FormulasforRiemannSumsAssumef（x）isagiven：对黎曼假..

In particular:
If f (x) is increasing on [xi−1, xi] then Maxi = f (xi).
If f (x) is increasing on the entire interval [a, b], then Upn = Rn,
i.e., the upper Riemann sum equals the right end point Riemann sum.
Step 7. To compute the upper Riemann sum, Upn, you must find the absolute maximum of the function f (x) on each of the subintervals [xi−1, xi]. Call this absolute maximum Maxi. Then the upper Riemann sum is given by the formula
Formulaห้องสมุดไป่ตู้ for Riemann Sums
Assume f (x) is a given function that is continuous on a closed interval [a, b]. Let n be a given positive integer. Use the following steps to draw and compute the left, right and mid-point rectangles and Riemann sums with n terms for f (x).
Although in general it is hard to compute the upper and lower Riemann sums, they do have the advantages of always giving upper and lower estimates of the areas or integrals:
Thus you can get an error bound on the actual area (or, more generally, the integral):
Lown ≤
b
f (x) dx ≤ Upn, for every n.
a
3
Step 3. The ith interval in the partition is [xi−1, xi], and so its
left end point is xi−1
right end point is xi
mid-point
is
xi−1 + xi 2
=
xi−1
+
1 2
∆x
Step 4. The left endpoints Riemann sum is
To compute the upper and lower Riemann sums for a partition as above with n rectangles first do exactly the same as above to compute the end points of all the intervals. As before, the ith interval in the partition is [xi−1, xi], and this is the base of the ith rectangle.
Step 1. First compute that value of ∆x as ∆x = b − a . n
Step 2. The end points of the intervals in the partition are given by the formulas: x0 = a x1 = a + ∆x x2 = a + 2∆x (= x1 + ∆x) x3 = a + 3∆x (= x2 + ∆x) ... xi = a + i∆x (= xi−1 + ∆x) ... xn = a + n∆x = b (= xn−1 + ∆x)
Step 8. To compute the lower Riemann sum, Lown, you must find the absolute minimum of the function f (x) on each of the subintervals [xi−1, xi]. Call this absolute minimum Mini. Then the lower Riemann sum is given by the formula
In particular, if the graph of f (x) is above the x-axis on the interval [a, b], then
The area under f (x) is always ≤ Upn, for every n and the area under f (x) is always ≥ Lown, for every n.
In certain cases it is easy to compute the maximum values: For example, if f (x) is increasing on the ith interval [xi−1, xi], then the maximum occurs at the right hand end point and so
Ln = ∆x f (x0) + f (x1) + · · · + f (xn−1)
n
= ∆x f a + (i−1)∆x
i=1
Step 5. The right endpoints Riemann sum is
Rn = ∆x f (x1) + f (x2) + · · · + f (xn)
n
= ∆x
f a + i∆x
i=1
Step 6. The mid-points Riemann sum is
Mn = ∆x
f (x0
+
1 2
∆x)
+
f (x1
+
1 2
∆x)
+
·
·
·
+
f (xn−1
+
1 2
∆x)
n
= ∆x
f
a
+
(i−
1 2
)∆x
i=1
Upper and Lower Riemann Sums, Upn and Lown
Lown = ∆x(Min1 + Min2 + · · · + Minn)
n
= ∆x Minn .
i=1
As with the upper sums, to compute these lower sums you must (separately) find the minimum of f (x) on each subinterval. This could take a lot of work for big n, because there is no general formula that tells you what this value is. If f (x) is differentiable, then you can again find the absolute minimum by the methods we developed in Section 4.1. In certain cases it is easy to compute the minimum values: For example, if f (x) is increasing on the ith interval [xi−1, xi], then the minimum occurs at the left hand end point and so
Upn = ∆x(Max1 + Max2 + · · · + Maxn)
n
= ∆x Maxn .
i=1
Note that to compute these upper sums you must (separately) find the absolute maximum value of f (x) on each subinterval. This could take a lot of work for big n, because there is no general formula that tells you what this value is. If f (x) is differentiable, then you can find the maximum by the methods we developed in Section 4.1.
2
In particular:
If f (x) is increasing on [xi−1, xi] then Mini = f (xi−1).
If f (x) is increasing on the entire interval [a, b], then Lown = Ln, i.e., the upper Riemann sum equals the left end point Riemann sum.