My question is simple, but I am not getting the answers for some reason.
The question is: Consider the area enclosed between the graph of $y = 1 – x^2 $and the $x$ axis. Which line parallel to the $x$ axis divides the area into two equal parts?
My solutions are as follows,
Which gives answers $k = 1/2$ and two complex number answers, which are all wrong. Where am I making the mistake?
Best Answer
Another approach would be to integrate with respect to y, so
$\hspace{.3 in}\displaystyle2\int_b^1\sqrt{1-y}\;\;dy=\frac{1}{2}\left(2\int_0^1\sqrt{1-y}\;dy\right)=\left[-\frac{2}{3}(1-y)^{3/2}\right]_0^1=\frac{2}{3}\implies$
$\hspace{.3 in}\displaystyle\left[-\frac{2}{3}(1-y)^{3/2}\right]_b^1=\frac{1}{3}\implies\frac{2}{3}\left(1-b\right)^{3/2}=\frac{1}{3}\implies (1-b)^{3/2}=\frac{1}{2}\implies$
$\hspace{.3 in}\displaystyle1-b=\frac{1}{2^{2/3}}\implies b=1-\frac{1}{2^{2/3}}=1-\frac{1}{\sqrt[3]{4}}$.