Q ²+5) ³ dQ
Now you didn't expand your cubic properly.
(2Q ²+5)(2Q ²+5)(2Q ²+5)
= (2Q ²+5)(4Q ⁴+20Q ²+25)
= 8Q ⁶+40Q ⁴+50Q ²+20Q ⁴+100Q ²+125
= 8Q ⁶+60Q ⁴+150Q ²+125
Note: If I did this for myself I would use a binomial expansion but I thought you might not know how to do that. So I did it the long way.

Now you have:
TR = ∫ 12Q (8Q ⁶+60Q ⁴+150Q ²+125) dQ
TR = ∫ 96Q ⁷+720Q ⁵+1800Q ³+1500Q dQ

TR = 96/8 Q ⁸+720/6 Q ⁶ + 1800/4 Q ⁴ + 1500/2 Q ² + k
TR = 12Q ⁸ + 120Q ⁶ + 450Q ⁴ + 750Q ² + k (where k is a constant)

I think that this is correct now.

Dear Melody,

Thank you very much for your answer, is it correct if i only state --> TR = 12Q ⁸ + 120Q ⁶ + 450Q ⁴ + 750Q ²

because we don't know the value of "k" ?

guest:

Dear Melody,

Thank you very much for your answer, is it correct if i only state --> TR = 12Q⁸ + 120Q⁶ + 450Q⁴ + 750Q²

because we don't know the value of "k" ?

It might be because it stands to reason that if Q=0 then TR will equal 0
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I think I would actually write,
"Since when Q=0, TR will also equal 0, k must be 0."
And then I would rewrite it leaving out the k.

(My logic behind this is that you cannot receive any revenue if you don't sell anything)

Dear Melody,

Thanks a lot for your answer..