Question

Compute the definite integral as the limit of LEFT RIEMANN SUMS Problem posted below

Answer 1

\[\int\limits_{-3}^{3} x^3 dx\]

Answer 2

\[\lim_{n\to~\infty}\sum_{i=0}^{n}\frac1{b-a}f(...)\]
hmm i have to remember the middle part

Answer 3

a + i/(b-a) seems to be the left sums parts to me

start at a and work your way by increments of the division of the interval

Answer 4

\[\lim_{n\to~\infty}~\sum_{i=0}^{n}\frac1{b-a}f(a+i\frac{1}{b-a})\]
does that ring a bell?

Answer 5

\[\lim_{n\to~\infty}~\sum_{i=0}^{n}\frac{b-a}{n}f(a+i\frac{b-a}n)\]
\[\lim_{n\to~\infty}~\sum_{i=0}^{n}\frac{6}{n}(-3+i\frac{6}n)^3\]

Answer 6

might be n-1 up top the summation tho

Answer 7

wouldn't the answer be 0??

Answer 8

i think it does my biggest problem with this is not the solving part but figuring out how many intervals there are to find N, but either way it is 0

Answer 9

subintervals*

Answer 10

I believe the interval flips from\[\int\limits_{3}^{0} \] to \[-\int\limits_{0}^{3}\] (BTW That's a 3 on both, not a 2 even though it looks like a 2)....Oh yeah, is this what you mean???

Answer 11

but where does the -3 in \[\int\limits_{-3}^{3}\] go. does it just change to a 0 because its negative?

Answer 12

finding the area of a rectangle deals with width and height

the width of the rectangles has to do with the division of the internal into smaller parts:

interval = b-a , divided into n parts. as n tends to infinity these widths turn into an integration.

the height of the rectangle is the value of the function of the left side of the given function for each smaller division that we iterate thru

the first height is at a, the second is at a+(b-a)/n, the next as at a+2(b-a)/n and so forth, giving us f(a+i(b-a)/n)

so, where to end the division process, and thereby the iterations?

1 2 3 4 5 6
o------o--------o-------o------o-------o------o
a x1 x2 x3 x4 x5 b
first last
height height

if we divide the interval up into 6 equal parts, we have 5 iterations to sum up

if we divide the interval up into 23 equal parts, we have 22 iterations to sum up

if we divide the interval up into n equal parts, we have (n-1) iterations to sum up