Verified Answer :

LammettHash

\(y=25x^2\implies x=\pm5\sqrt y\)

The area between the parabola and the line is

\(\displaystyle\int_0^4(5\sqrt y-(-5\sqrt y))\,\mathrm dy=10\int_0^4\sqrt y\,\mathrm dy=\dfrac{10}{\frac32}y^{3/2}\bigg|_{y=0}^{y=4}=\dfrac{20}3\times8=\frac{160}3\)

But you're looking for the line \(y=b\) such that the areas of each subdivision are equal. This means you want \(b\) to satisfy

\(\displaystyle10\int_0^b\sqrt y\,\mathrm dy=\dfrac{80}3\)

You have

\(\displaystyle\int_0^b\sqrt y\,\mathrm dy=\dfrac83\)
\(\dfrac23y^{3/2}\bigg|_{y=0}^{y=b}=\dfrac83\)
\(\dfrac23b^{3/2}=\dfrac83\)
\(b^{3/2}=4\)
\(b=4^{2/3}=2\sqrt[3]2\approx2.52\)

vijayhalder031

The correct answer is \(2 \sqrt[3]{2}\) ≈ 2.52

What is a Parabola?

  • A symmetrical plane curve that forms when a cone intersects with a plane parallel to its side.

How to Solve the Problem?

The problem can be solved by following steps.

\(y = 25x^2\) (Given)

y = 4 into two regions with equal area (Given)

We need to find b in the equation

So, The area of region given by

\(\int\limits^2_ {-2} (4-x^{2} ) \, dx \\= 2\int\limits^2_0 (4-x^{2} )\, dx\\\)

= \(2(4x-\frac{x^{3} }{3})^0_2\)

=\(\frac{32}{3}\)

Second, y=b intersects the curve

x=±\(\sqrt{b}\)

We want to find b such that

\(\int\limits^a_b {(b-x^2)} \, dx\)

Whereas , a = \(\sqrt{b}\) and b = \(-\sqrt{b}\)

= \(\frac{16}{3}\)

This will only occur if and only if

\(\int\limits^a_b {(b-x^2)} \, dx\)  Where the limits are a =\(\sqrt{b}\) and b =0

= 8/3

Integrating it we will get

\({(bx - \frac{x^3}{3}) }\limits^a_b\) = 8/3

Where a = \(\sqrt{b}\), b = 0

\(b\sqrt{b}-\frac{b\sqrt{b}}{3}\\\) = 8/3

Therefore,  \(\frac{2b\sqrt{b} }{3}\) = 8/3

\(b\sqrt{b} =4\)

\(b^{\frac{3}{2}} = 4\)

\(b=\sqrt[3]{16}\)

\(b=2\sqrt[3]{2}\)

Round answer in two decimal places ≈2.52

Hence the number b ≈ 2.52

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