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Riemann Sums and Integrals
Definition: Let f (x) be a continuous function on the interval [a, b]. Then the definite integral
of f from a to b is defined as
b
Z
f (x) dx = lim
a
n
X
n→∞
f (x∗i )∆x
i=1
b−a
where ∆x =
, xi = a + i∆x, and xi−1 ≤ x∗i ≤ xi .
n ∗
But what is this xi ? Once we have chopped our interval [a, b] into n rectangles of equal width
(∆x = (b − a)/n), our ith rectangle lies on top of the sub-interval [xi−1 ,xi ]. x∗i represents
any point in the interval [xi−1 ,xi ]. By using this x∗i notation, we include left-hand sums,
right-hand sums, midpoint sums, and even the possibility that we choose a random x-value
within each subinterval.
Riemann Sums
Sums of the form
n
X
f (x∗i )∆x are called Riemann sums.
i=1
• When x∗i = xi−1 , we have a Left Hand Sum.
• When x∗i = xi , we have a Right Hand Sum.
• When x∗i = 12 (xi−1 + xi ), we have a Midpoint Sum.
Based off of our definition, we expect these Riemann sums to approach the same value, as
we send the number of rectangles to infinity.
Monotonic Functions
1. When f (x) is strictly increasing on [a, b], we always have
b
Z
LHS(n) ≤
f (x)dx ≤ RHS(n)
a
Z
Example: Suppose we want to approximate
xdx using a Riemann sum with 4 rect0
angles.
1
4
(a) LHS(4)
(b) RHS(4)
Figure 1: Riemann Sums
2. When f (x) is strictly decreasing on [a, b], we always have
b
Z
RHS(n) ≤
f (x)dx ≤ LHS(n)
a
(a) LHS(4)
(b) RHS(4)
Figure 2: Riemann Sums
So, in the case that f (x) is either always increasing, or always decreasing, we can use LHS(n)
and RHS(n) as the best bounds on the definite integral of f (x) over [a, b]. In particular, if
we want to guarantee that our estimate is within some given level of accuracy, p, of the true
answer, it suffices to ensure that
|LHS(n) − RHS(n)| < p ⇔ |f (b) − f (a)|
2
b−a
<p
n
But what if f (x) is neither always increasing, nor always decreasing?
Example: From the Riemann Sums Lab, Part 4. We want to use Riemann sums with 4
Z 4
4
x+
rectangles to estimate
dx.
x
1
4−1
3
Here, ∆x =
= , and xi = 1 + i × ∆x.
4
4
• Largest sum: in each interval, choose the value of x that maximizes f (x). Note that
this is not one of the named Riemann sums, as the choice of x∗i changes in each interval.
4
Z
1
4
x+
dx ≈ (f (1) + f (10/4) + f (13/4) + f (4))(3/4) = 13.9356
x
• Smallest sum: in each interval, choose the value of x that minimizes f (x). Note that
this is not one of the named Riemann sums, as the choice of x∗i changes in each interval.
Z
1
4
4
x+
dx ≈ (f (7/4) + f (2) + f (10/4) + f (13/4))(3/4) = 12.4624
x
(a) Largest possible Riemann sum (b) Smallest possible Riemann sum
Figure 3: Riemann Sums
Practice Problems
Z
1. Suppose you want to estimate
appropriate sum in Σ notation.
5
ln(x)dx. For each of the following, write down an
2
(a) LHS(n)
(b) RHS(n)
3
(c) MPS(n)
2. Compute the following by recognizing them as definite integrals, and using the Fundamental Theorem of Calculus.
2
n X
10i 10
(a) lim
n→∞
n
n
i=1
2
n X
10i 10
(b) lim
2+
n→∞
n
n
i=1
2 !
n
X
10
10i
(c) lim
2+
n→∞
n
n
i=1
n X
1
1
(d) lim
n→∞
1 + (k/n)2 n
i=1
n X
πk
πk π
(e) lim
cos
+
n→∞
2n
2n 2n
i=1
4