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
Nov 26, 2017

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 [a,b][a,b] using nn rectangles is:
sum_(i=1)^n f(a+iDeltax)Deltax
where Deltax=(b-a)/n

Plugging in the numbers, we get:
Deltax=(7-1)/3=6/3=2
sum_(i=1)^3 2(3(1+2*i)^2+2(1+2*i)+5)

We can multiply out the 2:
sum_(i=1)^3 6(1+2*i)^2+4(1+2*i)+10

Now, let's evaluate it:
6(1+2*1)^2+4(1+2*1)+10+6(1+2*2)^2+4(1+2*2)+10+6(1+2*3)^2+4(1+2*3)+10=

=6(3)^2+4(3)+10+6(5)^2+4(5)+10+6(7)^2+4(7)+10

=6*9+12+10+6*25+20+10+6*49+28+10

=588

So, a 3-part right-sided rectangle Riemann approximation on the interval [1,7] of 3x^2+2x+5 gives an area of 588.

We can compare this to the actual answer, which would be computed using the anti-derivative:
int_1^7 3x^2+2x+5=420

If we look at the difference, we get 588-420=168, which is an error of 40%.

Nov 26, 2017

LRS = 276

Explanation:

Let:

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

We want to estimate int \ f(x) \ dx the interval [1,7] with 3 strips; thus:

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 ExcelSteve M using Microsoft Excel

Left Riemann Sum

http://mathworld.wolfram.com/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