MHB Calc Help: Find b to Divide Region into 2 Equal Areas

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To find the value of b that divides the area between the curves y = 16x^2 and y = 25 into two equal parts, the area A can be expressed as A = 2∫₀²⁵ x dy. By solving for x from y = 16x², it is determined that A = (1/2)∫₀²⁵ y^(1/2) dy. The goal is to find b such that ∫₀ᵇ y^(1/2) dy equals ∫ᵇ²⁵ y^(1/2) dy, leading to the equation b^(3/2) = 125/2. Ultimately, the solution for b is b = 25/√[3]{4}.
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Here is the question:

Calc help! One try left!?


Find the number b such that the line y = b divides the region bounded by the curves y = 16x^2 and y = 25 into two regions with equal area.

I have posted a link there to this thread so the OP can see my work.
 
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Hello eyeheartglitter,

The given bounded area $A$ may be expressed as:

$$A=2\int_0^{25} x\,dy$$

From $$y=16x^2$$ we find (by taking the positive root):

$$x=\frac{y^{\frac{1}{2}}}{4}$$

Hence:

$$A=\frac{1}{2}\int_0^{25} y^{\frac{1}{2}}\,dy$$

Now, we wish to find some number $b$ such that:

$$\frac{1}{2}\int_0^{b} y^{\frac{1}{2}}\,dy=\frac{1}{2}\int_b^{25} y^{\frac{1}{2}}\,dy$$

Multiply through by 2:

$$\int_0^{b} y^{\frac{1}{2}}\,dy=\int_b^{25} y^{\frac{1}{2}}\,dy$$

Apply the FTOC:

$$\frac{2}{3}\left[y^{\frac{3}{2}} \right]_0^b=\frac{2}{3}\left[y^{\frac{3}{2}} \right]_b^{25}$$

Multiply through by $$\frac{3}{2}$$:

$$\left[y^{\frac{3}{2}} \right]_0^b=\left[y^{\frac{3}{2}} \right]_b^{25}$$

$$b^{\frac{3}{2}}=5^3-b^{\frac{3}{2}}$$

$$2b^{\frac{3}{2}}=5^3$$

$$b^{\frac{3}{2}}=\frac{5^3}{2}$$

$$b=\frac{25}{\sqrt[3]{4}}$$
 
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