Question

How do you calculate the left and right Riemann sum for the given function over the interval [2,6], for `f(x)=5x^2+3x+2`?

Answer 1

`\trianglex=1`
Right: 320
Left: 492

Explanation:

First, we have to determine how we want to split each part. The more you split it, the more accurate it will be. To make things simple we are going to make each part one unit long. To find how many parts we are going to have we solve the for the following
`\trianglex=\frac{b-a}{n}`
`1=\frac{6-2}{n}`
`n=4`

We now have to get the points we want to plug in. We use our range adding our `\trianglex` value to each one to get the following points.
`x_0=2,x_1=3,x_2=4,x_3=5,x_4=6`

For the right hand rule we use the following
`\trianglex\sum_{i=0}^{n-1}f(x_i)`
`1\cdot\sum_{i=0}^3f(x_i)`

We can then solve this
`f(x_0)+f(x_1)+f(x_2)+f(x_3)`
`=f(2)+f(3)+f(4)+f(5)`
`=28+56+94+142`
`=320`

For the left hand rule we use this instead
`\trianglex\sum_{i=1}^{n}f(x_i)`
`1\cdot\sum_{i=1}^4f(x_i)`

We then solve
`f(x_1)+f(x_2)+f(x_3)+f(x_4)`
`=f(3)+f(4)+f(5)+f(6)`
`=56+94+142+200`
`=492`

Increasing `\trianglex` will produce values that are closer to the actual value of `1208/3=402.\bar{66}`.