Question

Where did I mess up?

Answer 1

One sec.

Answer 2

Find the number b such that the line y = b divides the region bounded by the curves y = x^2 and y = 4 into two regions with equal area. Give your answer correct to 3 decimal places.

Answer 3

\(\int_{-2}^{2}\left(-x^{2}+4\right)dx\) \(\left[-\frac{x^{3}}{3}+4x\right]_{-2}^{2}=\frac{32}{3}\) \(\int_{-\sqrt{b}}^{\sqrt{b}}\left(b-x^{2}\right)dx=\frac{16}{3}\) \(\left[bx-\frac{x^{3}}{3}\right]_{-\sqrt{b}}^{\sqrt{b}}=\frac{16}{3}\) \(\left[\left(\left(b\sqrt{b}-\frac{b\sqrt{b}}{3}\right)-\left(-b\sqrt{b}+\frac{b\sqrt{b}}{3}\right)\right]=\frac{16}{3}\right]\) \(b\sqrt{b}=4\) so b = 2.52 but I got that wrong.

Answer 4

ignore those extra brackets in the 2nd to last part, don't know how they showed up..

Answer 5

It seems to me that your work it totally correct and \(b = 4^{2/3}\) is the actually correct answer. Also, Mathematica seems to agree as well.
Do you think you got it wrong because you only have 2 decimal places on your approximation?

Answer 6

Maybe I only put 2 decimals when it asked for 3. Otherwise thanks man