James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

386 Chapter 5 Integrals
Property 1 says that the integral of a constant function f sxd − c is the constant times
the length of the interval. If c . 0 and a , b, this is to be expected because csb 2 ad is
the area of the shaded rectangle in Figure 13.
y
c
y=c
y b a
FIGURE 13
c dx − csb 2 ad
0
a
area=c(b-a)
b
x
y
g
f+g
0 a
b x
FIGURE 14
f fsxd 1 tsxdgdx −
y b a
y b a
f
fsxddx 1 y b a
tsxddx
Property 3 seems intuitively reasonable
because we know that multiplying
a function by a positive number c
stretches or shrinks its graph vertically
by a factor of c. So it stretches or
shrinks each approximating rectangle
by a factor c and therefore it has the
effect of multiplying the area by c.
Property 2 says that the integral of a sum is the sum of the integrals. For positive
functions it says that the area under f 1 t is the area under f plus the area under t.
Figure 14 helps us understand why this is true: In view of how graphical addition works,
the corresponding vertical line segments have equal height.
In general, Property 2 follows from Theorem 4 and the fact that the limit of a sum is
the sum of the limits:
y b
f f sxd 1 tsxdg dx − lim
a
n l ` o n
f f sx i d 1 tsx i dg Dx
i−1
− lim
n l `Fo n
f sx i d Dx 1 o n
tsx i d DxG
i−1
i−1
− lim
n l ` o n
f sx i d Dx 1 lim
i−1
n l ` o n
tsx i d Dx
i−1
− y b
f sxd dx 1 y b
tsxd dx
a
Property 3 can be proved in a similar manner and says that the integral of a constant
times a function is the constant times the integral of the function. In other words, a constant
(but only a constant) can be taken in front of an integral sign. Property 4 is proved
by writing f 2 t − f 1 s2td and using Properties 2 and 3 with c − 21.
Example 6 Use the properties of integrals to evaluate y 1
s4 1 3x 2 d dx.
SOLUTION Using Properties 2 and 3 of integrals, we have
a
0
y 1
0
s4 1 3x 2 d dx − y 1
4 dx 1 y 1
3x 2 dx − y 1
4 dx 1 3 y 1
x 2 dx
0
0
0
0
We know from Property 1 that
y 1
4 dx − 4s1 2 0d − 4
0
and we found in Example 5.1.2 that y 1
x 2 dx − 1 3 . So
y 1
0
0
s4 1 3x 2 d dx − y 1
4 dx 1 3 y 1
x 2 dx
0
0
− 4 1 3 ∙ 1 3 − 5 n
The next property tells us how to combine integrals of the same function over adjacent
intervals.
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