how to express this problem as integral

Question

I am given this word problem:

Find a straight line which goes through the center of x, y coordinates so that the area between this straight line and graph of $f(x)=x^2$ is exactly $\frac{1}{6}$

I thought of this:

$$ \int \mathrm{x^2+(a+bx)}\,\mathrm{d}x = \frac{1}{6} $$

i know, this is wrong, but where am i wrong and how can I put this problem in integral form?

Answer 1

A straight line that passes through the center of $x,y$ is of the form $$y=mx$$

Now let us try to visualize an example.

So the first thing you want to do is finding out where the two curves intersect. $(0,0)$ is a trivial point and the other point can be found out by solving $x^2=mx$. And the second point is thus $(m,m^2)$.

Now the area between the curve is given by $$ \int\limits_{0}^{m}{(mx-x^2)\, dx}=\frac{m^3}{6}=\frac{1}{6}$$

And hence $m=1$.

As you might have noticed, we have found out the answer silently assuming $m>0$, similarly solve for the case where $m<0$ and get $m=-1$.

Answer 2

The line passes through the origin $(0,0)$, so it has the equation $y = g(x) = mx$ for some slope $m$. The area between this line and the parabola $y = f(x) = x^2$ is given by $$\int_{x = a}^b |g(x) - f(x)| \, dx,$$ where $a < b$ are the roots of the equation $f(x) - g(x) = x^2 - mx = 0$.