Kenneth Kuttler

16.4. AN APPROACH TO THE INTEGRAL 449

Pass to a limit as n→ ∞ and use the definition to get 2. Also note that∫ b

b f (x)dx = 0follows from the definition.

Next consider 3. Let h ̸= 0 and let x be in the open interval determined by a and b. Thenfor small h,

F (x+h)−F (x)h

=1h

∫ x+h

xf (t)dt = f (xh)

where xh is between x and x+h. Let h→ 0. By continuity of f , it follows that the limit ofthe right side exists and so

limh→0

F (x+h)−F (x)h

= limh→0

f (xh) = f (x)

If x is either end point, the argument is the same except you have to pay attention to thesign of h so that both x and x+h are in [a,b]. Thus F is continuous on [a,b] and F ′ existson (a,b) so if G is an antiderivative,∫ b

af (t)dt ≡ F (b) = F (b)−F (a) = G(b)−G(a)

The claim that the integral is linear is obvious from this. Indeed, if F ′ = f ,G′ = g,∫ b

a(α f (t)+βg(t))dt = αF (b)+βG(b)− (αF (a)+βG(a))

= α (F (b)−F (a))+β (G(b)−G(a))

= α

∫ b

af (t)dt +β

∫ b

ag(t)dt

If f ≥ 0, then the mean value theorem implies that for some

t ∈ (a,b) ,F (b)−F (a) =∫ b

af dx = f (t)(b−a)≥ 0.

Thus∫ b

a (| f |− f )dx≥ 0,∫ b

a (| f |+ f )dx≥ 0 and so∫ b

a| f |dx≥

∫ b

af dx,

∫ b

a| f |dx≥−

∫ b

af dx

so this proves∣∣∣∫ b

a f dx∣∣∣ ≤ ∫ b

a | f |dx. This, along with part 2 implies the other claim that∣∣∣∫ ba f dx

∣∣∣≤ ∣∣∣∫ ba | f |dx

∣∣∣.The last claim is obvious because an antiderivative of 1 is F (x) = x. ■Note also that the usual change of variables theorem is available because if F ′ = f , then

f (g(x))g′ (x) = ddx F (g(x)) so that, from the above proposition,

F (g(b))−F (g(a)) =∫ g(b)

g(a)f (y)dy =

∫ b

af (g(x))g′ (x)dx.

We usually let y = g(x) and dy = g′ (x)dx and then change the limits as indicated above,equivalently we massage the expression to look like the above. Integration by parts alsofollows from differentiation rules.

16.4. AN APPROACH TO THE INTEGRAL 449Pass to a limit as n — oo and use the definition to get 2. Also note that iP f (x)dx =0follows from the definition.Next consider 3. Let h £ 0 and let x be in the open interval determined by a and b. Thenfor small h,F h) —F 1 opxthPerhaFo if f (t)dt = f (xn)where x; is between x and x+h. Let h — 0. By continuity of f, it follows that the limit ofthe right side exists and solim F (x+h) —F (x)tim =A = him f (an) =F 0)If x is either end point, the argument is the same except you have to pay attention to thesign of h so that both x and x+/ are in [a,b]. Thus F is continuous on [a,b] and F’ existson (a,b) so if G is an antiderivative,[roa =F (b) =F (b)—F (a) =G(b)—G(a)The claim that the integral is linear is obvious from this. Indeed, if F’ = f,G’ = g,[Car +Betar = a (0) +B6(0)—(aF (a) + BG(a))a (F (b) —F (a)) +B (G(b) —G(a))b b- a | f(t)ar+B | a(t)dtIf f > 0, then the mean value theorem implies that for somebt € (a,b) ,F (b) —F (a) -| fdx = f(t) (b—a) >0.aThus J? (|f|—f) dx > 0, [? (|f| +f) dx > 0 and so[inare [tae [irae — [fasfe fas| < fe |f|dx. This, along with part 2 implies the other claim thatso this provesb bI? fax| <2 \flax|.The last claim is obvious because an antiderivative of 1 is F (x) = x.Note also that the usual change of variables theorem is available because if F’ = f, thenf(g (x)) 9’ (x) = 4F (g(x)) so that, from the above proposition,(b)F(@(d)) Flea) =f" Fo)ay= [ Flets))e! aWe usually let y = g(x) and dy = g’ (x) dx and then change the limits as indicated above,equivalently we massage the expression to look like the above. Integration by parts alsofollows from differentiation rules.