Problems

The function \(\mathrm{f}\) satisfies the condition \(\mathrm{f}'(x)>0\) for \(a\leqslant x\leqslant b\), and \(\mathrm{g}\) is the inverse of \(\mathrm{f}.\) By making a suitable change of variable, prove that \[ \int_{a}^{b}\mathrm{f}(x)\,\mathrm{d}x=b\beta-a\alpha-\int_{\alpha}^{\beta}\mathrm{g}(y)\,\mathrm{d}y, \] where \(\alpha=\mathrm{f}(a)\) and \(\beta=\mathrm{f}(b)\). Interpret this formula geometrically, in the case where \(\alpha\) and \(a\) are both positive. Prove similarly and interpret (for \(\alpha>0\) and \(a>0\)) the formula \[ 2\pi\int_{a}^{b}x\mathrm{f}(x)\,\mathrm{d}x=\pi(b^{2}\beta-a^{2}\alpha)-\pi\int_{\alpha}^{\beta}\left[\mathrm{g}(y)\right]^{2}\,\mathrm{d}y. \]

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Solution: Let \(u = f(x)\) then \(\frac{\d u}{\d x} = f'(x)\) and \begin{align*} \int_a^b f(x) \d x &\underbrace{=}_{\text{IBP}} \left [ xf(x) \right]_a^b - \int_a^b x f'(x) \d x \\ &\underbrace{=}_{u = f(x)} b \beta - a \alpha - \int_{u = f(a) = \alpha}^{u = f(b) = \beta} g(u) \d u \\ &= b \beta - a \alpha - \int_{\alpha}^{\beta} g(u) \d u \end{align*}

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\[ \underbrace{\int_{a}^{b}\mathrm{f}(x)\,\mathrm{d}x}_{\text{red area}}=\underbrace{b\beta}_{\text{whole area}}-\underbrace{a\alpha}_{\text{area in green}}-\underbrace{\int_{\alpha}^{\beta}\mathrm{g}(y)\,\mathrm{d}y}_{\text{area in blue}}, \] \begin{align*} 2\pi \int_a^b x f(x) \d x &\underbrace{=}_{\text{IBP}}\pi \left [ x^2 f(x) \right]_a^b - \pi \int_a^b x^2 f'(x) \d x \\ &\underbrace{=}_{x = g(u)} \pi (b^2 \beta - a^2 \alpha) - \pi \int_{u = f(a) = \alpha}^{u = f(b) = \beta} [g(u)]^2 \d u \\ &= \pi(b^2 \beta - a^2 \alpha) - \pi \int_\alpha^\beta [g(u)]^2 \d u \end{align*} This is the volume outside the function in the volume of revolution about the \(y\) axis between \( \alpha\) and \(\beta\).

Let \(y=\mathrm{f}(x)\), \((0\leqslant x\leqslant a)\), be a continuous curve lying in the first quadrant and passing through the origin. Suppose that, for each non-negative value of \(y\) with \(0\leqslant y\leqslant\mathrm{f}(a)\), there is exactly one value of \(x\) such that \(\mathrm{f}(x)=y\); thus we may write \(x=\mathrm{g}(y)\), for a suitable function \(\mathrm{g}.\) For \(0\leqslant s\leqslant a,\) \(0\leqslant t\leqslant \mathrm{f}(a)\), define \[ \mathrm{F}(s)=\int_{0}^{s}\mathrm{f}(x)\,\mathrm{d}x,\qquad\mathrm{G}(t)=\int_{0}^{t}\mathrm{g}(y)\,\mathrm{d}y. \] By a geometrical argument, show that \[ \mathrm{F}(s)+\mathrm{G}(t)\geqslant st.\tag{*} \] When does equality occur in \((*)\)? Suppose that \(y=\sin x\) and that the ranges of \(x,y,s,t\) are restricted to \(0\leqslant x\leqslant s\leqslant\frac{1}{2}\pi,\) \(0\leqslant y\leqslant t\leqslant1\). By considering \(s\) such that the equality holds in \((*)\), show that \[ \int_{0}^{t}\sin^{-1}y\,\mathrm{d}y=t\sin^{-1}t-\left(1-\cos(\sin^{-1}t)\right). \] Check this result by differentiating both sides with respect to \(t\).

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Solution:

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The blue area is \(F(s)\) the red area is \(G(t)\), the dashed rectangle (which is a subset of the red and blue areas) has area \(st\) therefore \(F(s) + G(t) \geq st\). Equality holds if \(f(s) = t\). \begin{align*} && \int_0^t \sin^{-1} y \d y + \int_0^{\sin^{-1} t} \sin x \d x &= t \sin^{-1} t \\ \Rightarrow && \int_0^t \sin^{-1} y \d y &= t \sin^{-1} t - \left [ -\cos (x) \right]_0^{\sin^{-1} t} \\ &&&= t \sin^{-1} t - (1- \cos (\sin^{-1} t)) \end{align*} Let \(y = t \sin^{-1} t - (1- \cos (\sin^{-1} t))\) then, \begin{align*} \frac{\d y}{\d t} &= \sin^{-1} t +t \frac{\d}{\d t} \l \sin^{-1} (t) \r - \sin ( \sin^{-1} t) \frac{\d}{\d t} \l \sin^{-1} (t) \r \\ &= \sin^{-1} t \end{align*} as required