# Mathlab integral(area)

• May 4th 2008, 03:56 AM
gruvan
Mathlab integral(area)
Hello I have Mathlab installed at my computer and want to compute the integral (area) between to functions and I don´t know how and what commands to use. My two functions are (x/((11.08^0.138*4202463^0.235*5508000^0.4*1000*10^2. 806)/0.8515))^(1/-0.444)
and x= 10

And then is it possible to plot it?
• May 4th 2008, 04:10 AM
CaptainBlack
Quote:

Originally Posted by gruvan
Hello I have Mathlab installed at my computer and want to compute the integral (area) between to functions and I don´t know how and what commands to use. My two functions are (x/((11.08^0.138*4202463^0.235*5508000^0.4*1000*10^2. 806)/0.8515))^(1/-0.444)
and x= 10

And then is it possible to plot it?

Why have you not simplifed your function:

$f(x)=\frac{1.37127 \times 10^{23}}{ x^{2.25225}}$

Now you need to be more precise about what area you are talking about, because you have not specified an area yet.
(sketch the curves)

RonL
• May 4th 2008, 04:19 AM
CaptainBlack
This is the second time you have posted a question with long unsimplified products of numeric constants.

Why is that, what is stopping you from evaluating these things?

RonL
• May 4th 2008, 05:04 AM
gruvan
Hello captain black,thanks for your help.
I want to find the "producer surplus" (green area in the pic below).The equation (1.37127*10^23)/x^2.25225 is the demand function for driving cars and P is the price of gasoline. So how do I compute this in mathlab or any freeware program that you could suggest?
• May 4th 2008, 05:09 AM
mr fantastic
Quote:

Originally Posted by gruvan
Hello captain black,thanks for your help.
I want to find the "producer surplus" (green area in the pic below).The equation (1.37127*10^23)/x^2.25225 is the demand function for driving cars and P is the price of gasoline. So how do I compute this in mathlab or any freeware program that you could suggest?

You could integrate the inverse function between the two prices.
• May 4th 2008, 06:34 AM
gruvan
Hello I used a program called "Graph 4.3" and calulated the green area by dividing areas and my result turned out to be 8.83843E+10 (Muscle)(Rofl).
Can that be right?
• May 4th 2008, 07:28 AM
CaptainBlack
Quote:

Originally Posted by gruvan
Hello I used a program called "Graph 4.3" and calulated the green area by dividing areas and my result turned out to be 8.83843E+10 (Muscle)(Rofl).
Can that be right?

Maxima gives $8.8370091\times 10^{+10}$ for the integral, which when we dispose of the spurious prescission is $8.837\times 10^{+10}$.

RonL