Question

Find the number b such that the line y = b divides the region bounded by the curves y = x2 and y = 4 into two regions with equal area. Give your answer correct to 3 decimal places.

Answer 1

first step is to sketch the setup!

Answer 2

https://www.desmos.com/calculator/6ynett4qyx

\(\color{#000000 }{ \displaystyle \int\limits_{-2}^{2}4~dx-\int\limits_{-2}^{2} x^2~dx\quad \Longrightarrow \quad \int\limits_{-2}^{2}\left(4-x^2\right)~dx }\)

that would be the (full) area of the region.

There is some y=b (note: 0>b>4) that
splits the area into two equal regions.

That means that,

\(\color{#000000 }{ \displaystyle \frac{1}{2}\int\limits_{-2}^{2}\left(4-x^2\right)~dx= \int\limits_{\large ?}^{\large ?}\left(b-x^2\right)~dx}\)

We don't know the limits for the second integral.
However, once you picture y=b anywhere in that
region, you will know that they are going to be the
two x-values at which y=b and y=x^2 intersect.

\(\color{#000000 }{ \displaystyle y^2=b\quad \Rightarrow \quad \sqrt{y^2}=\sqrt{b~} \quad \Rightarrow \quad |y|=\sqrt{b~} \quad \Rightarrow \quad y=\pm\sqrt{b~}}\)

So your limits are x=√b and x=-√b

And therefore you have:

\(\color{#000000 }{ \displaystyle \frac{1}{2}\int\limits_{-2}^{2}\left(4-x^2\right)~dx= \int\limits_{-b}^{b}\left(b-x^2\right)~dx}\)

Answer 3

hope this helps.