# How do you Use a Riemann sum to find area?

Oct 10, 2014

The area A of the region under the graph of $f$ above the $x$-axis from $x = a$ to $b$ can be found by

$A = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$,

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

Let us find the area of the region under the graph of $y = 2 x + 1$ from $x = 1$ to $3$.

By definition,

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

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.