4.2. RIEMANN SUMS
225
setting where f (x) ≥ 0 on [a, b]. Throughout, unless otherwise indicated, we also assume
that f is continuous on [a, b].
The first choice we make in any such approximation is the number of rectangles. If we
x 0
a
x 1
x 2
· · ·
x i x i+1
△x
· · ·
x n−1 x n
b
Figure 4.13: Subdividing the interval [a, b] into n subintervals of equal length △x.
say that the total number of rectangles is n, and we desire n rectangles of equal width to
subdivide the interval [a, b], then each rectangle must have width △x =
b−a
n . We observe
further that x 1 = x 0 + △x, x 2 = x 0 + 2△x, and thus in general x i = a + i△x, as pictured in
Figure 4.13.
We use each subinterval [x i , x i+1 ] as the base of a rectangle, and next must choose
how to decide the height of the rectangle that will be used to approximate the area under
y = f (x) on the subinterval. There are three standard choices: use the left endpoint of
each subinterval, the right endpoint of each subinterval, or the midpoint of each. These
are precisely the options encountered in Preview Activity 4.2 and seen in Figure 4.12. We
next explore how these choices can be reflected in sigma notation.
If we now consider an arbitrary positive function f on [a, b] with the interval subdivided as shown in Figure 4.13, and choose to use left endpoints, then on each interval of
the form [x i , x i+1 ], the area of the rectangle formed is given by
A i+1 = f (x i ) · △x,
as seen in Figure 4.14. If we let L n denote the sum of the areas of rectangles whose heights
are given by the function value at each respective left endpoint, then we see that
L n = A 1 + A 2 + · · · + A i+1 + · · · + A n
= f (x 0 ) · △x + f (x 1 ) · △x + · · · + f (x i ) · △x + · · · + f (x n−1 ) · △x.
In the more compact sigma notation, we have
L n =
n−1
i=0
f (x i )△x.
Note particularly that since the index of summation begins at 0 and ends at n − 1, there
are indeed n terms in this sum. We call L n the left Riemann sum for the function f on the
interval [a, b].
225
setting where f (x) ≥ 0 on [a, b]. Throughout, unless otherwise indicated, we also assume
that f is continuous on [a, b].
The first choice we make in any such approximation is the number of rectangles. If we
x 0
a
x 1
x 2
· · ·
x i x i+1
△x
· · ·
x n−1 x n
b
Figure 4.13: Subdividing the interval [a, b] into n subintervals of equal length △x.
say that the total number of rectangles is n, and we desire n rectangles of equal width to
subdivide the interval [a, b], then each rectangle must have width △x =
b−a
n . We observe
further that x 1 = x 0 + △x, x 2 = x 0 + 2△x, and thus in general x i = a + i△x, as pictured in
Figure 4.13.
We use each subinterval [x i , x i+1 ] as the base of a rectangle, and next must choose
how to decide the height of the rectangle that will be used to approximate the area under
y = f (x) on the subinterval. There are three standard choices: use the left endpoint of
each subinterval, the right endpoint of each subinterval, or the midpoint of each. These
are precisely the options encountered in Preview Activity 4.2 and seen in Figure 4.12. We
next explore how these choices can be reflected in sigma notation.
If we now consider an arbitrary positive function f on [a, b] with the interval subdivided as shown in Figure 4.13, and choose to use left endpoints, then on each interval of
the form [x i , x i+1 ], the area of the rectangle formed is given by
A i+1 = f (x i ) · △x,
as seen in Figure 4.14. If we let L n denote the sum of the areas of rectangles whose heights
are given by the function value at each respective left endpoint, then we see that
L n = A 1 + A 2 + · · · + A i+1 + · · · + A n
= f (x 0 ) · △x + f (x 1 ) · △x + · · · + f (x i ) · △x + · · · + f (x n−1 ) · △x.
In the more compact sigma notation, we have
L n =
n−1
i=0
f (x i )△x.
Note particularly that since the index of summation begins at 0 and ends at n − 1, there
are indeed n terms in this sum. We call L n the left Riemann sum for the function f on the
interval [a, b].