Left Endpoint Integration Example - # 3

[Jason76]

Evaluate the Riemann sum for $\displaystyle f(x) = 1 - \dfrac{1}{2}x$ with six subintervals, taking the sample points to be left endpoints.

$\displaystyle f(x) = 1 - \dfrac{1}{2}x$ on interval $\displaystyle [2,14]$ given $\displaystyle [a, b]$

$\displaystyle \sum\limits_{i=4}^n \Delta x [f(a + i \Delta x - \Delta x)]$

$\displaystyle \Delta x = \dfrac{b - a}{n}$

$\displaystyle \Delta x = \dfrac{14 - 2}{6} = 2$

$\displaystyle n = 6$

$\displaystyle \sum\limits_{i=6}^n (2)[[f(2 + (1)(2) - (2))] + [f(2 + (2)(2) - (2))] + [f(2 + (3)(2) - (2))] + [f(2 + (4)(2) - (2))] + [f(2 + (5)(2) - (2))] + [f(2 + (6)(2) - (2))]$

$\displaystyle \sum\limits_{i=6}^n (2)[[f(2 + 2 - (2))] + [f(2 + 4 - (2))] + [f(2 + 6 - (2)]) + [f(2 + 8 - (2))] + [f(2 + 10 - (2))] + [f(2 + 12 - (2))]$

$\displaystyle \sum\limits_{i=6}^n (2)[[f(2)] + [f(4)] + [f(6)] + [f(8)] + [f(10)] + [f(12)]$

$\displaystyle \sum\limits_{i=6}^n (2)[[1 - \dfrac{1}{2}(2)] + [1 - \dfrac{1}{2}(4)] + [1 - \dfrac{1}{2}(6)] + [1 - \dfrac{1}{2}(8)] + [1 - \dfrac{1}{2} (10)] + [1 - \dfrac{1}{2}(12)]$

$\displaystyle \sum\limits_{i=6}^n (2)[[1 - (1)] + [1 - (2)] + [1 - (3)] + [1 - (4)] + [1 - (5)] + [1 - (6)]$

$\displaystyle \sum\limits_{i=6}^n (2)[[0] + [-1] + [-2] + [-3] + [-4] + [-5]$

$\displaystyle \sum\limits_{i=6}^n [0] + [-2] + [-4] + [-6] + [-8] + [-10]$ On the right track?

 

[HallsofIvy]

I think you are writing far too much in too abstract away, possibly because you are copying the general formula from the text rather than thinking and using the idea of the sum.

You are given the interval [2, 14] (it really makes no sense to then write "given [a, b]" after that!) and you want to divide it into 6 equal intervals. (14- 2)/6= 12/6= 2 so the endpoints of the intervals are 2, 4, 6, 8, 10, 12, 14. The left endpoints are, of course, 2, 4, 6, 8, 10, and 12. You need to calculate f(2)= 1- (1/2)(2)= 0, f(4)= 1- (1/2)(4)= -1, f(6)= 1- (1/2)(6)= -2, f(8)= 1- (1/2)(8)= -3, f(10)= 1- (1/2)(10)= -4, and f(12)= 1- (1/2)(12)= -5, the height of each "rectangle" in the Riemann sum. The "base" of each rectangle is 2 so each (signed) "area" is 2(0)= 0, 2(-1)= -2, 2(-2)= -4, 2(-3)= -6, 2(-4)= -8, and 2(-5)= -10.

The Riemann sum is 0- 2- 4- 6- 8- 10= -30. That is, essentially, the last sum you have although, again, it is written incorrectly- you do not use "$\displaystyle \sum$" when you are writing out the entire sum.

 

[Jason76]

The answer of $\displaystyle =-30$ is correct.