MATA30 Instructor: Dr. Ken Butler 1 Contact information (on - - PowerPoint PPT Presentation

Welcome to

MATA30

Instructor: Dr. Ken Butler

1

Contact information

(on Intranet: intranet.utsc.utoronto.ca, My Courses)

• E-mail: butler@utsc.utoronto.ca
• Office: B 426
• Office hours: to be announced
• Phone: 5654 (416-287-5654)

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Improper integrals

So far, have found integrals

b

a f(x) dx over finite interval,

assuming function continuous on [a, b]. But in applications, have kinds of “infinity” to worry about:

• one of the limits is infinite
• the function becomes infinite in [a, b]

Not all integrals with infinities have a finite answer; those that do said to converge, others diverge.

235

Infinite limits

Consider

∞

1 (1/x2) dx.

Idea: replace infinite limit by finite b, let b tend to infinity:

b

1

1 x2 dx = [−x−1]b

1 = −1

b + 1.

As b → ∞, 1/b → 0, so integral heads for 1. It seems to make sense to say

∞

1 (1/x2) dx = 1.

Seems strange that infinitely long region can have finite area. But

f(x) → 0 fast enough: region gets less tall faster than it gets wider.

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Simply having f(x) → 0 is not enough – eg. f(x) = 1/x:

b

1

1 x dx = [ln x]b

1 = ln b − ln 1.

As b increases, ln b also increases, so no finite value for integral. Integral

∞

1 (1/x2) dx converges,

∞

1 (1/x) dx diverges.

More examples:

∞

0 e−2x dx;

∞

0 xe−2x dx.

If lower limit is −∞, replace it by a and see what happens as

a → −∞. Thus to find

−∞ ex dx, find a

ex dx = [ex]0

a = 1 − ea.

Since a negative. ea → 0, and integral converges to 1.

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If both limits are infinite, split into two parts, at x = c (value of c not important). Thus to find

∞

−∞ xe−x2/2 dx, find a

xe−x2/2 dx + b xe−x2/2 dx = [−e−x2/2]0

a + [−e−x2/2]b

= −1 + e−a2/2 − e−b2/2 + 1.

As a and b decrease and increase, both exponentials head for 0, so integral converges to 0. (Could have guessed because xe−x2/2 is odd function.) If either integral had diverged, whole integral divergent.

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When integrand becomes infinite

Another kind of improper integral: value of function becomes infinite in interval. First case: infinity at end of interval. Example:

2

0 (1/√x) dx.

Can’t evaluate function at 0, so replace lower limit by a:

2

a

1 √x dx = [2x1/2]2

a = 2

√ 2 − 2√a.

Now let a → 0:

2

0 (1/√x) dx = 2

√

• 2. (Converges.)

Compare

2

0 {1/(x − 2)2} dx, trouble at 2. Diverges.

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If infinity of function inside interval, split integral in two so infinity on end of each new integral. Example:

2

−1(1/x2) dx. Trouble at 0, so write as

b

−1

1 x2 dx + 2

a

1 x2 dx = [−x−1]b

−1 + [−x−1]2 a

= −1 b + 1 − 1 2 + 1 a.

The two terms involving a and b get larger closer to 0, so the integral diverges. More examples:

6

0 {1/(x − 4)2/3} dx;

∞

0 (1/x2) dx.

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Approximating definite integrals

Many functions don’t have elementary antiderivatives, so only way to find their integrals is numerically. Already saw left and right-hand sums (take enough rectangles to get accurate enough answer). But can require a lot of work. Idea: use function values more efficiently to get more accurate answer with less work.

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Midpoint rule

With decreasing function, left-hand sum is too big, right-hand sum too small. How about evaluating function at midpoint each time? Example:

2

1 (1/x) dx. Know correct answer: ln 2 = 0.6931, so

can compare approximations. Use 2 rectangles, width 0.5:

x

1 1.25 1.5 1.75 2

1/x

1 0.80 0.67 0.57 0.50 Left

(1)(0.5) + (0.67)(0.5) 0.83

Right

(0.67)(0.5) + (0.50)(0.5) 0.58

Mid

(0.80)(0.5) + (0.57)(0.5) 0.68

Midpoint rule closest to right answer.

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Trapezoid rule

A trapezoid is a rectangle with a triangle on top of it. Drawing trapezoid under function and finding its area should be more accurate than left and right sums. Turns out that answer for trapezoid rule is average of results from left and right rules. For

2

1 (1/x) dx, this gives 0.71. Close, but not

as close as midpoint rule. Function here is concave up, so trapezoid rule overestimates (top of each trapezoid above curve). At same time, midpoint rule underestimates. Generally true for concave-up functions. For concave-down functions, trapezoid underestimates, midpoint overestimates.

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Approximation errors and Simpson’s rule

With left and right sums, found that doubling number of rectangles halved error. So each additional decimal place requires 10 times work! Left and right sums not practical for finding integrals accurately. Trapezoid and midpoint rules give two more decimal places of accuracy for 10 times work, so approach right answer much faster.

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Curious fact: error from midpoint rule about half error from trapezoid rule, and of opposite sign, eg. for

2

1 (1/x) dx:

n

Trapezoid Midpoint 2

• 0.0152

0.0074 10

• 0.00062

0.00031 50

• 0.0000250

0.0000125 250

• 0.0000010

0.0000005

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Suggests even better rule: Simpson’s rule, estimating area as Simpson = 2 Midpoint + Trapezoid

3 .

Turns out that 10 times the work gives four more digits of accuracy in the answer, so practical to get highly accurate answers. Example:

6

0 x2 dx using Simpson with 4 rectangles. Need to

evaluate x2 at these places:

x

1.5 3 4.5 6

x2

2.25 9 20.25 36

x

0.75 2.25 3.75 5.25

x2

0.5625 5.0625 14.0625 27.5625

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Then: left

= (0 + 2.25 + 9 + 20.25)(1.5) = 47.25

right

= (2.25 + 9 + 20.25 + 36)(1.5) = 101.25

trapezoid

= (47.25 + 101.25)/2 = 74.25

midpoint

= (0.56 + 5.06 + 14.06 + 27.56)(1.5) = 70.875

Simpson

= (2)(70.875) + 74.25 3 = 72

Turns out that correct answer is exactly 72, so Simpson exactly

• right. Function is concave up, so trapezoid is overestimate and

midpoint underestimate.

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Computational issues

Most general-purpose numerical integration routines (eg. on calculators) use Simpson’s rule. Not only because of accuracy, but also because of deciding when to stop.

• Eg. calculating Simpson for 2 rectangles, don’t generally know how

accurate answer is. So compare answer for 4 rectangles – stop if close enough, otherwise continue to 8, 16, . . . rectangles as necessary.

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Nice computational feature of Simpson: when doubling number of rectangles, can re-use all previous work.

• Eg. 2 rectangles on [0, 1]: evaluate function at 0, 0.5, 1 for left and

right, 0.25 and 0.75 for midpoint. For 4 rectangles, evaluate function at 0, 0.25, 0.5, 0.75, 1 for left and right, using previous values for left, right and midpoint. Only new work is for 4-rectangle midpoint.

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