25.3 Substitution to Alter the Form of an Integral 801 Let u = 3x 2 . du = 3 2 dx so dx = 2 3 du 3 4 1 1 + 3x 2 2 dx = 3 4 1 1 + u 2 · 2 3 du = 3 4 · 2 3 1 1 + u 2 du = 1 2 arctan u + C = 1 2 arctan 3 2 x + C (b) tan √ x √ x dx Use substitution to get rid of the √ x. Let u = √ x. du = 1 2 1 √ x dx so 2du = 1 √ x dx tan √ x · 1 √ x dx = tan u · 2du =2 tan udu =2 sin u cos u du (This looks essentially like dW W .) Let W = cos u. dW =−sin udu so − dW =sin udu 2 sin u cos u du =2 −dW W =−2ln|W|+C Replace W by cos u. =−2ln|cos u|+C Replace u by √ x. =−2ln|cos √ x|+C (c) e 2x e 2x +2e x +1 dx We want to use substitution to get rid of e x and e 2x .Ifweletu=e x ,then e 2x = (e x ) 2 = u 2 . If, instead, we were to let u = e 2x , then e x = (e 2x ) 1/2 = √ u. The first option looks much more appealing. Let’s try it. Let u = e x . du =e x dx e 2x e 2x + 2e x + 1 dx 802 CHAPTER 25 Finding Antiderivatives—An Introduction to Indefinite Integration Either we can express this as e x e 2x + 2e x + 1 e x dx = u u 2 + 2u + 1 du or (e x ) 2 (e x ) 2 + 2e x + 1 dx = u 2 u 2 + 2u + 1 · 1 u du because dx = 1 e x du = 1 u du The two are equivalent. u u 2 + 2u + 1 du = u (u + 1) 2 du 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 + 1sou=W−1. dW =du u (u + 1) 2 du = W −1 W 2 dW = W W 2 − 1 W 2 dW = 1 W dW − W −2 dW =ln |W |−(−W −1 )+C Replace W by u + 1. = ln |u + 1|+ 1 u + 1 + C Replace u by e x . = ln(e x + 1) + 1 e x + 1 + C ◆ EXERCISE 25.3 Check the answers to Example 25.7 by differentiating. PROBLEMS FOR SECTION 25.3 For Problems 1 through 11, find the given indefinite integral. 1. x x+1 dx 2. x+3 x−7 dx 3. x √ 3x + 5 dx 4. 2x 3+x dx 5. 3t √ t 2 + 5 dt 6. t √ 2t +5 dt 25.3 Substitution to Alter the Form of an Integral 803 7. x π 5 dx 8. cot(3x) dx 9. (1.5) (1−t) dt 10. sin 4 t cos tdt 11. 2t √ 2t+6 dt 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 − x 2 dx. We can’t let u = x 2 or 1 − x 2 because u (x) is nowhere in sight. Instead we take a different approach. We’ll replace x by something that makes 1 − x 2 a perfect square. We’ll exploit the trigonometric identity sin 2 θ + cos 2 θ = 1. We know 1 − sin 2 θ = cos 2 θ, 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 √ c 2 − x 2 , √ x 2 − c 2 , and x 2 + c 2 , where c is a constant.) 13. Evaluate 2 x(x+2) dx by rewriting the integrand in the form A x + B x + 2 , where A and B are constants. In other words, find A and B such that 2 x(x +2) = A x + B x + 2 . (This is an example of an integration technique known as partial fractions.) 14. Evaluate 1 x 2 −1 dx by factoring the denominator of the integrand and rewriting the integrand in the form A x − 1 + B x + 1 , where A and B are constants. 26 CHAPTER Numerical Methods of Approximating Definite Integrals 26.1 APPROXIMATING SUMS: L n , R n , T n , AND M n 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. 1 From 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 −x 2 , or sin(x 2 ), or sin( 1 x ). 2 Knowing 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 xdx=xln 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 xdx= sec x · sec x+tan x sec x+tan x dx = sec 2 x+sec x tan x sec x+tan x dx = ln | sec x + tan x|+C. 2 That is, if we want to obtain an antiderivative that is a finite sum, product, or composition of the elementary functions. 805 806 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 Q approaches P we determine the exact slope of the tangent line at P . y = f(x) P Q x = ax = 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: L n , R n , T n , and M n 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 Approximate 5 1 1 x dx. 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. 26.1 Approximating Sums: L n , R n , T n , and M n 807 We’ ll construct three Riemann sums: the left- hand sum, denoted by L n ; the right-hand sum, denoted by R n ; and the midpoint sum, denoted by M n . (We are already familiar with the first two.) We describe these sums as follows. L n : The height of the rectangle on each subinterval is given by the value of f at the left-hand endpoint of the subinterval. R n : The height of the rectangle on each subinterval is given by the value of f at the right-hand endpoint of the subinterval. M n : The height of the rectangle on each subinterval is given by the value of f at the midpoint of the subinterval. On the i th interval, [x i−1 , x i ], the height of the rectangle is f(c),where c is midway between x i−1 and x i ; c = x i−1 +x i 2 . 4 Subdivisions Let n = 4; we chop [1, 5] into 4 equal subintervals each of length x = 5−1 4 = 1. ∆ x = 1 12345 Below are the left- hand sum, the right-hand sum, and the midpoint sum. 123 45 f x L 4 = 1 1 • 1 + • 1• 1 + • 1 + 1 2 1 3 1 4 = = 2.083 1 1 +++ 1 2 1 3 1 4 123 45 f x R 4 = 1 2 • 1 + • 1• 1 + • 1 + 1 3 1 4 1 5 = = 1.283 1 2 +++ 1 3 1 4 1 5 12 1.5 2.5 3.5 4.5 345 f x M 4 = 2 3 • 1 + • 1• 1 + • 1 + 2 5 2 7 2 9 = = 1.574 3 2 3 +++ 2 5 2 7 2 9 Figure 26.2 3 By “ ”we mean there are more decimal places but we have stopped recording them. 808 CHAPTER 26 Numerical Methods of Approximating Definite Integrals The function f(x)= 1 x is decreasing; therefore the left-hand sums provide an upper bound and the right-hand sums a lower bound for 5 1 1 x dx. 1.283 =R 4 < 5 1 1 x dx < L 4 = 2.083 Suppose we take the average of R 4 and L 4 , L 4 +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. 123 45 f AD BC EF x x i–1 x i (a) (b) 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, L n +R n 2 , as the trapezoidal sum (or the Trapezoidal rule) and denote it by T n . T 4 = 1.283 +2.083 2 = 1.683 Because f is concave up 4 on [1, 5] we know that on each subinterval the area under the trapezoid is larger than that under the curve. Thus T 4 gives an upper bound for the integral, a better upper bound than that provided by L 4 . R 4 < 5 1 1 x dx < T 4 <L 4 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 T n will 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 t ,sof (t) =− 1 t 2 and f (t) = 2 t 3 . The latter is positive on [1, 5], so f is concave up on the interval under discussion. 26.1 Approximating Sums: L n , R n , T n , and M n 809 x x x x x x 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 b a f(t)dt <T n if f is concave up on [a, b], and T n < b 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 T n and M n are a complementary pair. We will show that if f is concave up on [a, b], then M n < b a f(t)dt <T n while if f is concave down on [a, b], then T n < b a f(t)dt <M n . 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 [x i , x i+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 [x i , x i+1 ]. We conclude that triangles ACD and BCE are congruent and hence have the same area. Therefore, rectangle x i ABx i+1 and trapezoid x i DEx i+1 (Figures 26.5(i) and (ii), respectively) have the same area; we can interpret the midpoint sum as the midpoint tangent sum. 810 CHAPTER 26 Numerical Methods of Approximating Definite Integrals AB C (i) x i x i +1 midpt. D EE C (ii) x i x i +1 midpt. D BA C (iii) x i x i +1 midpt. congruent triangles as long as the oblique line passes through midpoint C. E D B A C Figure 26.5 Below is a picture of M 4 using 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, M 4 gives a lower bound for the integral. f x f x x 0 x 1 x 2 x 3 12345 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 M n < b a f(t)dt;iff is concave down on [a, b], then b a f(t)dt <M n . x 1 x 1 x 1 x 2 x 2 x 2 x 3 x 3 x 3 x 1 x 2 Figure 26.7 How do L n , R n , T n , and M n improve as we increase n? 8 Subdivisions Let n = 8; we chop [1, 5] into 8 equal pieces each of length x = 5−1 8 = 1 2 . . by factoring the denominator of the integrand and rewriting the integrand in the form A x − 1 + B x + 1 , where A and B are constants. 26 CHAPTER Numerical Methods of Approximating Definite Integrals 26.1. 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. T n , and M n 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