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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 :)

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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.

$$\int r\ dr = \frac{r^2}{2} + C$$

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  • $\begingroup$ Yes, here you took a definite integral using the reimann sum method. I know about this technique, however I'm looking at something that doesn't require an introduction of bounds. Is there any way to do that? $\endgroup$ – Dude156 Jan 28 at 4:10
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    $\begingroup$ You can't do a Riemann approximation without bounds, no. The indefinite integral function is defined in terms of the definite integral. $$\int f(x) dx \equiv \left( \int_0^x f(u)du \right) + C$$ $\endgroup$ – Trevor Kafka Jan 28 at 4:20
  • $\begingroup$ ahh, now I understand what you mean by "be more specific". Thanks man! $\endgroup$ – Dude156 Jan 28 at 4:38

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