# How do you use a Riemann sum to calculate a definite integral?

Oct 5, 2014

Definition of Definite Integral

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$,

where ${x}_{i} = a + i \Delta x$ and $\Delta x = \frac{b - a}{n}$.

Let us look at the following example.

${\int}_{1}^{3} \left(2 x + 1\right) \mathrm{dx}$

by definition,

$= {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \left[2 \left(1 + \frac{2}{n} i\right) + 1\right] \frac{2}{n}$

by simplifying the expression inside the summation,

$= {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \left(\frac{8}{n} ^ 2 i + \frac{6}{n}\right)$

by splitting the summation and pulling out constants,

$= {\lim}_{n \to \infty} \left(\frac{8}{n} ^ 2 {\sum}_{i = 1}^{n} i + \frac{6}{n} {\sum}_{i = 1}^{n} 1\right)$

by the summation formulas ${\sum}_{i = 1}^{n} i = \frac{n \left(n + 1\right)}{2}$ and ${\sum}_{i = 1}^{n} 1 = n$,

$= {\lim}_{n \to \infty} \left(\frac{8}{n} ^ 2 \cdot \frac{n \left(n + 1\right)}{2} + \frac{6}{n} \cdot n\right)$

by cancelling out $n$'s,

$= {\lim}_{n \to \infty} \left[4 \left(1 + \frac{1}{n}\right) + 6\right] = 4 \left(1 + 0\right) + 6 = 10$

I hope that this was helpful.