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Pay it Forward - September 8, 2011
Rating: 4.8/5 (131 votes cast)
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Spiked Math Comic - Pay it Forward



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14 Comments

Just a thought: you ought to set the default rating so it doesn't default to 0/5 when there are no votes. Be subversive and set it to default to 5...

How do you do that integral?

I'm fairly sure that you write 2^x as e^(x ln 2), then integrate by parts twice.

seems like \frac{(2^x)(lg2)(sin x)-(2^x)(cos x)}{1+lg^2 2} + c

I know the "+c" is correct and necessary, as mentioned yesterday.

Isn't it just two integrations by parts, and a linear equation?

... which is a lot more than the average student can do :(

Solve this for the integral and you got your answer :-)
PS: I bet that 90% of the readers immediately grabbed pen and paper when they saw this comic.

And then you simply move the ln(2)^2 * integral of 2^x sin(x)dx to the left side, and divide both sides by (1+ln(2)^2). Really simple once you get to that point, but getting there would take me some time, I've almost completely forgot all things calculus.

No, I realized that two partial integrations would reduce it to a linear equation => the problem has a (closed-form, analytical) solution => the problem is trivial.

Go for the imaginary part of the integral of exp((ln(2)+i)x). None of that silly "integrating by parts" business.

i grabbed immediately wolfram alpha when is saw this comic :D

Ja, just express the sin(x) through the exponential and you can solve it in a few lines...

I think subsituting u=2^x and du=2^x*ln(x) is a good place to start since you can then replace x with ln(u)/ln(2). so you get 1/ln(2)Int(sin(ln(u)/ln(2)))du then sub r=ln(u) dr=(1/u)du or du=(e^r)dr and you get e^r*sin(r/ln(2)) inside the Int and I forget the formula to Int that but if you know that it should just be sub back in from there.

why not integration by parts?

Just wondering, is Mike(the character, not the cartoonist)a kid or a professor?

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