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I know this isn't a specific math question, but I need some help coming up with an integral that evaluates to 42 for a calc shirt. Obviously, I'm striving for something more complicated than $\int_{10}^{11} 4x dx$ or $\int_{0}^{42} 1 dx$. Try and stick to integral content that would be covered in the BC calc curriculum, because that is the class I'm trying to design the shirt for. Thanks!

Sapphira
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  • This'll get closed for the same reason this question was closed. Sorry, but it's too broad and not subjectively answerable. – Simply Beautiful Art Jun 01 '17 at 13:24
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    Hopefully it gets a few answers before that happens... – Sapphira Jun 01 '17 at 13:28
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    One thing you can try is to use a computer algebra system (or WolframAlpha) for definite integrals that you know are "BC computable" and which contain a parameter, and then set the answer you get equal to $42$ and then use the computer algebra system to solve for the parameter. Of course, if you get an equation like $ae^{-2a}+5a = 42,$ skip the solving step and try another integral possibility. – Dave L. Renfro Jun 01 '17 at 13:54
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    Do you really want something more complicated than the examples you present? If you're putting it on a t-shirt, then most people who see it will have only a short time to think about it. If it's not something that your intended audience will be able to work out in their heads, then the joke will probably be lost on them. – John Bollinger Jun 01 '17 at 16:30
  • @DaveL.Renfro Your comment is, indeed, the right answer !!!. – Felix Marin Jun 03 '17 at 02:41

4 Answers4

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A nice and simple one is $$\int_0^\infty (2x^4-x^3)e^{-x}\, dx$$ This works, since for any natural number $n$, $$\int_0^{\infty} x^ne^{-x}\,dx = n!$$

florence
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  • Excellent - the gamma function isn't in the AP Calculus syllabus but they can prove it! – Sean Roberson Jun 01 '17 at 13:36
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    What about $$\dfrac{\int_0^\infty x^7e^{-x}dx}{\int_0^\infty x^5e^{-x}dx}$$ –  Jun 01 '17 at 15:29
  • @florence In general, for all integers $n \ge 2$, we have $$\int_0^\infty \left(\sum_{k=2}^n t_kx^k + \left(m-\sum_{k=2}^n k!t_k\right)x\right)e^{-x} , dx=m,$$ where $m,t_k$ ($k=2,\ldots,n$) are real numbers. – Cookie Jan 12 '20 at 05:00
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Because $42$ is the fifth Catalan number then using the known integral representation of the Catalan's numbers we have that

$$C_5=\frac1{2\pi}\int_0^4x^5\sqrt{\frac{4-x}x}\,\mathrm dx=42$$

Playing with some change of variable above you can get different integral expressions. By example with the change of variable $(4-x)/x=t$ we can build the following improper integral

$$C_5=\frac{16}{\pi}\int_0^\infty\sqrt t\left(\frac12+\frac{t}2\right)^{-7}\mathrm dt=42$$

Maybe more interesting integral representations can be achieved considering first the integral representations for $21$, what is a number with more interesting properties.

Masacroso
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You can try: \begin{align} \int^4_0 \frac{945 } {2(x^2+9)^{3/2} } dx \end{align}

Solvable by suitable substitution $x=a\tan z$. Or you can try: \begin{align} \int_0^{2\pi}\frac{63 }{5\pi+4\pi\cos x}dx \end{align} Solvable by $z=\tan(x/2)$

I think both techniques is taught in BC Calculus.

Shashi
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  • I don't think both are (especially the latter), but I think recent changes include the former technique. – Sean Roberson Jun 01 '17 at 16:49
  • @SeanRoberson good to hear that they are not standard. I think it depends on the university, the country etc. I have been taught both of them in my first year bachelor Calculus. – Shashi Jun 02 '17 at 06:36
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Partial fractions for this one... $\int_1^2 \frac{336}{\ln(\frac{5}{2}) x(x^2+4)} dx \\\ $

I cheated...I played with the constant multiples to get 42 for the answer to the definite integral. :p

randomgirl
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