There was a question I was doing:
Find $y = b$ (find $b$) so that it splits the area between $y=4$ and $y = x^2$ into 2 equal areas.
I found the solution which they integrate it to y axis. And discard the other half since it's symmetrical.
That made sense.
What I did was find the area between the two curves first, which gave me $32/3$.
I halved that which gave me $16/3$. Shouldn't this be half the area?
Then I integrated $\int^2_{-2}(4-b)dx$ and set that to $16/3$.
So that gave me $(8-2b)-(-8+2b) = 16/3$
I solved that and got $8/3$. The answer is different from the other method.
Shouldn't the way I did it give me the correct answer?
The correct answer is $4^{(2/3)}$.
Best Answer
So your initial evaluation of the area is correct. $\int_{-2}^2(4-x^2)dx$ is indeed $32/3$. However it looks like you made a mistake setting up your second integral. If you look at the integral $\int_{-2}^2(4-b)dx$, you'll see that the area that it is evaluating is the rectangle of height $4-b$ and length $4$ ($-2$ to $2$). This isn't the area you want to find, since the edges of the area lie on the parabola $x^2$ (thus the area can't be a rectangle). You can solve this issue by evaluating the area bounded below by $y=x^2$ and above by $y=b$. Setting the integral equal to $16/3$ and solving for $b$ will give the correct answer, which is in fact $4^{2/3}$.
The calculations: