# Thread: Using Riemann Sum for intergrals

1. ## Using Riemann Sum for intergrals

Evaluate the definite integral (x^3-x)dx with bounds 1 to 5 by the limit definition of a Riemann Sum.

2. Originally Posted by Hibijibi
Evaluate the definite integral (x^3-x)dx with bounds 1 to 5 by the limit definition of a Riemann Sum.
We break the interval into n steps:
$\Delta x = \frac{5 - 1}{n} = \frac{4}{n}$

We then form each little rectangle for the area. The area of the ith rectangle will be
$\Delta A = ( f(x_{i + 1}) - f(x_i) ) \Delta x$
where $x_i = 1 + i \cdot \Delta x$

So this is
$\Delta A = \left ( \left [ \left ( 1 + (i + 1) \cdot \frac{4}{n} \right ) ^3 - \left ( 1 + (i + 1) \cdot \frac{4}{n} \right ) \right ] \right .$ $\left . ~- \left [ \left ( 1 + i \cdot \frac{4}{n} \right ) ^3 - \left ( 1 + i \cdot \frac{4}{n} \right ) \right ] \right ) \cdot \frac{4}{n}$

So the estimate of the area will be
$\sum_{i = 0}^n \Delta A$

which is exact when taking the limit:
$A = \lim_{n \to \infty} \sum_{i = 0}^n \Delta A$

There's a lot of simplifying to do but the process is otherwise simple.

-Dan