# I X^2 graph

1. Mar 31, 2017

### Brage Eidsvik

Hello,

If I have an x^2 graph that goes from 0 to a point a. Is there a general solution to where the area of the left side is equal to the area of the right?

2. Mar 31, 2017

### Math_QED

I do not fully understand your question I believe.

Because $f(x) = x^2$ is symmetric relative to the $y$-axis, the area between the curve and the $x$-axis in the interval $[-a,0]$ is equal to the area under the curve and the $x$-axis in the interval $[0,a]$

In integral terms:

$\int\limits_0^a x^2 dx = \int\limits_{-a}^{0} x^2 dx$

3. Mar 31, 2017

### Staff: Mentor

On the left side of what?

With integrals that should be easy to do.

4. Mar 31, 2017

### Brage Eidsvik

I want to cut the graph in two. And yeah, the integral of the left piece should be equal the integral of the right.

I think I solved it and got 2/3 b^3 = 1/3 c^3.
This is if I take the integral from 0 to c, then b will be the center. I was wondering if this works as a general solution?

5. Mar 31, 2017

### Staff: Mentor

For some interpretation of the areas you consider, that works as general solution.

6. Mar 31, 2017

### Staff: Mentor

It would be helpful if you stated the problem more clearly. Here is what I think you were trying to say.

Find a point in the interval [0, a] that divides the area under the graph of $y = x^2$ and above the x axis into two equal parts. In other words, find c for with $\int_0^c x^2 dx = \int_c^a x^2 dx$.​