(^218) Integrals
by abandoning the good old idea that the elementary functions (polynomials,
trigonometric functions, etcetera) are always the simplest and most useful func-
tions. There are very many situations in which step functions are the simplest
and most useful functions. Our first problem is to follow instructions to prove
that
(1)
rrn n
RS = 4 f(tk) At, = 7 At,,
k=1 k=1
Draw a figure showing a partition P of the interval 2 < x 5 S into subintervals
having lengths At,, Ot2, , At. andobserve that
n++
(2) k=1G Atk = 3.
Observe that the integral in (1) involves the function f for which f (x) is always 7.
Show that, with the notation of the text,
(3)
f25 7 dt= 21.
n
=7 1 Atk=7.3 = 21.
k=1
Since each Riemann sum is 21, it is quite apparent that RS is near 21 whenever
the norm of P is near 0. This proves the formula (1).
6 Supposing that a < b and k is a constant, prove that
(1) fab k dt= k(b - a).
Remark: We are (or soon will be) authorities on areas of rectangular regions.
We can observe that if k > 0, then the right side of (1) is the area of a rectangu-
lar region having base length (b - a) and height k and hence is the area of the
region of Figure 4.291 which is bounded by the graphs of the equations x = a,
k
Y
0
OI a
it
x
Y
a b
0 x
b -k
Figure 4.291 Figure 4.292
x = b, y = 0, and y = k. In case k < 0, we can put (1) in the form
(2) fabkdt= -(-k)(b - a),
where -k > 0, and observe that the right side is the negative of the area of the
region in Figure 4.292 bounded by the graphs of the equations x = a, x = b,
y = 0, and y = k. We must always know that areas of rectangles are positive
The idea that rectangles below the x axis have negative areas is as absurd as the
idea that cities south of the equator have negative populations.