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) (3x2+2x+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;
Steve M using Microsoft Excel
Left Riemann Sum
http://mathworld.wolfram.com/
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