# Marginal Revenue with definite integral

• December 7th 2009, 07:33 PM
mellowdano
Marginal Revenue with definite integral
Hello,

I was given a graph of a marginal revenue function which i have attached. (If for some reason it does not come up, these are main points on the graph (0,100), (500,0) and (1000,-100).)
The first thing i needed to do is find the equation of the marginal revenue function

What i have done so far:
found slope which is -0.2
y intercept is 100
Therefore marginal revenue function is R'(x)=--0.2x + 100

However my prof tells me she wants me to find it as a definite integral with limit 0 to 1000.

Would I start the integral as
∫(100 - 0.2x)dx [0,1000]
and finish with
= 100x -0.1x^2 + C [0,1000]
• December 7th 2009, 08:34 PM
VonNemo19
Quote:

Originally Posted by mellowdano
Hello,

I was given a graph of a marginal revenue function which i have attached. (If for some reason it does not come up, these are main points on the graph (0,100), (500,0) and (1000,-100).)
The first thing i needed to do is find the equation of the marginal revenue function

What i have done so far:
found slope which is -0.2
y intercept is 100
Therefore marginal revenue function is R'(x)=--0.2x + 100

However my prof tells me she wants me to find it as a definite integral with limit 0 to 1000.

Would I start the integral as
∫(100 - 0.2x)dx [0,1000]
and finish with
= 100x -0.1x^2 + C [0,1000]

$\int_0^{1000}(-0.2x+100)dx=(-\frac{1}{10}x^2+100x)|_0^{1000}$

Now evaluate...

This being a definte integral, there will be no constant C.
• December 8th 2009, 10:41 AM
mellowdano
when i evaluates the answer is 0.
= ∫1000 (100 – 0.2x) dx
0
= (100x – 0.1x2)|1000
0
= [100(1000) – 0.1(1000)2] – [100(0) – 0.1(0)2]
=[100000 – 100000] – [0 – 0]
= 0-0
= 0

Would I conclude from this that the 2 x intercepts are therefore x=0 and x=1000 without graphing it?