Archive for the ‘Calculus’ Category

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Math 324 Mid-Term 1 Solution

四月 18, 2009

1) (a) V(S) = \int_{-1}^{1}\int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} \int_{x^2 + y^2}^{\sqrt{x^2 + y^2}}1 \, dz \, dy \, dx.

(b and c) \begin{aligned} V(S) &= \int_{0}^{2\pi}\int_{0}^{1}\int_{r^2}^{r}r \, dz \, dr \, d\theta\\ &= 2\pi \int_{0}^{1} r^2 - r^3 dr \\ &= 2\pi (\frac{1}{3} -\frac{1}{4}) = \frac{\pi}{6}.\end{aligned} Read the rest of this entry ?

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Tricks of Integration (1) – The Koop Subs

四月 8, 2009

In this article, I will talk about some of my favorite substitutions for integration, which I will named them as the “Koop subs”. They are particularly useful in mathematics competitions such as the Putnam competition.

Example 1) Compute the integral: I = \int_{0}^{1} \frac{x^3}{1 - 3x + 3x^2} dx. Read the rest of this entry ?

Posted in Calculus, Teaching | 9 意見»