Numerical Methods of Approximating Definite Integrals 26.1 APPROXIMATING SUMS: Ln, Rn, Tn, AND Mn Introduction Not only can we differentiate all the basic functions we’ve encountered, po
Trang 125.3 Substitution to Alter the Form of an Integral 801
2 .
3 4
1
1 +3x22
4
1
1 + u2·2
4 ·2 3
1 + u2du
=12 arctan u + C
2 arctan
3
2x
+ C
(b) tan√√x
x dx
Use substitution to get rid of the√
x
Let u =√x
2
1
x ·√1
tan u · 2du
= 2
tan u du
= 2
cos udu (This looks essentially like
Let W = cos u
2
cos udu = 2
−dW W
x
= −2 ln | cos√x| + C
e2x+2e x +1dx
We want to use substitution to get rid of ex and e2x If we let u = ex, then
e2x= (ex)2= u2 If, instead, we were to let u = e2x, then ex
= (e2x)1/2=√u The first option looks much more appealing Let’s try it
Let u = ex
du = exdx
e2x+ 2ex+ 1dx
Trang 2802 CHAPTER 25 Finding Antiderivatives—An Introduction to Indefinite Integration
Either we can express this as
e2x+ 2ex+ 1e
x
dx =
u2+ 2u + 1du or
(ex)2+ 2ex+ 1 dx =
u2+ 2u + 1·
1
udu
exdu = 1
udu
The two are equivalent
u2+ 2u + 1du =
(u + 1)2du This is a simpler integral than the one we began with, but it is awkward If we could get the sum in the numerator as opposed to the denominator, we’d be happier We’ll accomplish this with another substitution
Let W = u + 1 so u = W − 1
dW = du
(u + 1)2du =
W − 1
=
=
x
= ln(ex+ 1) +ex1
EXERCISE 25.3 Check the answers to Example 25.7 by differentiating
P R O B L E M S F O R S E C T I O N 2 5 3
For Problems 1 through 11, find the given indefinite integral.
x+1dx
2. x+3
x−7dx
3. x√ 3x + 5 dx
4. 3+x2x dx
5. 3t√
t2+ 5 dt
6. t√ 2t + 5 dt
Trang 325.3 Substitution to Alter the Form of an Integral 803
7. x5π dx
8. cot(3x) dx
9.(1.5)(1−t)dt
10. sin4tcos t dt
11. √2t
2t+6dt
12 Suppose you’ve forgotten the antiderivative of √1
1−x 2 In this problem you will use a sophisticated substitution that will help you proceed The goal is to find
1/2 0
1
√
1 − x2dx
We can’t let u = x2or 1 − x2because u(x)is nowhere in sight Instead we take a different approach We’ll replace x by something that makes 1 − x2a perfect square We’ll exploit the trigonometric identity sin2θ + cos2θ = 1 We know 1 − sin2θ = cos2θ, which is a perfect square Let x = sin θ Rewrite the integral (including the endpoints) in terms of θ and evaluate
Check your answer by using the antiderivative of √1
1−x 2 (Look it up in the table
on page 784 if you have forgotten it.)
(This is an example of an integration technique known as trigonometric
substi-tution It involves exploiting trigonometric identities to simplify expressions such as
√
c2− x2,√
x2− c2, and x2+ c2, where c is a constant.)
13 Evaluate x(x+2)2 dxby rewriting the integrand in the form
A
x + 2, where A and B are constants In other words, find A and B such that
2
A
x + 2.
(This is an example of an integration technique known as partial fractions.)
x 2 −1dx by factoring the denominator of the integrand and rewriting the integrand in the form
A
B
x + 1, where A and B are constants
Trang 5Numerical Methods of Approximating Definite Integrals
26.1 APPROXIMATING SUMS: Ln, Rn, Tn, AND Mn
Introduction
Not only can we differentiate all the basic functions we’ve encountered, polynomials, exponential and logarithmic functions, trigonometric and inverse trigonometric functions, and rational functions, but armed with the Product and Chain Rules, we can happily differentiate any new function constructed by multiplication, division, addition, subtraction,
or composition of these functions This gives us a sense of competence and satisfaction Although at this point we can integrate many functions, there are basic functions (such
as ln x and sec x) that we have not yet tackled.1From the Chain Rule for differentiation
we get the technique of substitution for antidifferentiation; from the Product Rule for differentiation we get a technique of antidifferentiation known as Integration by Parts (The latter is something to look forward to learning in Chapter 29.) Learning more sophisticated methods of substitution and algebraic manipulation will enlarge the collection of functions
we can antidifferentiate Tables of integrals and high-powered computer packages can provide assistance if we’re in dire straits Nevertheless, there are some very innocent-looking functions that cannot be dealt with easily For instance, all the technical skill in the world won’t help us find an antiderivative for e−x2, or sin(x2), or sin(1x).2Knowing that there is
no guarantee that we can antidifferentiate can be unnerving This chapter will restore our
sense of having things under control when we are faced with a definite integral
Suppose we are interested in evaluating a definite integral and we have not found an antiderivative for the integrand In any practical situation we’ll need to evaluate the definite
1
ln x dx = x ln x − x + C (verify this for yourself) This can be found either by some serious guess-work, methods given
in Section 27.3, or using the technique called Integration by Parts.
With some work we can integrate sec x.sec x dx =sec x · sec x+tan x dx = sec2sec x+tan xx+sec x tan xdx = ln | sec x + tan x| + C.
2That is, if we want to obtain an antiderivative that is a finite sum, product, or composition of the elementary functions.
805
Trang 6806 CHAPTER 26 Numerical Methods of Approximating Definite Integrals
integral with a certain degree of accuracy Provided that an approximation is satisfactory,
it is not necessary to be able to find an antiderivative of the integrand Instead, we return to the basic ideas that led us to the limit definition of the definite integral
The theoretical underpinnings of calculus involve the method of successive approxi-mations followed by a limiting process
Differential Calculus
Let f be a differentiable function We obtain numerical approximations of the slope of
the tangent to the graph of f at point P by looking at the slope of secant lines through
P and Q, where Q is a point on the graph of f very close to P By taking the limit as
Qapproaches P we determine the exact slope of the tangent line at P
y = f(x)
P Q
x = a x = b
Figure 26.1
Integral Calculus
Let f be an integrable function We obtain numerical approximations of the signed
area between the graph of f and the x-axis on [a, b] by partitioning the interval [a, b] into many equal subintervals, treating f as if it is constant on each tiny subinterval, and
approximating the signed area with a Riemann sum By taking the limit as the number
of subintervals increases without bound we determine the definite integral
In this chapter we return to Riemann sums to obtain approximations of definite in-tegrals The numerical methods discussed here are often used in practice Computers or programmable calculators are ideal for performing the otherwise tedious calculations
Approximating Sums: Ln, Rn, Tn, and Mn
In the context of the following example we’ll discuss left-hand, right-hand, midpoint, and trapezoidal sums, sums that can be used to approximate a definite integral In order to be able to compare our approximations with the actual value, we’ll look at an integral we can evaluate exactly
EXAMPLE 26.1 Approximate15x1dx Keep improving upon the approximation until you know the value
to four decimal places
SOLUTION (i) To approximate the integral we chop up the interval of integration into n equal
subinter-vals, (ii) approximate the area under the curve on each subinterval by the area of a rectangle, and (iii) sum the areas of these rectangles
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We’ll construct three Riemann sums: the left- hand sum, denoted by Ln; the right-hand sum, denoted by Rn; and the midpoint sum, denoted by Mn (We are already familiar with the first two.) We describe these sums as follows
Ln: The height of the rectangle on each subinterval is given by the value of f at the left-hand endpoint of the subinterval
Rn: The height of the rectangle on each subinterval is given by the value of f at the right-hand endpoint of the subinterval
Mn: The height of the rectangle on each subinterval is given by the value of f at the midpoint of the subinterval On the it hinterval, [xi−1, xi], the height of the rectangle is f (c), where c is midway between xi−1and xi; c =xi−1 +x i
2
4 Subdivisions
Let n = 4; we chop [1, 5] into 4 equal subintervals each of length x =5−14 = 1
∆ x = 1
Below are the left- hand sum, the right-hand sum, and the midpoint sum
f
x
L4 = 1
1 2
1 3
1 4 =
= 2.083
1
1 2
1 3
1 4
f
x
R4 = 1
1 3
1 4
1 5 =
= 1.283
1
1 3
1 4
1 5
1.5 2.5 3.5 4.5
f
x
M4 = 2
2 5
2 7
2 9 =
= 1.574 3
2
2 5
2 7
2 9
Figure 26.2
3
Trang 8808 CHAPTER 26 Numerical Methods of Approximating Definite Integrals
The function f (x) = 1x is decreasing; therefore the left-hand sums provide an upper bound and the right-hand sums a lower bound for15x1dx
1.283 = R4<
1
1
x dx < L4= 2.083 Suppose we take the average of R4and L4, L4 +R 4
2 This is closer to the value of the integral than either the right- or left-hand sums Geometrically this average is equivalent
to approximating the area on each interval by a trapezoid instead of a rectangle, as shown below in Figure 26.3
f
x
x i–1 x i
Figure 26.3
Averaging the area of rectangles of ABCD and AEFD in Figure 26.3(a) gives the area of the trapezoid AECD
We refer to the average of the left- and right- hand sums, Ln +R n
sum(or the Trapezoidal rule) and denote it by Tn
T4=1.283 + 2.083
Because f is concave up4on [1, 5] we know that on each subinterval the area under the trapezoid is larger than that under the curve Thus T4gives an upper bound for the integral,
a better upper bound than that provided by L4
R4<
1
1
x dx < T4< L4 1.283 < 1
x dx <1.683 < 2.083
At this point the mind of the critical reader should be buzzing with questions Perhaps they include the following
What are the conditions under which Tnwill be larger than the value of the integral? Smaller?
Where does the midpoint sum fit into the picture?
Let’s investigate the first question by looking at some graphs
4 f (t ) =1, so f (t ) = −1and f (t ) = 2 The latter is positive on [1, 5], so f is concave up on the interval under discussion.
Trang 926.1 Approximating Sums: Ln, Rn, Tn, and Mn 809
Figure 26.4
If f is concave up on [a, b], then every secant line joining two points on the graph of
f on [a, b] lies above the graph.
If f is concave down on [a, b], then every secant line joining two points on the graph
of f on [a, b] lies below the graph.
We conclude that
a
f (t ) dt < Tn if f is concave up on [a, b],
and
Tn<
a
f (t ) dt if f is concave down on [a, b]
Where does the midpoint sum fit into the picture? It turns out that Tn and Mnare a complementary pair We will show that
if f is concave up on [a, b], then Mn<
b a
f (t ) dt < Tn while
if f is concave down on [a, b], then Tn<
a
f (t ) dt < Mn
To do this we’ll give an alternative graphical interpretation of the midpoint sum Figure 26.5(i) illustrates the midpoint approximation, approximating the area under the graph of
f on [xi, xi+1] by the area of a rectangle whose height is the value of f at the midpoint of the interval In Figure 26.5(ii) we approximate the area under f by the area of the trapezoid formed by the tangent line to the graph of f at the midpoint of the interval We claim that these areas are identical; pivoting the line through A, B, and C about the midpoint C does not change the area bounded below
Look at Figure 26.5(iii) The triangles ACD and BCE are congruent We argue this as follows Angles CAD and CBE are right angles Angles ACD and BCE are equal Therefore the two triangles in question are similar But AC = CB because C is the midpoint of the interval [xi, xi+1] We conclude that triangles ACD and BCE are congruent and hence have the same area Therefore, rectangle xiABxi+1and trapezoid xiDExi+1(Figures 26.5(i) and (ii), respectively) have the same area; we can interpret the midpoint sum as the midpoint tangent sum
Trang 10810 CHAPTER 26 Numerical Methods of Approximating Definite Integrals
(i )
D
C
(ii )
D
B
(iii )
x i midpt. x i +1 as long as the oblique linecongruent triangles
passes through midpoint C.
E
D
B
Figure 26.5
Below is a picture of M4using the midpoint tangent line interpretation f is concave
up, so the tangent lines lie below the curve We know that on each subinterval the area under the midpoint tangent line is less than the area under the curve; therefore, M4gives a lower bound for the integral
f
x
f
x
x0 x1 x2 x3
Figure 26.6
Where a function is concave up its tangent line lies below the curve; where a function is concave down its tangent line lies above the curve It follows that if f is concave up on [a, b], then Mn<abf (t ) dt; if f is concave down on [a, b], thenabf (t ) dt < Mn
x1 x2 x3 x1 x2 x3 x1 x2 x1 x2 x3
Figure 26.7
How do Ln, Rn, Tn, and Mnimprove as we increase n?
8 Subdivisions
Let n = 8; we chop [1, 5] into 8 equal pieces each of length x =5−18 =12