While I was watching some physics lectures, I saw a professor write down the $\int r*dr$. The writing multiplication sign (normally just implied) prompted me to attempt to solve this integral without the use of the fundamental theorem. So I wrote as follows:
$r(\Delta r) +(r+\Delta r)(\Delta r)+(r+2\Delta r)(\Delta r)...$
However, if I take this $\Delta r$ to $0$, I will end up with 0. What am I doing wrong?
Thanks :)
You need to be more specific, group terms where warranted, and take a limit.
Let's add bounds to the integral.
$$\int_a^b r\ dr$$
Now, say we are going to divide the interval into $n$ chunks, so that each interval is of width $\Delta r = (b-a)/n$.
Starting at $r = a$, our Riemann approximation becomes...
$$a \left( \frac{b-a}{n} \right) + \left( a + \frac{b-a}{n} \right) \left( \frac{b-a}{n} \right) + \left( a + 2 \cdot \frac{b-a}{n} \right) \left( \frac{b-a}{n} \right) + \cdots + \left( a + (n-1)\cdot \frac{b-a}{n} \right) \left( \frac{b-a}{n} \right)$$
Simplfy the expression.
$$\left[ a + \left( a + \frac{b-a}{n} \right) + \left( a + 2 \cdot \frac{b-a}{n} \right) + \cdots + \left( a + (n-1)\cdot \frac{b-a}{n} \right) \right] \left( \frac{b-a}{n} \right)$$
$$\left[ na + \left(\frac{b-a}{n} \right) + \left(2 \cdot \frac{b-a}{n} \right) + \cdots + \left((n-1)\cdot \frac{b-a}{n} \right) \right] \left( \frac{b-a}{n} \right)$$
Use the fact that $1 + 2 + \cdots + (n-1) = \frac{n(n-1)}{2}$.
$$\left[ na + \left( \frac{n(n-1)}{2} \right) \left(\frac{b-a}{n} \right) \right] \left( \frac{b-a}{n} \right)$$
Continue simplifying...
$$na \left( \frac{b-a}{n} \right) + \left( \frac{n(n-1)}{2} \right) \left(\frac{b-a}{n} \right)^2 $$
$$a(b-a) + \frac{(n-1)(b-a)^2}{2n}$$
$$a(b-a) + \left( \frac{1}{2} - \frac{1}{2n}\right)(b^2 - 2ab + a^2)$$
$$ab - a^2 + \frac{a^2}{2} - ab + \frac{b^2}{2} - \frac{b^2 - 2ab + a^2}{2n}$$
$$\frac{b^2}{2} -\frac{a^2}{2} - \frac{b^2 - 2ab + a^2}{2n}$$
Now, take the limit as $\Delta r \rightarrow \infty$.
$$\lim_{n \rightarrow \infty} \left[ \frac{b^2}{2} -\frac{a^2}{2} - \frac{b^2 - 2ab + a^2}{2n} \right] = \frac{b^2}{2} -\frac{a^2}{2}$$
Thus,
$$\int_a^b r\ dr = \frac{b^2}{2} - \frac{a^2}{2},$$
which implies the indefinite integral takes the following form.
$$\int r\ dr = \frac{r^2}{2} + C$$