Product Integral

To start, the answer to this question is wrong. It should be $\lfloor e^{\frac{5}{2}(41\ln(205)-\ln(5)-40)}\rfloor$ , which is MUCH too big. The desired answer to get points is the floor of the exponent on that monster. I guessed that that was what the problem writer meant.

So, I bet you're all wondering how I know the answer. That's the fun part, though. :D

Look at that limit. I don't think I know anything much about infinite products. Woe is me! I guess there's nothing we can do in this horrid non-Newtonian calculus.

...Nope. We have a special way of converting products to sums, and we know a LOT about sums. Indeed, for my fellow algebraists, we have a homomorphism of groups from the multiplicative group of positive real numbers to the additive group of real numbers. We call this homomorphism...the logarithm! Because the natural logarithm commutes with the limit operator here (Remember that $\ln:\mathbb{R}_+\to\mathbb{R}$ is a continuous function everywhere on its domain, so we can switch the places of the limit operator and the natural logarithm in this case), we can write that

$\displaystyle \lim_{n\rightarrow +\infty}\prod_{1\leq i\leq n}f(x_i)^{\Delta x_i}=e^{\displaystyle \ln\left(\lim_{n\rightarrow +\infty}\prod_{1\leq i\leq n}f(x_i)^{\Delta x_i}\right)}=e^{\displaystyle \left(\lim_{n\rightarrow +\infty}\sum_{1\leq i\leq n}\ln(f(x_i))\Delta x_i\right)}=e^{\displaystyle \left(\, \int\limits_{[a,b]}(\ln\circ f)~\mathrm{d}x \!\right)}$ .

Thus, we just need to integrate $\ln(2x+5)$ from 0 to 100 AND EXPONENTIATE to get our answer. I'll leave that trivial exercise for fun.