Induction and Integral Question (sum of 5th powers, integral of $x^5$)

I am quite lost on this question:

(a) For $n\in \mathbb{N}$, use induction to show that

$$\sum_{k=1}^{n}k^{5}=\frac{2n^{6}+6n^{5}+5n^4-n^2}{12}$$

(b) Fix $b>0$. Use the definition of the definite integral together with the Riemann partition $$P_{n}:=\left \{ (x_{k},\left [ x_{k-1},x_{k} \right ]) \right \}$$ where $x_{k}:=\frac{kb}{n}$ for $0\leq k\leq n$, to compute $\int_{0}^{b}x^{5}dx$

I know with induction that I should be starting with a base case, like $n=1$. My textbook has 1-2 pages about induction but I am not able to properly understand and apply it here. Part (a) must help with the solving of part (b)

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(a): To use induction, you need to prove the equation for the base case (in this case $n=1$), then prove that given the equation is true for $n=m$, it is true for $n=m+1$ as well. In other words, you show that (eqn. is true for $n=m$) implies (eqn. is true for $n=m+1$). Then, since the equation is true for $n=1$, it must be true for $n=2$ ($m=1$ and $m+1=2$). By the same logic, it is true for 2, 3, 4, etc., and therefore it is true for all $n\in\mathbb{N}$.

Here, the base case is clearly true: the left hand side is 1 and the right hand side is $\frac{2+6+5-1}{12}=1$. For the inductive step, you know that $\sum^{m+1}_{k=1} k^5=\sum^{m}_{k=1} k^5+(m+1)^5$. So you need to show that $[2(m+1)^6+6(m+1)^5+5(m+1)^4-(m+1)^2]/12=(2m^6+6m^5+5m^4-m^2)/12+(m+1)^5$. Unfortunately, this requires a lot of messy algebra.

(b): To integrate a function $f$, you need to add up all the $f(x_k)(\Delta x)_k$'s. (For each sample rectangle $k$, $f(x_k)$ is the value of $f$ somewhere in the rectangle, and $(\Delta x)_k$ is the width of the rectangle). In this case, $f(x)=x^5$ and $x_k=kb/n$, so $f(x_k)=(b/n)^5 * k^5$. You are cutting up the interval $[0,b]$ into $n$ rectangles, so the width of each one is $b/n$. The Riemann sum is $\sum^n_{k=1} f(x_k)(\Delta x)_k=\sum^n_{k=1} (b/n)^5 * k^5 * (b/n)=(b/n)^6 \sum^n_{k=1} k^5=(\frac{b}{n})^6 \frac{2n^6+6n^5+5n^4-n^2}{12}=b^6 \frac{2n^6+6n^5+5n^4-n^2}{12n^6}$. The integral is the limit of the Riemann sum as the number of sample rectangles ($n$) goes to infinity. The $n^5,n^4,n^2$ terms become negligible compared to the $n^6$ term as $n$ becomes very large, so the final answer is $\frac{1}{6}b^6$.

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You are being asked (in (a)) to prove a statement about an integer $n$. As you say, you first establish it for $n=1$. Then you write down the statement for $n$, and you write down the statement for $n+1$, and you show that you can deduce the second statement from the first. By induction, that will prove the statement for all (positive integers) $n$. Have a go at that and then we'll get back to the integral.

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$$\sum_{k=1}^{n}k^{5}=1^{5}+2^{5}+3^{5}+\cdots +(n-1)^{5}+n^{5}$$ $$\sum_{k=1}^{n+1 }k^{5}=1^{5}+2^{5}+3^{5}+\cdots +n^{5}+(n+1)^{5}$$ $$=\frac{2n^{6}+6n^{5}+5n^4-n^2}{12}+(n+1)^{5}$$ $$=\frac{2n^{6}+6n^{5}+5n^4-n^2+12(n+1)^{5}}{12}$$ – Malthus Nov 24 '11 at 6:49
$$=\frac{2n^{6}+18n^{5}+65n^{4}+120n^{3}+119n^{2}+60n+12}{12}$$ The last is as far as I am able to go. I cannot figure out how to manipulate it into: $$=\frac{2(n+1)^{6}+18(n+1)^{5}+65(n+1)^{4}+120(n+1)^{3}+119(n+1)^{2}+60(n+1)+12‌​}{12}$$ – Malthus Nov 24 '11 at 7:06
That last expression is not where you want to go. You want to go to the statement for $n+1$, which is the equation in your question but with $n$ replaced everywhere by $n+1$. – Gerry Myerson Nov 24 '11 at 12:05