Find the number $b$ $b$ such that the line y=b divides the region...?

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Find the number $b$ $b$ such that the line y=b divides the region...?

Find the number $b$ $b$ such that the line y=b divides the region bounded by the curves $y = x^{2}$ $y=x^2$ and $y = 4$ $y=4$ into two regions with equal area.

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Answer 1 · 2017-09-19T17:56:52

Find the area of the region first.

Explanation:

The area of the region is given by
$\int_{- 2}^{2} (4 - x^{2}) d x = 2 \int_{0}^{2} (4 - x^{2}) d x$ $int_-2^2(4-x^2)dx = 2int_0^2(4-x^2)dx$
$= 2 {(4 x - \frac{x^{3}}{3})}_{0}^{2}$ $= 2(4x-x^3/3)_0^2$
$= \frac{32}{3}$ $= 32/3$

Second, $y = b$ $y = b$ intersects the curve $y = x^{2}$ $y = x^2$ when
$x = \pm \sqrt{b}$ $x = +-sqrtb$ .
Third, we want to find b such that
$\int_{- \sqrt{b}}^{\sqrt{b}} (b - x^{2}) d x = \frac{16}{3}$ $int_-sqrtb^sqrtb(b-x^2)dx = 16/3$
This will occur if and only if
$\int_{0}^{\sqrt{b}} (b - x^{2}) d x = \frac{8}{3}$ $int_0^sqrtb(b-x^2)dx = 8/3$
Integrate:
$\int_{0}^{\sqrt{b}} (b - x^{2}) d x = {(b x - \frac{x^{3}}{3})}_{0}^{\sqrt{b}}$ $int_0^sqrtb(b-x^2)dx =(bx-x^3/3)_0^sqrtb$
$= b \sqrt{b} - \frac{b \sqrt{b}}{3}$ $= bsqrtb-(bsqrtb)/3$
$= \frac{2 b \sqrt{b}}{3}$ $= (2bsqrtb)/3$

Now set this equal to $\frac{8}{3}$ $8/3$ .
$\frac{2 b \sqrt{b}}{3} = \frac{8}{3}$ $(2bsqrtb)/3 = 8/3$
$b \sqrt{b} = 4$ $bsqrtb = 4$
$b^{\frac{3}{2}} = 4$ $b^(3/2) = 4$
$b = \sqrt[3]{16}$ $b = root(3)(16)$
or $b = 2 \sqrt[3]{2}$ $b = 2root(3)(2)$

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Find the number $b$ $b$ such that the line y=b divides the region...?