7. TECHNIQUES OF INTEGRATION. TECHNIQUES OF INTEGRATION. There are two situations in which it is impossible to find the exact value of a definite integral. . TECHNIQUES OF INTEGRATION.
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TECHNIQUES OF INTEGRATION
There are two situations in which it is impossible to find the exact value of a definite integral.
The first situation arises from the fact that, in order to evaluate using the Fundamental Theorem of Calculus (FTC), we need to know an antiderivative of f.
However, sometimes, it is difficult, or even impossible, to find an antiderivative (Section 7.5).
The second situation arises when the function is determined from a scientific experiment through instrument readings or collected data.
In both cases, we need to find approximate values of definite integrals.
In this section, we will learn:
How to find approximate values
of definite integrals.
We already know one method for approximate integration.
If we divide [a, b] into n subintervals of equal length ∆x = (b – a)/n, we have:
where xi* is any point in the i th subinterval [xi -1, xi].
If xi* is chosen to be the left endpoint of the interval, then xi* = xi -1 and we have:
If f(x) ≥ 0, the integral represents an area and Equation 1 represents an approximation of this area by the rectangles shown here.
If we choose xi* to be the right endpoint, xi* = xi and we have:
In Section 5.2, we also considered the case where xi* is chosen to be the midpoint of the subinterval [xi -1, xi].
The figure shows the midpoint approximation Mn.
Mn appears to be better than either Ln or Rn.
Another approximation—called the Trapezoidal Rule—results from averaging the approximations in Equations 1 and 2, as follows.
where ∆x = (b – a)/n and xi = a + i ∆x
The reason for the name can be seen from the figure, which illustrates the case f(x) ≥ 0.
The area of the trapezoid that lies above the i th subinterval is:
Approximate the integral with n = 5, using: a. Trapezoidal Rule
b. Midpoint Rule
Example 1 a
With n = 5, a = 1 and b = 2, we have: ∆x = (2 – 1)/5 = 0.2
Example 1 a
The approximation is illustrated here.
Example 1 b
The midpoints of the five subintervals are: 1.1, 1.3, 1.5, 1.7, 1.9
Example 1 b
So, the Midpoint Rule gives:
In Example 1, we deliberately chose an integral whose value can be computed explicitly so that we can see how accurate the Trapezoidal and Midpoint Rules are.
The error in using an approximation is defined as the amount that needs to be added to the approximation to make it exact.
From the values in Example 1, we see that the errors in the Trapezoidal and Midpoint Rule approximations for n = 5 are:ET≈ – 0.002488 EM≈ 0.001239
In general, we have:
The tables show the results of calculations similar to those in Example 1.
We can make several observations from these tables.
In all the methods. we get more accurate approximations when we increase n.
The errors in the left and right endpoint approximations are:
The Trapezoidal and Midpoint Rules are much more accurate than the endpoint approximations.
The errors in the Trapezoidal and Midpoint Rules are:
The size of the error in the Midpoint Rule is about half that in the Trapezoidal Rule.
The figure shows why we can usually expect the Midpoint Rule to be more accurate than the Trapezoidal Rule.
The area of a typical rectangle in the Midpoint Rule is the same as the area of the trapezoid ABCD whose upper side is tangent to the graph at P.
The area of this trapezoid is closer to the area under the graph than is the area of that used in the Trapezoidal Rule.
The midpoint error (shaded red) is smaller than the trapezoidal error (shaded blue).
These observations are corroborated in the following error estimates—which are proved in books on numerical analysis.
Notice that Observation 4 corresponds to the n2 in each denominator because: (2n)2 = 4n2
That the estimates depend on the size of the second derivative is not surprising if you look at the figure.
Suppose | f’’(x) | ≤ Kfor a≤ x ≤ b.
If ET and EM are the errors in the Trapezoidal and Midpoint Rules, then
Let’s apply this error estimate to the Trapezoidal Rule approximation in Example 1.
So, taking K = 2, a = 1, b = 2, and n = 5 in the error estimate (3), we see:
Comparing this estimate with the actual error of about 0.002488, we see that it can happen that the actual error is substantially less than the upper bound for the error given by (3).
How large should we take n in order to guarantee that the Trapezoidal and Midpoint Rule approximations for are accurate to within 0.0001?
We saw in the preceding calculation that | f’’(x) | ≤ 2 for 1 ≤ x ≤ 2
Accuracy to within 0.0001 means that the size of the error should be less than 0.0001
Solving the inequality for n, we get
It’s quite possible that a lower value for n would suffice.
For the same accuracy with the Midpoint Rule, we choose n so that:
Example 3 a
As a = 0, b = 1, and n = 10, the Midpoint Rule gives:
Example 3 a
The approximation is illustrated.
Example 3 b
As f(x) = ex2, we have: f’(x) = 2xex2 and f’’(x) = (2 + 4x2)ex2
Also, since 0 ≤ x ≤ 1, we have x2≤ 1.
Hence, 0 ≤ f’’(x) = (2 + 4x2) ex2≤ 6e
Example 3 b
Taking K = 6e, a = 0, b = 1, and n = 10 in the error estimate (3), we see that an upper bound for the error is:
Error estimates give upper bounds for the error.
Another rule for approximate integration results from using parabolas instead of straight line segments to approximate a curve.
As before, we divide [a, b] into n subintervals of equal length h = ∆x = (b – a)/n.
However, this time, we assume n is an evennumber.
Then, on each consecutive pair of intervals, we approximate the curve y = f(x) ≥ 0 by a parabola, as shown.
If yi = f(xi), then Pi(xi, yi) is the point on the curve lying above xi.
To simplify our calculations, we first consider the case where: x0 = -h, x1 = 0, x2 = h
We know that the equation of the parabola through P0, P1, and P2is of the form y = Ax2 + Bx + C
Therefore, the area under the parabola from x = - h to x = h is:
However, as the parabola passes through P0(- h, y0), P1(0, y1), and P2(h, y2), we have: y0 = A(– h)2 + B(- h) + C = Ah2 – Bh + C
y1 = C
y2 = Ah2 + Bh + C
Therefore, y0 + 4y1 + y2 = 2Ah2 + 6C
So, we can rewrite the area under the parabola as:
Now, by shifting this parabola horizontally, we do not change the area under it.
This means that the area under the parabola through P0, P1, and P2 from x = x0 to x = x2is still:
Similarly, the area under the parabola through P2, P3, and P4 from x = x2 to x = x4is:
Thus, if we compute the areas under all the parabolas and add the results, we get:
Though we have derived this approximation for the case in which f(x) ≥ 0, it is a reasonable approximation for any continuous function f .
This is called Simpson’s Rule—after the English the English mathematician Thomas Simpson (1710–1761).
where n is even and ∆x = (b – a)/n.
Use Simpson’s Rule with n = 10 to approximate
Putting f(x) = 1/x, n = 10, and ∆x = 0.1 in Simpson’s Rule, we obtain:
In Example 4, notice that Simpson’s Rule gives a much better approximation (S10 ≈ 0.693150) to the true value of the integral (ln 2 ≈ 0.693147) than does either:
It turns out that the approximations in Simpson’s Rule are weighted averages of those in the Trapezoidal and Midpoint Rules:
In many applications of calculus, we need to evaluate an integral even if no explicit formula is known for y as a function of x.
If there is evidence that the values are not changing rapidly, then the Trapezoidal Rule or Simpson’s Rule can still be used to find an approximate value for .
The figure shows data traffic on the link from the U.S. to SWITCH, the Swiss academic and research network, on February 10, 1998.
Use Simpson’s Rule to estimate the total amount of data transmitted on the link up to noon on that day.
Since we want the units to be consistent and D(t) is measured in Mb/s, we convert the units for t from hours to seconds.
If we let A(t) be the amount of data (in Mb) transmitted by time t, where t is measured in seconds, then A’(t) = D(t).
We estimate the values of D(t) at hourly intervals from the graph and compile them here.
Then, we use Simpson’s Rule with n = 12 and ∆t = 3600 to estimate the integral, as follows.
The table shows how Simpson’s Rule compares with the Midpoint Rule for the integral , whose true value is about 0.69314718
This table shows how the error Esin Simpson’s Rule decreases by a factor of about 16 when n is doubled.
That is consistent with the appearance of n4in the denominator of the following error estimate for Simpson’s Rule.
Suppose that | f (4)(x) | ≤ Kfor a≤ x ≤ b.
If Es is the error involved in using Simpson’s Rule, then
How large should we take n to guarantee that the Simpson’s Rule approximation for is accurate to within 0.0001?
If f(x) = 1/x, then f (4)(x) = 24/x5.
Since x≥ 1, we have 1/x≤ 1, and so
So, for an error less than 0.0001, we should choose n so that:
Therefore, n = 8 (n must be even) gives the desired accuracy.
Example 7 a
If n =10, then ∆x = 0.1 and the rule gives:
Example 7 b
The fourth derivative of f(x) = ex2 is: f(4)(x) = (12 + 48x2 + 16x4)ex2
So, since 0 ≤ x ≤ 1, we have: 0 ≤ f(4)(x) ≤ (12 + 48 +16)e1 = 76e
Example 7 b
Putting K = 76e, a = 0, b = 1, and n = 10 in (4), we see that the error is at most:
Example 7 b
Thus, correct to three decimal places, we have: