FDWK_3ed_Ch05_pp262-319

Section 5.5 Trapezoidal Rule 307
y
P 4
The Trapezoidal Rule
To approximate b f x dx, use
a
T h 2 ( y 0 2y 1 2y 2 … 2y n1 y n) ,
where a, b is partitioned into n subintervals of equal length h b an.
Equivalently,
T LRAM n
2 RRAM n
,
P 3
where LRAM n and RRAM n are the Riemann sums using the left and right
endpoints, respectively, for f for the partition.
y x 2 2
P 2
0 1
P 1
P 0
25
—
1 16
5
–
4
36
—
16
6
–
4
49
—
16
7
–
4
Figure 5.32 The trapezoidal approximation
of the area under the graph of
y x 2 from x 1 to x 2 is a slight
overestimate. (Example 1)
Table 5.4
4
x y x 2
1 1
5 4 2 5
16
6 4 3 6
16
7 4 4 9
16
2 4
x
EXAMPLE 1
Applying the Trapezoidal Rule
Use the Trapezoidal Rule with n 4 to estimate 2
1 x2 dx. Compare the estimate
with the value of NINT x 2 , x, 1,2 and with the exact value.
SOLUTION
Partition 1, 2 into four subintervals of equal length (Figure 5.32). Then evaluate y x 2
at each partition point (Table 5.4).
Using these y values, n 4, and h 2 14 14 in the Trapezoidal Rule, we
have
T h 2 ( y 0 2y 1 2y 2 2y 3 y 4)
1 8 ( ( 1 2 2 5
16
) ( 2 3 6
16
) ( 2 4 9
16
) ) 4
7 5
2.34375.
32
The value of NINT x 2 , x, 1,2 is 2.333333333.
The exact value of the integral is
2
1
x 2 dx x 3
3
]
2
1
8 3 1 3 7 3 .
The T approximation overestimates the integral by about half a percent of its true value of
73. The percentage error is 2.34375 73730.446%. Now try Exercise 3.
We could have predicted that the Trapezoidal Rule would overestimate the integral in
Example 1 by considering the geometry of the graph in Figure 5.32. Since the parabola is
concave up, the approximating segments lie above the curve, giving each trapezoid slightly
more area than the corresponding strip under the curve. In Figure 5.31 we see that the
straight segments lie under the curve on those intervals where the curve is concave down,