# Thread: Area of the curve between 2 functions

1. ## Area of the curve between 2 functions

I need to find the area of the curve between 2 functions:

$\displaystyle y= x^2$ and $\displaystyle y= 2x-x^2$

Taking the integral of both is easy:
$\displaystyle \frac{1}{3}x^3$ and $\displaystyle x^2-\frac{1}{3}x^3$

At this point I know I need to find the intersects of the functions so I can then find the definitive integral.

I begin the process like so:
$\displaystyle x^2=2x-x^2$

Then:
$\displaystyle 2x-x^4$

Then:
$\displaystyle 2x(1-x^3)$

At this point I think I am just having a brain spaz as I don't know how to get the 2 points I need so I can plug them into the functions. Am I doing something wrong and in general, do I have the correct idea?

2. Originally Posted by x0ne
I need to find the area of the curve between 2 functions:

$\displaystyle y= x^2$ and $\displaystyle y= 2x-x^2$

Taking the integral of both is easy:
$\displaystyle \frac{1}{3}x^3$ and $\displaystyle x^2-\frac{1}{3}x^3$

At this point I know I need to find the intersects of the functions so I can then find the definitive integral.

I begin the process like so:
$\displaystyle x^2=2x-x^2$

Then:
$\displaystyle 2x-x^4$

Then:
$\displaystyle 2x(1-x^3)$

At this point I think I am just having a brain spaz as I don't know how to get the 2 points I need so I can plug them into the functions. Am I doing something wrong and in general, do I have the correct idea?
How did you get from $\displaystyle x^2=2x-x^2$ to $\displaystyle 2x-x^4$

I get $\displaystyle x^2-x=0$ which gives x=0 and x=1 which will be your limits

3. Originally Posted by e^(i*pi)
How did you get from $\displaystyle x^2=2x-x^2$ to $\displaystyle 2x-x^4$

I get $\displaystyle x^2-x=0$ which gives x=0 and x=1 which will be your limits
I subtracted the x^2 to the other side and that gave me x^4. Could you break out exactly what you did? For some reason I am not seeing it.

4. Originally Posted by x0ne
I subtracted the x^2 to the other side and that gave me x^4. Could you break out exactly what you did? For some reason I am not seeing it.
I added x^2 to both sides. When adding you change the coefficient rather than the exponent. x^4=(x^2)(x^2) whereas 2x^2 = x^2+x^2

5. Originally Posted by e^(i*pi)
I added x^2 to both sides. When adding you change the coefficient rather than the exponent. x^4=(x^2)(x^2) whereas 2x^2 = x^2+x^2
Got it!

Now to be sure I am on the right path:

I thought I would have ended up with some room between the points. I now need to do the definite integral of both the above functions, but I only have points that go from 0 to 1. How would I proceed now?

6. Originally Posted by x0ne
Got it!

Now to be sure I am on the right path:

I thought I would have ended up with some room between the points. I now need to do the definite integral of both the above functions, but I only have points that go from 0 to 1.

$\displaystyle \frac{1}{3}x^3 + x^2-\frac{1}{3}x^3$

How would I proceed now?
Normally you would do one function subtract the other one:

$\displaystyle \frac{1}{3}x^3 (\frac x^2-\frac{1}{3}x^3) - (\frac{1}{3}x^3)$

$\displaystyle = [1 - \frac{1}{3} - \frac{1}{3}] - 0 = \frac{1}{3}$

Then evaluate for a number.