Quiz from Friday, Oct 21

Problem (2 points): Sketch the region of integration and evaluate the integral

$\int_1^4\int_\sqrt{y}^y x^2y^3\,dx\,dy$

Solution: I’ll draw the region in class. (It isn’t a trivial region, but it also isn’t very challenging.) Calculating the integral, we have

\begin{align*}\int_1^4\int_{\sqrt{y}}^y x^2y^3\,dx\,dy &= \int_1^4\left(\frac{1}{3}x^3y^3\right)\Big|_{x=\sqrt{y}}^{x=y}\,dy\\ &=\frac{1}{3}\int_1^4 y^6-y^{9/2}\,dy\\ &= \frac{1}{3}\left(\frac{y^7}{7}-\frac{2y^{11/2}}{11}\right)\Big|_{y=1}^{y=4}\\ &= \frac{1}{3}\left(\frac{4^7}{7}-\frac{2(4)^{11/2}}{11}-\frac{1}{7}+\frac{2}{11}\right)\\ &=\frac{1}{3}\cdot\frac{1}{7}\cdot\frac{1}{11}\left(11\cdot 2^{14}-2^{12}\cdot 7+3\right) \end{align*}

That answer, or something similar, is perfectly fine if you don’t happen to have a calculator. Otherwise, you can compute that as

$\frac{5\cdot 7\cdot 1783}{3\cdot 7\cdot 11}\approx 656.082251$

Replacement Question: How many official GOP debates have there been up until now?

Answer: I’ll take multiple answers on this, either 8 or 9. CNN said that this recent event in Las Vegas was the 8th debate, and lots of other news agencies were saying the same thing. By the official schedule, it should have been number 10, but number 3 (also in Las Vegas) was cancelled back on July 10. I’m counting 9: May 5, June 13, August 11, September 5, September 7, September 12, September 22, October 11 and October 18.