# Math Help - Mathlab integral(area)

1. ## 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?

2. 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

3. 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

4. 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?

5. 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.

6. 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 .
Can that be right?

7. 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 .
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