# Reading the Comics, March 19, 2016: I Do Some Calculus Edition

It’s been a normal cluster of mathematically-themed jokes this past week. But one of them lets me show off my ability to do introductory calculus.

Norm Feuti’s Gil for the 15th of March is a resisting-the-word-problems joke. It’s also a rerun, sad to say. King Features syndicated Feuti’s strip for a couple of years, but couldn’t make a go of it. GoComics.com is reprinting what ran and that’s something, at least.

Justin Boyd’s Invisible Bread for the 16th plays on alarm clocks that make you solve problems. I’ve heard of these things, and I suppose they exist or something. The idea is that making you do a bit of arithmetic proves you’ve gotten up enough to not fall right back asleep. The clockmakers are underestimating my ability to get back to sleep. Anyway, I like the escalation of this.

The integral that has to be solved, $\int_0^{\infty} \left(1 + 2x\right)e^{-x} dx$, is a good problem for people taking their first calculus course. Let me spoil it as a homework problem by saying how I’d solve it. If you haven’t got the first idea what calculus is about and don’t wish to know, go ahead and skip to the bit about Rudy Park. Or just enjoy the parts of the sentences below that aren’t mathematics.

The first thing I notice is the integrand, the thing inside the integral. That’s $\left(1 + 2x\right)e^{-x}$, which is the same as $e^{-x} + 2xe^{-x}$. Distributive law, as if you didn’t know. That strikes me as worth doing because, if the integral converges, the integral of the sum of two things is the same as the sum of the integral of two things. I’m willing to suppose it converges until given evidence otherwise. So this integral is the same as $\int_{0}^{\infty} e^{-x} dx + \int_{0}^{\infty} 2xe^{-x} dx$.

I think that’s worth doing because that first integral is incredibly easy. It’ll be a number equal to whatever -e-x is, when x is infinitely large, minus what -e-x is when x is zero. When x is infinitely large, -e-x is zero. When x is 0, -e-x is -1. So 0 minus -1 is … 1.

$\int_{0}^{\infty} 2xe^{-x} dx$ is harder. But it suggests how to evaluate it. The integrand is the quantity 2x times the quantity e-x. 2x is easy to take the derivative of. e-x is easy to integrate. (It’s also easy to take the derivative of, but it’s easier to integrate.) This suggests trying out integration by parts.

When you integrate by parts, you notice the original integral is the product of a part that’s easy to differentiate and a part that’s easy to integrate. My Intro Calculus textbooks generically label the easy-to-differentiate part u, and the easy-to-integrate part dv. Then the derivatie of the easy-to-differentiate part is du, and the integral of the easy-to-integrate part is v. When you integrate by parts, the integral of u times dv turns out to be equal to u times v (no integral signs there) minus the integral of v du. This may sound like we’ve just turned one integral into another. So we have. But we’ve often made it into an easier integral to evaluate. This is why we ever try it.

So if u equals 2x, then its derivative du is equal to 2 dx. If dv is equal to e-xdx (we want to carry those little d’s along), then v is equal to -e-x. And this means we have this:

$\int_{0}^{\infty} 2xe^{-x} dx = -2xe^{-x}|_{0}^{\infty} - \int_{0}^{\infty} -2e^{-x} dx$.

That middle part, $-2xe^{-x}|_{0}^{\infty}$, is not an integral. It’s been integrated. The notation there means to evaluate the thing when x is infinitely large, and evaluate the thing when x is zero. Then subtract the x-is-zero value from the x-is-infinitely-large value. The x-is-zero value of this expression turns out to be zero, as you realize when you start writing “2 times 0 times oh wait we’re done here”. The x-is-infinitely-large value of this expression takes longer to get done. If you want to do it right you have to invoke l’Hôpital’s Rule. But it’s also zero.

The right-hand part, $- \int_{0}^{\infty} -2e^{-x} dx$, is equal to $\int_{0}^{\infty} 2e^{-x} dx$, and that’s equal to $-2 e^{-x}|_{0}^{\infty}$. Which will be 0 minus a -2. Or 2 altogether.

So the integral is 1 plus 2, or in total, 3. The strip got its integration right.

Darrin Bell and Theron Heir’s Rudy Park for the 16th speaks of some architect who said the job didn’t demand being good at mathematics. I hadn’t heard the original claim and didn’t feel my constitution up to finding it. It was hard enough reading the comments at GoComics.com.

Ruben Bolling’s Super-Fun-Pak Comix for the 17th has found a weakness in my policy of “we’ve maybe done enough Chaos Butterfly and Schrödinger’s Cat mantions”.

Mark Anderson’s Andertoons for the 18th mentions circle and radius and that’s all Mark Anderson needs to get my publicity.

David L Hoyt and Jeff Knurek’s Jumble for the 18th has an arithmetic theme. Note the quote marks in the final answer. They’re a warning that the punch line is a pun or wordplay.

## Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.

## 3 thoughts on “Reading the Comics, March 19, 2016: I Do Some Calculus Edition”

1. Glad you liked. I do wonder sometimes whether I don’t do enough mathematics around these parts.

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