if I have an area between two curves, f(x)=x^2 and g(x)=x^(1/2) would there be a way to algebraically calculate 1/4 of the area and show it in two different parts of the area starting from the bottom intersect? enter image description here Hopefully this image helps with understanding the question I'm asking? each line represents 1/8th
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Since the two lines intersect after 4 ticks, I assume each line represents 1/4, not 1/8, right? – 5xum Jul 31 '17 at 11:59
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the question asks us to make the area 1/4 of the shaded colour and if there are two different equal shaded parts it will be equal to 1/8 for each of them – flynn swift Jul 31 '17 at 12:04
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No idea what that last comment is supposed to mean. – 5xum Jul 31 '17 at 12:05
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so the question asked us to have 1/4th of the total area to be shaded, and if the shaded are is split into two equal parts it would be 1/8? – flynn swift Jul 31 '17 at 12:10
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If I understand you correctly, you want to solve $$ \int_0^y x^{\frac{1}{2}} - ax = \frac{A}{8} $$ where A denote the total area between both curves, and y is the root of $x^1/2-ax$, i.e $y=\frac{1}{a^2}$.
If I'm not mistaken, you should obtain $a=\sqrt[3]{4}$.
BgB
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Welcome to MathSE. The motive of this website is to lead people towards a possible solution by giving out hints. So it would be better if you can show how your reached the answer rather than the final answer itself. – Naive Jul 31 '17 at 13:02
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If I understood well you want that the area between the line and the parabola is $\dfrac{1}{8}$ of the area between the two parabolas.
We first compute the area between the two parabolas $$A=\int_0^1 \left(\sqrt{x}-x^2\right) \, dx=\left[\frac{2 x^{3/2}}{3}-\frac{x^3}{3}\right]_0^1=\dfrac{1}{3}$$
Now we want to find a line $y=mx$ such that the area between the line and the parabola is $\dfrac{1}{8}\,A=\dfrac{1}{24}$.
First we find the intersection point between the line and the parabola
$x^2=mx\to x=m$ so we must have
$$\int_0^m \left(m x-x^2\right) \, dx=\frac{1}{24}\to \frac{m^3}{6}=\frac{1}{24}\to m=\frac{1}{\sqrt[3]{4}}$$
In a similar way we find the other line $y=px$ which intersects the other parabola in $x= \frac{1}{m^2}$
$$\int_0^{\frac{1}{m^2}} \left(\sqrt{x}-m x\right) \, dx=\frac{1}{24}\to \frac{1}{6 m^3}=\frac{1}{24}\to m=\sqrt[3]{4}$$
Hope this helps
Raffaele
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@flynnswift It's just a remake of your picture https://i.stack.imgur.com/kR1Oa.jpg I tried to understand: it's YOU who need to help ME :) – Raffaele Jul 31 '17 at 13:37
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$\int \left(m x-x^2\right) =\frac{m x^2}{2}-\frac{x^3}{3}$ then plug $m$ and you get the definite integral $\frac{m^3}{2}-\frac{m^3}{3}=\frac{m^3}{6}$ – Raffaele Jul 31 '17 at 13:40
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what if it was to the power of 4 instead of power of 2? how would that be done? – flynn swift Jul 31 '17 at 14:00
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@flynnswift the line intersects at $x^4=mx\to x=\sqrt[3]{m}$ the area must be $1/8$ of $\dfrac{3}{5}$ and the result is $m=\dfrac{1}{2 \sqrt[5]{2}}$ – Raffaele Jul 31 '17 at 15:03
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I'm having trouble understanding the two integrals to get 1/cuberoot 4 and cuberoot 4. would you be able to go into depth with them for me please? – flynn swift Aug 01 '17 at 16:08
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@flynnswift $ \frac{1}{6 m^3}=\frac{1}{24}\to 6m^3=24\to m^3=4\to m=\sqrt[3]{4}$ – Raffaele Aug 01 '17 at 16:20
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