I am sorry for overlooking it earlier, but in the evaluation of $\sum\limits_{i=1}^n f(x_i)\Delta_{x_i} = \sum\limits_{i=1}^n (1+(\frac{5i}{n})^3).(\frac{5}{n}) = 5 + \frac{625}{n^4}(\sum\limits_{i=1}^n i^3)$,
So you wont be needing $\sum\limits_{i=1}^n i^4$ as such.
But if you are still interested in knowing how to find the summation without having to apply mathematical induction I know a fast way of arguing combinatorially for $n>6$,
Count the number of quintuples $(x_1,x_2,x_3, x_4,x_5)$ with the property $x_5> \max\limits_{i\in S}$ from the set $S=\{1,2,...,n,n+1\}$,
then as $x_5$ varies from $2$ to $n+1$, the positions $(x_1,x_2,x_3, x_4)$
the positions can be filled in $\sum\limits_{i=1}^n i^4$ ways.
Counting it according to the cases: $5 = 4 + 1 = 2 + 2 + 1 = 3 + 1 + 1 = 2+1+1+1=1+1+1+1+1$ (where the numbers in partitions indicate the number of equal components in that partition, i.e. for example $5=4+1$ corresponds to the case $x_1=x_2=x_3=x_4<x_5$; and $2+2+1$ corresponds to the case two pairs have equal value but the value of the pair are different ,.. and so on). Count the number of ways in which each of the partitioning can happen: $1,14,36,24$ respectively (in the same order as I have presented the partitions).
You get the number of ways to be: ${n+1\choose 2}+14{n+1\choose 3}+36{n+1 \choose 4}+24{n+1\choose 5}$
Therefore, $\sum\limits_{i=1}^n i^4 = {n+1\choose 2}+14{n+1\choose 3}+36{n+1 \choose 4}+24{n+1\choose 5} = (1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n$.
The slope at $(x,y)$ is another way of saying the derivative of the curve, so the question can be read as find the equation of a line that satisfies
\begin{equation}
\frac{dy}{dx} = \frac{x}{y},
\end{equation}
and that passes through the point $(x,y) = (0,6)$.
Hint: this is a first order ODE that can be solved and will have one constant of integration that can be set such that the solution passes through the desired point.
Best Answer
Minor nitpick - you should have had $-3 \ln |t|$ instead of $+$.
As to the main question, what's going on here is that $f'(0)$ isn't defined, and the derivative of $f$ approaches $\pm \infty$ at this point. So you'll actually have a vertical asymptote here, and the values of $C$ on either side are allowed to be different, since it does not affect the derivative. Visually, this means that there are two different sides of the function separated by the asymptote, and you can move each side up and down without affecting the derivative. This is why you obtain two different values of $C$.