7.01: Prerequisites to Numerical Integration
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Lesson 1: Prerequisites to Numerical Integration
Learning Objectives
After successful completion of this lesson, you should be able to:
1) define an integral,
2) use Riemann’s sum to approximately calculate integrals,
3) use Riemann’s sum and its limit to find the exact expression of integrals, and
4) find exact integrals of a few functions.
What is integration?
The dictionary definition of integration is combining parts so that they work together or form a whole. Mathematically, integration stands for finding the area under a curve from one point to another. It is represented by
∫baf(x) dx(7.1.1.1)
where the symbol ∫ is the integral sign, and a and b are the lower and upper limits of integration, respectively, the function f(x) is the integrand of the integral, and x is the variable of integration. Figure 7.1.1.1 represents a graphical demonstration of the concept.

Riemann Sum
Let f be defined on the closed interval [a,b], and let Δ be an arbitrary partition of [a,b] such as: a=x0<x1<x2<…<xn−1<xn=b, where Δxi is the length of the ith subinterval (Figure 7.1.1.2).
If ci is any point in the ith subinterval, then the sum
n∑i=1f(ci)Δxi, xi−1≤ci≤xi(7.1.1.2)

is called a Riemann sum of the function f for the partition Δ on the interval [a,b]. For a given partition Δ, the length of the longest subinterval is called the norm of the partition. It is denoted by ‖Δ‖ (the norm of Δ). The following limit is used to define the definite integral.
lim‖Δ‖→0n∑i=1f(ci)Δxi=I(7.1.1.3)
This limit exists if and only if for any positive number ε, there exists a positive number δ such that for every partition Δ of [a,b] with ‖Δ‖<δ, it follows that
|I−n∑i=1f(ci)Δxi|<ε(7.1.1.4)
for any choice of ci in the ith subinterval of Δ.
If the limit of a Riemann sum of f exists, then the function f is said to be integrable over [a,b] and the Riemann sum of f on [a,b] approaches the number I.
lim‖Δ‖→0n∑i=1f(ci)Δxi=I(7.1.1.5)
where
I=∫baf(x) dx(7.1.1.6)
Audiovisual Lecture
Title: Accumulation Function
Summary: This video discusses how the accumulation function works to create a basis for numerical integration.
Find the area of the region between the parabola y=x2 and the x-axis on the interval [0,4.5]. Use the right Riemann sum with four partitions of equal segment width. Compare the value with the exact value.
Solution
We evaluate the integral for the area as a limit of Riemann sums. We sketch the region (Figure 7.1.1.3), and partition [0,4.5] into four subintervals of length
Δx=4.5−04=1.125.

The points of partition are
x0=0, x1=1.125, x2=2.25, x3=3.375, x4=4.5
Let us choose ci to be right-hand endpoint of its subinterval. Thus,
c1=x1=1.125, c2=x2=2.25, c3=x3=3.375, c4=x4=4.5
The rectangles defined by these choices have the following areas:
f(c1)Δx=f(1.125)×(1.125)=(1.125)2(1.125)=1.4238
f(c2)Δx=f(2.25)×(1.125)=(2.25)2(1.125)=5.6953
f(c3)Δx=f(3.375)×(1.125)=(3.375)2(1.125)=12.814
f(c4)Δx=f(4.5)×(1.125)=(4.5)2(1.125)=22.781
The sum of the areas then is the approximate value of the integral
∫4.50x2 dx≈4∑i=1f(ci)Δx,=1.4238+5.6953+12.814+22.781=42.715
How does this compare with the exact value of the integral ∫4.50x2 dx?
∫4.50x2 dx=[x33]4.50=4.533−033=30.375
Find the exact area of the region between the parabola y=x2 and the x-axis on the interval [0,b]. Use the right Riemann’s sum with equal segment widths.
Solution
Note that in Example 7.1.1.1 for y=x2 that
f(ci)Δx=(iΔx)2(Δx)=i2(Δx)3(7.1.1.E2.1)
Thus, the sum of these areas, if the interval is divided into n equal segments, is
Sn=n∑i=1f(ci)Δx=n∑i=1i2(Δx)3=(Δx)3n∑i=1i2(7.1.1.E2.2)
Since
Δx=bn, and(7.1.1.E2.3)
n∑i=1i2=n(n+1)(2n+1)6(7.1.1.E2.4)
Substituting Equations (7.1.1.E2.3) and (7.1.1.E2.4) in Equation (7.1.1.E2.1), we get
Sn=b3n3n(n+1)(2n+1)6=b362n2+n+2n+1n2=b36(2+3n+1n2)
The definition of a definite integral can now be used.
∫baf(x) dx=lim|Δx|→0n∑i=1f(ci)Δx
To find the area under the parabola from x=0 to x=b, we have
∫b0x2 dx=lim|Δ|→0n∑i=1f(ci)Δx=limn→∞Sn=limn→∞b36(2+3n+1n2)=b36(2+0+0)=b33
Mean Value of a Function
The mean value ¯f of a function f in an interval [a,b] is given by
¯f=1b−a∫baf(x) dx(7.1.1.7)
Find the average value of f(x)=(x−1)2 over the interval [0,3]. At what point in the given interval does the function assume this average value?
Solution
¯f=1b−a∫baf(x) dx=13−0∫ba(x−1)2 dx=13−0[(x−1)33]30=13−0[(3−1)33−(0−1)33]=1
The average value of the function f over the interval [0,3] is 1.

The function assumes the average value at the points where
(x−1)2=1
x2−2x+1=1
x2−2x=0
x(x−2)=0
x=0 and x=2
As evident from Figure 7.1.1.4, the function value is 1 at x=0 and x=2.
Table 7.1.1.1 lists a number of standard indefinite integral forms.
Integrals |
---|
∫dx=x+C |
∫af(x) dx=a∫f(x) dx+C |
∫[u(x)±v(x) dx]=∫u(x)dx±∫v(x) dx+C |
∫xn dx=xn+1n+1+C |
∫u dv=uv−∫v du+C |
∫dxax+b=1aln|ax+b|+C |
∫ax dx=axlna+C |
∫eax dx=eaxa+C |
∫cosx dx=sinx+C |
∫sinx dx=−cosx+C |
∫tanx dx=−ln|cosx|+C=ln|secx|+C |
∫sec(ax) dx=1aln|sec(ax)+tan(ax)|+C |
∫cotx dx=−ln|cscx|+C=ln|sinx|+C |
∫sec2ax dx=1atan(ax)+C |
∫sec(x)tan(x) dx=sec(x)+C |
∫csc(x)cot(x) dx=−csc(x)+C |
Multiple Choice Test
(1). Physically, integrating ∫baf(x) dx means finding the
(A) area under the curve from a to b
(D) area to the left of point a
(C) area to the right of point b
(D) area above the curve from a to b
(2). The mean value of a function f(x) from a to b is given by
(A) f(a)+f(b)2
(D) f(a)+2f(a+b2)+f(b)4
(C) ∫baf(x) dx
(D) ∫baf(x) dxb−a
(3). The exact value of ∫2.20.2xex dx is most nearly
(A) 7.8036
(D) 11.807
(C) 14.034
(D) 19.611
(4). ∫20.2f(x) dx for
f(x)={x,0≤x≤1.2x2,1.2<x≤2.4
is most nearly
(A) 1.9800
(D) 2.6640
(C) 2.7907
(D) 4.7520
(5). The area of a circle of radius a can be found by which of the following integrals?
(A) ∫a0(a2−x2) dx
(D) ∫2π0√a2−x2 dx
(C) 4∫a0√a2−x2 dx
(D) ∫a0√a2−x2 dx
(6). Velocity distribution of a fluid flow through a pipe varies along the radius and is given by v(r). The flow rate through the pipe of radius a is given by
(A) πv(a)a2
(D) πv(0)+v(a)2a2
(C) ∫a0v(r) dr
(D) 2π∫a0v(r)r dr
For complete solution, go to
http://nm.mathforcollege.com/mcquizzes/07int/quiz_07int_background_solution.pdf