9.7. A SIMPLE PROCEDURE FOR FINDING INTEGRALS 209

a = x0 < x1 < · · ·< xn = b which have the property that on [xi,xi+1] , the following functionis differentiable and has a continuous derivative.

Gi (x)≡

 F (x) on (xi,xi+1)F (xi+) when x = xiF (xi+1−) when x = xi+1

Also assume F (a) = F (a+) ,F (b) = F (b−). Then∫ b

af (x)dF =

n−1

∑j=0

∫ x j+1

x j

f (x)G′j (x)dx+n−1

∑i=1

f (xi)(F (xi+)−F (xi−))

Here is why this procedure works. Let δ be very small and consider the partition

a = x0 < x1−δ < x1 < x1 +δ < x2−δ < x2 < x2 +δ <

· · ·xn−1−δ < xn−1 < xn−1 +δ < xn−δ < xn = b

where δ is also small enough that whenever |x− y|< δ , it follows | f (x)− f (y)|< ε. Thenfrom the properties of the integral presented above,∫ x1−δ

af dF +

∫ x2−δ

x1+δ

f dF + · · ·+∫ b

xn−1+δ

f dF +n−1

∑i=1

( f (xi)− ε)(F (xi +δ )−F (xi−δ ))

≤∫ b

af dF ≤

∫ x1−δ

af dF +

∫ x2−δ

x1+δ

f dF + · · ·+∫ b

xn−1+δ

f dF

+n−1

∑i=1

( f (xi)+ ε)(F (xi +δ )−F (xi−δ ))

By Lemma 9.4.3 this implies∫ x1−δ

af G′0dx+

∫ x2−δ

x1+δ

f G′1dx+ · · ·+∫ b

xn−1+δ

f G′n−1dx

+n−1

∑i=1

( f (xi)− ε)(F (xi +δ )−F (xi−δ ))≤∫ b

af dF ≤

∫ x1−δ

af G′0dx+

∫ x2−δ

x1+δ

f G′1dx+ · · ·+∫ b

xn−1+δ

f G′n−1dx

+n−1

∑i=1

( f (xi)+ ε)(F (xi +δ )−F (xi−δ ))

Now let δ → 0 to obtain the desired integral is between

n−1

∑j=0

∫ x j+1

x j

f (x)G′j (x)dx+n−1

∑i=1

( f (xi)+ ε)(F (xi+)−F (xi−))

andn−1

∑j=0

∫ x j+1

x j

f (x)G′j (x)dx+n−1

∑i=1

( f (xi)− ε)(F (xi+)−F (xi−))

Since ε is arbitrary, this shows the procedure is valid. This yields the following right away.

9.7. A SIMPLE PROCEDURE FOR FINDING INTEGRALS 209a=X9 <x <+++ <X, =b which have the property that on |x;,xi+1] , the following functionis differentiable and has a continuous derivative.F (x) on (x;,X;41)+) when x = x;F(xF (xj41—) when x = xj41F(b—). ThenG; (x) =Also assume F (a) = F (a+), F (b) =n-1[roar =¥ [" reopeace¥ fey ro) Fs)Here is why this procedure works. Let 5 be very small and consider the partitiona= x <x-6 <x) <x +6 <m-b6 << m+d6<“Xp - 8 << Xp <Xp_-1 +6 <x, -— 86 <= dDwhere 6 is also small enough that whenever |x — y| < 6, it follows |f (x) — f (y)| < €. Thenfrom the properties of the integral presented above,Xx] n—-1| “far faP ot [ faP+¥ (f(s) —€) (Fi +8)—F (8)a i=1Xn— 1+6[ras [ “far fdF bo fdFXn—1 +6Su f (xi) + €) (F (xi + 6) — F (x — 6))By Lemma 9.4.3 this impliesXx] —6| fGhdx+ I, ° pGldxten + fGl_,dxa Xn—1+6n—1+¥ Ve) 6) +6) Faia) < [far <i=1x1 —-8 x28 b| fGhdx+ / fGldxte+ / fG)_ydxa x, +6 Xn—1 +6n—1+) (f (ai) +€) (F (1+) — F 3 — 8)i=lNow let 6 — 0 to obtain the desired integral is betweenXj+l , nalf(x) Gj (x)dxt+ Vf 0%) +) (F Git) —F Gi)=0°7%j i=1andMod pxjyt n—1& f° resvesioyar Yrs) £9 rt) Fe)Since € is arbitrary, this shows the procedure is valid. This yields the following right away.