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## Main Question or Discussion Point

Ok, so doing a physics problem the other day I had to find a point c on a graph such that there was an equal amount of area between 0 and c and c and a point b > c. I found that for the graph f(x) = kx, the point was b over the square root of 2. This intrigued me so I looked for a generalization. Does this look correct? I'm still a low level math student (only in Calculus 2) so I don't know anything about the correct way to solve this stuff. Is this a proof as well? If not, how would I prove it?

[tex]f(x) = kx^n[/tex]

[tex]\int_{0}^{c} kx^n dx = \int_{c}^{b} kx^n dx[/tex]

[tex]k\int_{0}^{c} x^n dx = k\int_{c}^{b} x^n dx[/tex]

[tex]\left[ \frac{x^{n+1}}{n+1} \right]_{0}^{c} = \left[ \frac{x^{n+1}}{n+1} \right]_{c}^{b}[/tex]

[tex]c^{n+1} = b^{n+1} - c^{n+1}[/tex]

[tex]2c^{n+1} = b^{n+1}[/tex]

[tex]c = \frac{b}{\sqrt[n+1]{2}}[/tex]

thanks!

[tex]f(x) = kx^n[/tex]

[tex]\int_{0}^{c} kx^n dx = \int_{c}^{b} kx^n dx[/tex]

[tex]k\int_{0}^{c} x^n dx = k\int_{c}^{b} x^n dx[/tex]

[tex]\left[ \frac{x^{n+1}}{n+1} \right]_{0}^{c} = \left[ \frac{x^{n+1}}{n+1} \right]_{c}^{b}[/tex]

[tex]c^{n+1} = b^{n+1} - c^{n+1}[/tex]

[tex]2c^{n+1} = b^{n+1}[/tex]

[tex]c = \frac{b}{\sqrt[n+1]{2}}[/tex]

thanks!