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Brain Teaser - Area enclosed by a parabola and a line
09-18-2015, 08:20 PM
Post: #29
RE: Brain Teaser - Area enclosed by a parabola and a line
We start with a slightly different definition of \(h(x)=g(x)-f(x)\) but now consider it as a function of two variables \(u\) and \(x\):
\[h(u,x)=\frac{u-x}{2u}+u^2-x^2\]
From this we define the indefinite integral:
\[H(u,x)=\int h(u,x)\, dx\]
For each \(u\) we find the lower limit \(a(u)\) and the upper limit \(b(u)\) as roots of \(h(u,x)=0\):
\[\begin{align*}
a(u) &= -(u+\frac{1}{2u})\\
b(u) &= u
\end{align*}\]
We evaluate the anti-derivative at these limits, calculate the derivative and set it equal to 0:
\[\frac{\mathrm{d} }{\mathrm{d} u}(H(u,b)-H(u,a))=0\]
To calculate the total differential we use partial differentials:
\[\frac{\mathrm{d} }{\mathrm{d} u}H(u,x)=\frac{\partial H}{\partial u}+\frac{\partial H}{\partial x}\frac{\partial x}{\partial u}\]
From the definition of \(H(u,x)\) we find:
\[\frac{\partial H(u,x)}{\partial x}=h(u,x)\]
However this function is 0 when evaluated at the lower or the upper limit. Thus this term vanishes.
This leaves only the first term:
\[\frac{\partial}{\partial u}H(u,x)=\frac{\partial}{\partial u}\int h(u,x)\,dx \]
But we can switch the order of integration and differentiation:
\[\begin{align*}
\frac{\partial}{\partial u}\int h(u,x)\,dx &=\int \frac{\partial}{\partial u}h(u,x)\,dx \\
&=\int\frac{\partial}{\partial u}(\frac{u-x}{2u}+u^2-x^2)\,dx \\
&=\int\frac{\partial}{\partial u}(\frac{1}{2}-\frac{x}{2u}+u^2-x^2)\,dx \\
&=\int\frac{x}{2u^2}+2u\,dx \\
&=\frac{x^2}{4u^2}+2ux
\end{align*}\]
We evaluate this expression at the upper and lower limit and find the difference. But we can do that for both terms separately:
\begin{align*}
\frac{b^2-a^2}{4u^2} + 2u(b-a) &= (b-a)\left(\frac{b+a}{4u^2}+2u\right)\\
&= \left(2u+\frac{1}{2u}\right)\left(2u-\frac{1}{8u^3}\right) \\
&= \frac{4u^2+1}{2u} \cdot \frac{16u^4-1}{8u^3} \\
&= \frac{(4u^2+1)(16u^4-1)}{16u^4} \\
&= \frac{(4u^2+1)(4u^2+1)(4u^2-1)}{16u^4} \\
&= \frac{(4u^2+1)^2(4u^2-1)}{16u^4} \\
&= \frac{(4u^2+1)^2(2u+1)(2u-1)}{16u^4} = 0 \\
\end{align*}
We conclude that \(2u-1=0\) which leads to the solution \(u=\tfrac{1}{2}\).

Kind regards
Thomas
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RE: Brain Teaser - Area enclosed by a parabola and a line - Thomas Klemm - 09-18-2015 08:20 PM



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