Okay, I've set

u = 1-y

du = -1dy

-du = dy

So

$\displaystyle \int k (1-y)^4 dy = k \int u^4 \cdot -du$

$\displaystyle -k 4 u^3 du$

The definite integrals are still zero though. Even if I assume there shouldn't be a negative in front of the final k, my larger integral is still smaller than the one that should be larger. (The rest of the problem has me comparing $\displaystyle \int_0^{0.4} k (1-y)^4 dy$ to $\displaystyle \int_0^{0.1} k (1-y)^4 dy$. The latter should be quite a bit larger but it is much smaller.)

I guess I'm still not really comfortable with u substitution. It seems like dy (or dx or whatever, from the original equation) was just a kind of marker meaning "with respect to" that trailed the problem. After substituting it for du, du become a factor in the original sense, something we have to multiply the rest of the problem by. Since it's okay in math to string factors together without multiplicaton signs, intending them to be multiplied, it's hard to say when du or dx or whatever is to be multiplied and when it just means "with respect to". I kind of miss multiplication signs.