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

Section 5.2 The Definite Integral 385
y
y= 1 x
mation given by the Midpoint Rule is the sum of the areas of the rectangles shown in
Figure 11.
n
0 1 2 x
FIGURE 11
At the moment we don’t know how accurate the approximation in Example 5 is,
but in Section 7.7 we will learn a method for estimating the error involved in using the
Midpoint Rule. At that time we will discuss other methods for approximating definite
integrals.
If we apply the Midpoint Rule to the integral in Example 2, we get the picture in Figure
12. The approximation M 40 < 26.7563 is much closer to the true value 26.75 than
the right endpoint approximation, R 40 < 26.3998, shown in Figure 7.
y
TEC In Visual 5.2 you can compare
left, right, and midpoint approximations
to the integral in Example 2 for
different values of n.
FIGURE 12
M 40 < 26.7563
5 y=˛-6x
0
3 x
Properties of the Definite Integral
When we defined the definite integral y b f sxd dx, we implicitly assumed that a , b. But
a
the definition as a limit of Riemann sums makes sense even if a . b. Notice that if we
reverse a and b, then Dx changes from sb 2 adyn to sa 2 bdyn. Therefore
y a
f sxd dx − 2y b
f sxd dx
b
a
If a − b, then Dx − 0 and so
y a
f sxd dx − 0
a
We now develop some basic properties of integrals that will help us to evaluate integrals
in a simple manner. We assume that f and t are continuous functions.
Properties of the Integral
1. y b
c dx − csb 2 ad, where c is any constant
a
2. y b
f f sxd 1 tsxdg dx − y b
f sxd dx 1 y b
tsxd dx
a
a
3. y b
cf sxd dx − c y b
f sxd dx, where c is any constant
a
a
4. y b
f f sxd 2 tsxdg dx − y b
f sxd dx 2 y b
tsxd dx
a
a
a
a
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