Advanced High-School Mathematics

300 CHAPTER 5 Series and Differential Equations

u=f(n)(t) dv=

(x−t)(n−1)

(n−1)!

dt.

du=f(n+1)(t)dt v=−

(x−t)n

n!

From the above, one obtains

∫x

a

f(n)(t)

(n−1)!

(x−t)(n−1)dt = −f(n)(t)

(x−t)n

n!

∣∣

∣∣

∣

x

a

+

∫x

a

f(n+1)(t)

n!

(x−t)ndt

=

f(n)(a)

n!

(x−a)n+

∫x

a

f(n+1)(t)

n!

(x−t)ndt

Plugging this into the induction hypothesis shows that the original
statement is correct for alln.
Next, note that if F =F(t) is a continuous function oft, then one
has that ∫b

a F(t)dt=F(c)(b−a)
for some numbercbetweenaandb. (IndeedF(c) is the average value
ofF on the interval [a,b].) Using the substitutionu= (x−t)(n+1), and
applying the above observation, we have

(n+ 1)

∫x

a F(t)(x−t)

ndt =

∫(x−a)(n+1)

0 F(x−

n+1√u)du

= F(x− n+1

√

α)(x−a)n+1,

whereαis between 0 and (x−a)(n+1).If we setc=x− n+1

√

α we see
thatcis betweenaandxand that

(n+ 1)

∫x

a F(t)(x−t)

ndt=F(c)(x−a)(n+1).

Now setF(t) =

f(n+1)(t)

n!

and apply the above to infer that

∫x

a

f(n+1)(t)

n!

(x−t)ndt=

f(n+1)(c)

(n+ 1)!

(x−a)n+1.

This completes the proof.