Brain Teaser  Area enclosed by a parabola and a line

09182015, 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 \(x\):
\[h(u,x)=\frac{ux}{2u}+u^2x^2\] 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 antiderivative 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: \[\begin{align*} \frac{\partial}{\partial u}\int h(u,x)\,dx &=\int \frac{\partial}{\partial u}h(u,x)\,dx \\ &=\int\frac{\partial}{\partial u}(\frac{ux}{2u}+u^2x^2)\,dx \\ &=\int\frac{\partial}{\partial u}(\frac{1}{2}\frac{x}{2u}+u^2x^2)\,dx \\ &=\int\frac{x}{2u^2}+2u\,dx \\ &=\frac{x^2}{4u^2}+2ux \end{align*}\] 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^2a^2}{4u^2} + 2u(ba) &= (ba)\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^41}{8u^3} \\ &= \frac{(4u^2+1)(16u^41)}{16u^4} \\ &= \frac{(4u^2+1)(4u^2+1)(4u^21)}{16u^4} \\ &= \frac{(4u^2+1)^2(4u^21)}{16u^4} \\ &= \frac{(4u^2+1)^2(2u+1)(2u1)}{16u^4} = 0 \\ \end{align*} We conclude that \(2u1=0\) which leads to the solution \(u=\tfrac{1}{2}\). Kind regards Thomas 

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