184 CHAPTER 7. THE INTEGRAL

Lemma 7.3.18 Let α ≥ 0. Then∫ b

a α f (x)dx = α∫ b

a f (x)dx.

Proof: It is routine to verify that U (α f ,P) = αU ( f ,P) ,L(α f ,P) = αL( f ,P). LetU ( f ,P)−L( f ,P)< ε . Therefore,∫ b

aα f (x)dx ∈ [L(α f ,P) ,U (α f ,P)] = [αL( f ,P) ,αU ( f ,P)]

α

∫ b

af (x)dx ∈ [αL( f ,P) ,αU ( f ,P)]

Thus both∫ b

a α f (x)dx and α∫ b

a f (x)dx are in an interval of length αε so since ε is arbi-trary, this proves the lemma.

Proposition 7.3.19 Let f ,g be integrable. Then if α,β are real numbers,then the fol-lowing holds for the integrals:

∫ ba (α f +βg)(x)dx = α

∫ ba f (x)dx+β

∫ ba g(x)dx.

Proof: I want to show that∫ b

a α f (x)dx = α∫ b

a f (x)dx regardless of whether α isnonnegative. By the first lemma,∫ b

a(|α|−α) f (x)dx+

∫ b

aα f (x)dx =

∫ b

a|α| f (x)dx

and now, by the second lemma and subtracting |α|∫ b

a f (x)dx,

(|α|−α)∫ b

af (x)dx+

∫ b

aα f (x)dx =

∫ b

a|α| f (x)dx

−α

∫ b

af (x)dx+

∫ b

aα f (x) = 0

so this shows one can factor out any real number.Now from what was just shown and the lemmas,∫ b

a(α f +βg)(x)dx =

∫ b

aα f (x)dx+

∫ b

aβ f (x)dx

= α

∫ b

af (x)dx+β

∫ b

af (x)dx

Proposition 7.3.20 If a < b, Then∫ b

a | f (x)|dx ≥∣∣∣∫ b

a f (x)dx∣∣∣ . Here f is assumed in-

tegrable.

Proof: From the definition, if a < b, then∫ b

a f (x)dx ≥ 0 if f (x) ≥ 0 for all x. Nowfrom Proposition 7.3.16, if f is integrable, so is | f | . Then∫ b

a(| f (x)|− f (x))dx ≥ 0 so

∫ b

a| f (x)|dx ≥

∫ b

af (x)dx∫ b

a(| f (x)|+ f (x))dx ≥ 0 so

∫ b

a| f (x)|dx ≥−

∫ b

af (x)dx

which implies that for a < b,∫ b

a | f (x)|dx ≥∣∣∣∫ b

a f (x)dx∣∣∣ .

Definition 7.3.21 Suppose f is a function and P is a partition P = x0 = a < x1 <· · · < xn = b. A Riemann sum is of the form ∑

nk=1 f (zk)(xk − xk−1) where zk ∈ [xk−1,xk].

Thus every such Riemann sum is between U ( f ,P) and L( f ,P) and so, if f is integrable,every such Riemann sum can be considered an approximation to

∫ ba f (x)dx.