(^234) Integrals
With the aid of these formulas, it is easy to verify (1) for the cases in whichs
is 1 and 2 and 3. In fact, Archimedes did it. The formulas of this problem
have tremendous importance in the history of science because they stimulated
interest in limits of sums that culminated in the invention of "the calculus"
by Leibniz and Newton.
13 Letting f(x) = (1 + x)', where s is a constant for which s s' -1, derive
the formula
lim (n+i)'+(n +2).+.
.. + (n+n)' 28+1 - 1
n-. n'+1 S + 1
Write the formulas to which this reduces when s has the values -2, 0, -,
1, and 2.
14 Letting f(x) = (1 + x)-1, derive the formula
lim(^1 +^1 + 1 ++ 1 ) log 2.
n-,.\n+1 n+2 n+3 n + n
(^15) Letting f(x) = 2x/(l + x22)8, where s is a constant for which s 76 -1,
derive the formula
2e_2^123
R w
n
[(n2 +12)' + (n2 +. 22)' + (n2 + 32)'+
- (n2 + n2)' 1 2(s 1 1) L1 21 1J
Write the formulas to which this reduces when s has the values -2, -1, - , 0,
and 1.
16 Letting f(x) = 2x/(1 + x2), derive the formula
li
in2[n2+12'"2+22+n2 3 32+ (^21) = log 2.
T 1J
17 Letting f (x) = 1
- x2and borrowing the fact that fo 1 + x2dx = 4
derive the implausible formula
1 1 11 1
n m n Gs +-V + n2+ 22+n2 +32+
+ n2+ n2\ = 4 r.
18 Persons are sometimes credited with substantial knowledge of calculus
when they can simplify
(1) d a e' dt.
dx
fo:
The problem can and should be solved by noticing that puttingf(t) = t-" enables
us to use the fundamental theorem of the calculus (Theorem 4.35) to obtain
(2) du
d
f ouf(t) it=
f(n).