# Thread: One more homework problem

2. ## Re: One more homework problem

What is "the area in the first quadrant bounded above by the curve from (0, 0) to (x, y)= (x, f(x)) and below by y the x-axis? What is "the area of the rectangle with opposite vertices at (0, 0) and (x, y)= (x, f(x))? Set the first equal to 1/3 the second and solve for f(x).

3. ## Re: One more homework problem

I'm still having a little trouble picturing what the question is asking.

4. ## Re: One more homework problem The shaded blue area is one third of the area of the rectangle.

Thus
$\displaystyle \int \limits_0^x f(t,c) \, \mathrm d t = \tfrac13xf(x,c)$

If you differentiate that and write $\displaystyle y=f(x,c)$, you should have a differential equation that you can solve for $\displaystyle y$

5. ## Re: One more homework problem

Thank you for the illustration. I was only imagining (x,y) as a point and didn't see it made a box. I wrote down (1/3)xf(x,c) but didn't understand where it came from, now I do.

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