THE DEFINITE INTEGRAL THE MIDPOINT RULE In the previous sections, we found that limits of sums could be computed using sample endpoints (which include right endpoints, left endpoints, or any number within the subinterval). In most cases, we choose to use right endpoints because it is considerably convenient for computing limits of sums. If, however, our goal is to find an estimate (i.e. an approximation) of a definite integral, it is better to choose the sample point to be the midpoint of the interval, which is denoted by xi. Note that ANY definite integral could be estimated using the Riemann sum, but if we decide to use midpoints, we follow this:
∫
b a
n
f(x) dx
∑
=
i=1
f(xi ) ∆x
=
∆x [f(x1) + f(x2) + ........ + f(xn)]
Where ∆x = (b – a)/n, and
xi
=
½ [xi-1
+
xi] is the midpoint of the subinterval [xi-1, xi]
So now, we’ll illustrate the use of the midpoint rule in the following examples. (In each example, we’ll use the given value of n to estimate the integral, which will then be rounded to four decimal places). Note that, when we use the midpoint rule, we are only making approximations. Therefore, we do not always know accurate our estimates are. We will, however, learn a method used for estimating the error involved while making an approximation using the midpoint rule. We will also learn a more efficient method for approximating definite integrals.
Example 1 ∫
10 0
(sin √x ) dx
n=5
SOLUTION Here, we have 5 subintervals, whose right end points, are 0,2,4,6,8 and 10. The figure below the graph of f(x) = sin √x plotted on the interval [0,10]:
We put 10 approximating rectangles in that region to give:From the graph, We find that the midpoints of the subintervals are 1, 3, 5, 7 and 9. Note that a = 10 and b = 10. Therefore,
∆x
= (b – a)/n
∆x
= (10 – 0)/5
∆x
=
10/5
=
2
So, the midpoint rule gives
∫
10 0
(sin √x ) dx
=
∆x [f(1) + f(3) + f(5) + f(7) + f(9)]
=
2 [sin √1 + sin √3 + sin √5 + sin √7 + sin √9]
=
2 [0.8415 + 0.9870 + 0.7868 + 0.4758 + 0.1411]
=
2 [3.2322]
=
6.4644
Since f(x) = sin √x > 0 for 0 ≤ x ≤ 10, the integral represents an approximation of the area, using the midpoint rule. So, we can say that an estimate of the area under the curve f(x) = sin √x from 0 to 10 is about 6.4644. In other words,
∫
10 0
(sin √x ) dx
=
6.4644
(For calculations involving trigonometric identities, like the one above , please note that you must work in Radians, NOT degrees).
Example 2 ∫
π 0
(sec x/3 ) dx
n=6
SOLUTION Recall that sec x = 1/(cos x), so that the integral becomes
∫
π 0
[1/cos (x/3)]
dx
(where a = 0 and b = π). Since n = 6, this means we have six subintervals. So,
∆x = (b – a)/n ∆x = (π – 0)/6 ∆x = π/6 Therefore, the right end points of the subintervals are
π/6,
0,
2π/6,
3π/6,
π/3,
π/2,
4π/6,
5π/6,
6π/6
which are equivalent to
0,
π/6,
2π/3,
5π/6,
π
From the right endpoints, the midpoint of each interval is
π/12,
π/4,
5π/6,
7π/6,
3π/4,
11π/12
Note that the integrand is
f(x) =
1/[cos (x/3)]
Therefore, using the midpoint rule, we have
∫ = =
π 0
[1/cos (x/3)]
dx
∆x [f(x1) + f(x2) + ........ + f(xn)]
=
∆x [ f(π/12) + f(π/4) + f(5π/6) + f(7π/6) + f(3π/4) + f(3π/4) + f(11π/12)]
[
π/6 (1/[cos(π/36)]) + (1/[cos(π/12)]) + (1/[cos(5π/18)]) + (1/[cos(7π/18)]) + (1/[cos(3π/4)]) +
]
(1/[cos(11π/36)])
≈ ≈
0. 5236
[1.0038 + 1.0353 + 1.5557 + 2.9238 + 1.0353 + 1.0535] [
]
0.5236 8.8074
≈
4.5068
So, an estimate of the definite integral of the curve y = sec(x/3) from 0 to p using the midpoint rule, is about 4.5068. In other words,
∫
π 0
(sec x/3 ) dx
=
∫
π 0
[1/cos (x/3)]
dx
=
4.5068
The graph below shows layout approximating rectangles used to estimate the area of the required region on the graph of f(x) = sec (x/3):
Example 3 ∫
2 1
n = 10
√(1 + x2 ) dx
SOLUTION Here, the integrand is y = √(1 + x2). Note that a = 1, b = 2 and n = 10. Thus, the subinterval width is
∆x ∆x
= =
(b – a)/n (2 – 1)/10
∆x
=
1/10
=
0.1
Therefore, the right endpoints of the 10 subintervals are
1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, and 2.0 Hence, the midpoints of the subintervals are
1.05, 1.15, 1.25, 1.35, 1.45, 1.55, 1.65, 1.75, 1.85, and 1.95 Using the midpoint rule gives
∫
2 1
√(1 + x2 ) dx
=
[
∆x f(1.05) + f(1.15) + f(1.25) + f(1.35) + f(1.45) + f(1.55) + f(1.65) +
]
f(1.75) + f(1.85) + f(1.55) =
[
0.1 1.4500 + 1.5420 + 1.6008 + 1.6800 +1.7614 + 1.8446 + 1.9294
]
+ 2.0156 + 2.1030 + 2.1915 =
0.1[18.1003] =
1.81003
The figure above shows the graph of y = √(1 + x2) ,plus the approximating rectangles used to estimate the area of the region. (The actual value of the integral is about 1.81009214039. Therefore, our estimate above is considerably accurate).
Example 4
∫
x x2 + 1
4 2
dx
n=4
SOLUTION For this integral, the integrand is y = x/(x2 + 1). With n = 4, we therefore have 4 subintervals to deal with. The width is given by
∆x = (b – a)/n ∆x = (4 – 2)/4 ∆x = 2/4
=
0.5
Therefore, the right endpoints of the subintervals are
2.0, 2.5, 3.0, 3.5, and 4.0 This means that the midpoints are
2.25, 2.75, 3.25, and 3.75 Using the midpoint rule, we have
∫
4 2
x x +1 2
dx
= =
∆x [f(2.25) + f(2.75) + f(3.25) + f(3.75)] 0.5 [0.371134 + 0.321168 + 0.281081 + 0.248963]
≈
0.6112
Hence, the area under the curve y = x/(x2 + 1) from 2 to 4 is about 0.6112. Alternatively, we say
∫
4 2
x x +1 2
dx
≈
0.6112
(The actual value of the integral is about 0.611887715811. So, the above approximation is fairly accurate).
So far, we have been able to make noticeably correct approximations of definite integrals using the midpoint rule. Like I mentioned earlier, this rule is used to make APPROXIMATIONS ONLY. In the following section, we’ll discuss the properties of the definite integral, and then we’ll use these properties to obtain precise values of definite integrals, instead of mere approximations.
Before we move on to the next section, try solving these using the midpoint rule:
(a) (b) (c)
∫ ∫ ∫
1 0 2 1 2 0
(d)
∫
1
(e)
∫
1
3
2
√(1 + x2 ) dx
n=5
(8x3 + 3x2) dx
n = 10
(x2 – x) dx
n=7
√(x2 + 3) dx
n = 10
(5x2 – 4x + 3) dx
n=8
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