Finding Area of Graphs with x^2: A General Solution?

[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?

 

[member 587159]

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:

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

 

[mfb]

On the left side of what?

With integrals that should be easy to do.

 

[Brage Eidsvik]

On the left side of what?

With integrals that should be easy to do.

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?

 

[mfb]

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

 

[Mark44]

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?

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 $\int_0^c x^2 dx = \int_c^a x^2 dx$.​