202310171726 Solution to 1988-CE-AMATH-II-6

Evaluate

$\displaystyle{\int_{0}^2}\frac{x^2\,\mathrm{d}x}{\sqrt{9-x^3}}$ .

Roughwork.

Let

$\begin{aligned} f(x) & =x^2 \\ g(x) & = \sqrt{9-x^3} \\ \end{aligned}$

s.t.

$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}x} & = 2x \\ \frac{\mathrm{d}g}{\mathrm{d}x} & = -\frac{3}{2}\bigg(\frac{x^2}{\sqrt{9-x^3}}\bigg) \\ & = -\frac{3}{2}\frac{f}{g} \\ -\frac{2}{3}\,\mathrm{d}g & = \frac{f}{g}\,\mathrm{d}x \\ \end{aligned}$

Then

$\begin{aligned} &\quad\enspace \int_{0}^{2}\frac{x^2\,\mathrm{d}x}{\sqrt{9-x^3}} \\ & = \int_{0}^{2}\frac{f(x)}{g(x)}\,\mathrm{d}x \\ & = -\frac{2}{3}\int_{0}^{2}\mathrm{d}g \\ & = -\frac{2}{3} \big[ g(x)\Big]\Big|_{0}^{2} \\ & = -\frac{2}{3} \Big[\sqrt{9-x^3}\Big]\Big|_{0}^{2} \\ & = -\frac{2}{3}(1-3) \\ & = \frac{4}{3} \\ \end{aligned}$

Ah, but no sense of physical meaning.

This problem is not to be attempted.

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