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Math 111.17 Fall 2002

The Deﬁnite Integral

Suppose that f (x) is a continuous, positive function deﬁned on the interval [ a, b]. We denote by

(1)

a

b

f (x) dx

the area under the curve f (x), above the x-axis, and between x = a and x = b. We call (1) the deﬁnite integral of f (x) from a to b.

Properties of the Deﬁnite Integral

Fact:

a

b

1 dx = b a

Fact: If a < b < c, then

a

c

f (x) dx =

a

b

f (x) dx + c f (x) dx.

b

Fact:

a

Fact:

a

b

b

f (x) dx = a f (x) dx

b

f (x) + g(x) dx =

a

b

f (x) dx +

a

b

g(x) dx

Fact:

a

b

cf (x) dx = c

a

b

f (x) dx

Exercise: Draw pictures of the associated areas to convince yourself that the above facts are true.

Riemann Sums

Only in rare cases is it possible to compute deﬁnite integrals exactly. Although the Fun- damental Theorem of Calculus, which we will study shortly, will greatly expand the class of so-called integrable functions, the majority of deﬁnite integrals can only be approximated.

We construct Riemann sums to approximate (1). Suppose that we partition (subdivide) the interval [a, b] into N equal subintervals each of length

x

= b a

N

.

If we denote the partition points by

x 0 = a, x 1 = a + ∆x, x 2 = a + 2∆x,

, x k = a + kx,

, x N = a + Nx = b,

then the left hand sum (LHS) and right hand sum (RHS) are computed as follows.

LHS: ∆x

RHS: ∆x

N1

k=0

N

k=1

f(x k ) = ∆x

f(x k ) = ∆x

N1

k=0

f(a + kx)

N

k=1

f(a + kx)

Exercise: Make sure you understand these sums.

Example: Suppose f (x) = 2x 3 +4. Compute the LHS and RHS to approximate 1 f (x) dx

using N = 200 subintervals.

1

Solution: We partition [1, 1] into 200 subintervals each of width ∆x = 1 1 =

200

Thus, the partition points are 1, 1 + 100 , 1 + 100 ,

1

2

, 1 + 199

100 , 1.

1

100 .

Hence,

LHS = ∆x

Similarly,

RHS = ∆x

N1

k=0

N

k=1

f(a + kx) =

f(a + kx) =

1

100

1

100

199

k=0 f 1 +

200

k=1 f 1 +

k

100

k

100

=

=

1

100

1

100

199

k=0 2 1 +

200

k=1 2 1 +

k

100

k

100

3 + 4 .

3 + 4 .

We can compute each of these on the TI-83 as follows:

LHS: sum(seq(1/100(2(-1+X/100)3 + 4), X, 0, 199, 1))= 7.98

RHS: sum(seq(1/100(2(-1+X/100)3 + 4), X, 1, 200, 1))= 8.02

Thus,

Note that sum is found as 2nd

[LIST] OPS 5:seq.

1

7.98 <

1

2x 3 + 4 dx < 8.02. 2nd

[LIST] MATH 5:sum and seq is found as

The Program RIEMANN

It is tedious to repeatedly enter such sums manually. The following simple program com-

putes the LHS and RHS for

a b f (x) dx using

N subintervals. You must enter f (x) as Y1.

Either input this program directly into your TI-83, or copy it from someone using the TI Graph Link.

 :Prompt A,B,N :(B-A)/N -> D

:A

->

X

 :0 -> S :For (I,0,N-1) :S+Y1*D -> S :X+D -> X :End :S -> L :S+Y1*D -> S :A -> X :S-Y1*D -> R
 :Disp "RIGHT SUM IS", R :Disp "LEFT SUM IS", L

One ﬁnal warning: Although this program will save you time on your homework and while you study, you will still be required to know how to set up Riemann sums on paper and how to compute simple ones manually, without the aid of a program.
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