Finding the slope of a line that cuts an area in half A region $A$ in the first quadrant is bounded by $y=x^2$, $y=25$ and the $y$-axis. Find the value of $m$ with the property that the line $y = mx$ divides $A$ into two regions with the same area.

First you can calculate the area of the region A, therefore note that 5^2=25.
Hence
$$
Area_A = \int_0^5 25- x^2 \; dx = \frac{250}3.
$$
Now consider the line $g(x)=5\cdot x$. This line goes through $(0/0)$ as well as $(5/25)$.
The area between $g$ the $y$-axis and $y=25$ is a triangle with area
$$
Area_{Triangle} = \frac{25 \cdot 5}2 =\frac{125}2> \frac{Area_A}2 =\frac{125}3.
$$
This means that the line we are looking for, $h(x)=m\cdot x$ has to satisfy m>5. Taking this into account, the are $B$ bounded by  $h$ the $y$-axis and $y=25$ is again a triangle and therefore
$$
Area_B=\frac{25\cdot \frac{25}m}2 =\frac{Area_A}2= \frac{125}3.
$$
We see that $m = \frac{15}2$.
