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(c_j)(x_j - x_j-1)

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

R = c*∑ (j=1,n) f(c_j)(x_j - x_j-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(c_j)(x_j - x_j-1) =

= c*lim(n → ∞) ∑ (j=1,n) f(c_j)(x_j - x_j-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 x_1,i be the set of intervals that make up P₁ [See Definition 1, here for definition of partition]

(6) Let x_2,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=x_1,0 less than x_1,1 less than ... less than x_1,n=c=x_2,0 less than x_2,1 less than ... less than x_2,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

• Edwards & Penney, Calculus and Analytic Geometry