A curiously convoluted integration problem

Calculus Level 4

$\large \int^{50}_0 \frac{50!}{49^{50}-1} \left(e^{x+(50-x)\ln(50-x)}-e^{50+x(\ln(x)-1)} \right) \, dx = \, ?$

The answer is 0.

1 solution

Jeremy Marquardt
May 18, 2017

Since the $\frac{50!}{49^{50}-1}$ is a red herring, I'll leave it out.

$\int_0^{50} \left ( e^{x}(50-x)^{(50-x)} - e^{(50-x)}x^x \right ) dx$

$\int_0^{50} e^{x}(50-x)^{(50-x)} dx - \int_0^{50} e^{(50-x)}x^x dx$

I'll focus on the second integral $u = 50-x$ so $du = -dx$

$-\int_{50}^0 e^{(u)}(50-u)^{(50-u)} -du$

$\int_0^{50} e^{(50-x)}x^x dx = -\int_{0}^{50} e^{(u)}(50-u)^{(50-u)} du = -\int_0^{50} e^{x}(50-x)^{(50-x)} dx$

therefore the second half of the integral subtracts the first half, and the result is zero

I think the integrand is very poorly written to reflect the solution.

For instance, $\displaystyle \large e^{x + \ln (50 - x)(50 - x)}$ should equal $\displaystyle \large e^x \cdot (50 - x)^2$ . If you had wanted to express $\displaystyle \large e^x \cdot (50 - x)^{(50 - x)}$ , your integrand in the problem should have been $\displaystyle \large e^{x + \ln (50 - x)^{(50 - x)}}$ .

Zach Abueg - 4 years ago

I've never understood the convention of omitting parenthesis when using natural log, but I'll move some stuff around so this confusion doesn't arise again.

Thanks for your comment!

Also, As for your last sentence, I plugged in 49 for x; I'm pretty sure e^49 does not equal 1. So, I'm pretty sure your last latex snippet should've read $e^{x+ln(50-x)^{(50-x)}}$ . And, at that point, raising the second (50-x) creates more confusion than it helps.

Jeremy marquardt - 4 years ago

Thanks buddy! And yeah I did notice that haha, I changed it.

Zach Abueg - 4 years ago

A curiously convoluted integration problem

The answer is 0.

1 solution

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