## Wednesday, September 20, 2006

### Properties of integrals

In today's blog, I use Reimann sums (see Definition 2, here) to establish some basic properties of integrals. I will later use these properties to present the fundamental theorem of calculus.

The content in today's blog is taken straight from Edwards and Penney's Calculus and Analytic Geometry.

Lemma 1: Integral of a Constant

∫ (b,a) c dx = c(b-a)

Proof:

In the case of a constant, one interval will do and the area is exactly c(b-a). [See here for a discussion of Riemann sums]

QED

Lemma 2: Constant Multiple Property

∫ (b,a) cf(x)dx = c ∫ (b,a) f(x)dx

Proof:

(1) Let R = the Riemann sum (see Definition 2, here) for this equation so that we have:

R = ∑ (j=1,n) c*f(cj)(xj - xj-1)

(2) Now, since c is constant, we have:

R = c*∑ (j=1,n) f(cj)(xj - xj-1)

(3) Now, using the definition of the integral as the limit of the Riemann sum R (see Definition 5, here), we get:

I = lim (n → ∞) R = lim (n → ∞) c*∑ (j=1,n) f(cj)(xj - xj-1) =

= c*lim(n → ∞) ∑ (j=1,n) f(cj)(xj - xj-1) = c*∫ (b,a) f(x)dx

QED

Lemma 3: Interval Union Property

If a is less than c is less than b, then:

∫ (a,b) f(x)dx = ∫ (c,a) f(x)dx + ∫ (b,c) f(x)dx

Proof:

(1) Let R1 = Riemann sum for f(x) on the interval [c,a] with Partition P1 and a set S1 of arbitrary points in each subinterval [See Definition 2, here for definition of Riemann sum]

(2) Let R2 = Riemann sum for f(x) on the interval [b,c] with Partition P2 and a set S2 of arbitrary points in each subinterval [See Definition 2, here for definition of Riemann sum]

(3) Let S = S1 ∪ S2

(4) Let P = P1 ∪ P2

(5) Let x1,i be the set of intervals that make up P1 [See Definition 1, here for definition of partition]

(6) Let x2,i be the set of intervals that make up P2 [See Definition 1, here for definition of partition]

(7) Since a is less than c which is less than b, we have:

a=x1,0 less than x1,1 less than ... less than x1,n=c=x2,0 less than x2,1 less than ... less than x2,n = b

(8) From this, we can see that P is a partition on [a,b] (See Definition 1, here for definition of partition)

(9) Since S1 is a set of arbitary points in each subinterval on [a,c] and S2 is a set of arbitrary points in each subinterval on [b,c], it follows that S is set of arbitrary points on [a,c] [See Definition 2, here on Riemann sum]

(10) So, R is a Riemann sum since it is defined on [a,c] using a partition P and a set S of arbitrary points in each subinterval.

(11) Since f(x) is continuous, we know that R, R1, and R2 have limits (that is, their integrals exist. (See Theorem, here)

(12) Therefore, it follows that:

lim (n → ∞) R = lim (n → ∞)R1 + lim(n → ∞) R2

which using the definition of definite integrals in terms of Riemann sums (see Definition 5, here) gives us:

∫ (a,b) f(x)dx = ∫ (c,a)f(x)dx + ∫ (b,c) f(x)dx.

QED

Lemma 4: Comparison Property

If m ≤ f(x) ≤ M for all x in [a,b], then:

m(b - a) ≤ ∫ (b,a) f(x)dx ≤ M(b-a)

Proof:

(1) Let m be minimum of f(x) on [a,b] (See Theorem, here for proof of the existence of the minimum)

(2) Let M be ≥ maximum of f(x) on [a,b] (See Lemma 3, here for proof of the existence of the maximum)

(3) For any partition P on [a,b], m(b-a) ≤ L(P) [Since L(P) = min of f(x) on [a,b], see Definition 2, here]

(4) Likewise, M(b-a) ≥ U(P) [Since U(P) = max of f(x) on [a,b], see Definition 2, here]

(5) Since L(P) ≤ I ≤ U(P) [See Lemma 1, here], we have:

m(b-a) ≤ L(P) ≤ I ≤ U(P) ≤ M(b-a)

QED

References

#### 1 comment :

Deepak Suwalka said...

Thank, my friend. It's a good post. You covered almost properties of integrals. It's really appreciable. But if you provide the sum and difference Properties of definite integral. Then it will be a better post.