# How do you calculate the left Riemann sum for the given function over the interval [1,7], using n=3 for #(3 x^2+2 x +5) #?

##### 2 Answers

588 (Using a right Riemann sum, oops)

#### Explanation:

I just realized the actual question was for a left Riemann sum, not a right one. Pardon my mistake. Below is the procedure for evaluating it with a right Riemann sum:

The general formula for a right-sided rectangle Riemann approximation on the interval

where

Plugging in the numbers, we get:

We can multiply out the

Now, let's evaluate it:

So, a

We can compare this to the actual answer, which would be computed using the anti-derivative:

If we look at the difference, we get

# LRS = 276 #

#### Explanation:

Let:

# f(x) = 3x^2+2x+5 #

We want to estimate

# Deltax = (7-1)/3 = 2#

Note that we have a **fixed** interval (strictly speaking a Riemann sum can have a varying sized partition width). The values of the function are tabulated as follows;

**Left Riemann Sum**

# LRS = sum_(r=0)^2 f(x_i) \ Deltax_i #

# " " = 2 * (10 + 38 + 90) #

# " " = 2 * (138) #

# " " = 276 #

**Actual Value**

For comparison of accuracy:

# Area = int_1^7 \ 3x^2+2x+5 \ dx #

# " " = [x^3+x^2+5x]_1^7 #

# " " = (343+49+35) - (1+1+5)#

# " " = 427-7#

# " " = 420#