EDIT: My question has been answered and I've solve the problem, thanks for the help.

I'm having some difficulty with a problem from the "Calculus, Early Transcendentals" textbook, section 5.2 number #24. The problem is as follows:

Use the form of the definition of the integral given in Theorem 4 to evaluate the integral: $$ \int_0^5 (1+2x^3) dx$$

Theorem 4 states that:

If f is integrable on [a, b] then: $$ \int_a^b f(x) dx = \lim_{n \to \infty} \sum\limits_{i=1}^n f(x_i) \Delta x $$ where $ \Delta x = \frac{b-a}{n} $ and $x_i = a + i \Delta x$

The problem I'm running into is that, when trying to solve it, I get stuff like $i^4$ which I don't know the summation of.

I'm attaching my work here as an image because I find it pretty tough to use latex.

My work

Thank you VERY much for any help! :)


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, $\sum\limits_{i=1}^n i^4 = {n+1\choose 2}+14{n+1\choose 3}+36{n+1 \choose 4}+24{n+1\choose 5} = (1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n$.

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  • $\begingroup$ How did you find this? Isn't there another way to solve this problem? The book provided formulas for $i, i^2$, and $i^3$ but not for $i^4$, which leads me to believe there might be another way of solving it that doesn't involve this... $\endgroup$ – GuiR Feb 23 '14 at 17:09
  • $\begingroup$ I Edited the answer:) $\endgroup$ – r9m Feb 23 '14 at 19:25
  • $\begingroup$ Thanks! I learned something new hehe. I found my error in my work and turns out I don't need $i^4$ at all, but thanks! $\endgroup$ – GuiR Feb 23 '14 at 20:40

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