Proof of integration power rule

[acegikmoqsuwy]

Hey, I was just wondering if there was a way to prove the power rule for integration using the definition of a definite integral. And I don't mean using the proof for the differentiation power rule, I mean is it possible to derive \displaystyle\large\int_a^b x^c=\frac{b^{c+1}-a^{c+1}} {c+1} using the formula \displaystyle\large\int_a^b f(x)=\lim\limits_{n\to\infty} \sum\limits_{i=1}^n f(x^*_i)\Delta x ?

I tried substituting the values for f(x^*_i) and \Delta x without any success.

 

[micromass]

Apostol does it in his book. Section I.23.

 

[shinobi20]

Hi micromass just want to ask why did apostol used this inequality to prove the integration of power rule? He just suddenly said "We begin with this inequality", I'm confused, why did he use that? And how did he come up with such an inequality (knowing it will work)?

INEQUALITY:
k=1 to k=n-1 ∑k^p < (n^p+1)/(p+1) < k=1 to k=n ∑k^p

 

[LAZYANGEL]

Take a simple definite integral like $f(x)=x$ and use simple limits.

$\displaystyle\large\int_0^x f(x) \: dx=\lim\limits_{n\to\infty} \sum\limits_{i=1}^n f(x^*_i) \Delta x$

$\displaystyle\large\int_0^x x \: dx=\lim\limits_{n\to\infty} \sum\limits_{i=1}^n (x_i) \frac{x}{n}$

$\displaystyle\large\int_0^x x \: dx=\lim\limits_{n\to\infty} \frac{x}{n} \sum\limits_{i=1}^n (x_i)$

Using left hand Reimann sum the width of the next interval is $x_i = \Delta x \cdot i$

$\displaystyle\large\int_0^x x \: dx=\lim\limits_{n\to\infty} \frac{x}{n} \sum\limits_{i=1}^n (\frac{x}{n} \cdot i)$

$\displaystyle\large\int_0^x x \: dx=\lim\limits_{n\to\infty} \frac{x^2}{n^2} \sum\limits_{i=1}^n (i)$

Use the partial sums formula $\sum\limits_{i=1}^n (i) = \frac {n (n+1)}{2}$

$\displaystyle\large\int_0^x x \: dx=\lim\limits_{n\to\infty} \frac{x^2}{n^2} \cdot \frac{n(n+1)}{2}$

$\displaystyle\large\int_0^x x \: dx=\lim\limits_{n\to\infty} \frac{ n^2 x^2 + nx^2}{2n^2}$

$\displaystyle\large\int_0^x x \: dx=\frac {x^2}{2}$

You can generalize it for $x^c$ but it will be hairier than a gorilla.