Сalculus Techniques of Integration

10 Techniques of Integration 10.1 Powers of sine and osine Functions consisting of products of the sine and cosine can be integrated by using substitution and trigonometric identities. These can sometimes be tedious, but the technique is straightforward. Some examples will suffice to explain the approach. EXAMPLE 10.1.1 Z 5 sin x dx = Z Evaluate Z 4 sin5 x dx. Rewrite the function: sin x sin x dx = Z 2 2 sin x(sin x) dx = Z sin x(1 − cos2 x)2 dx. Now use u = cos x, du = − sin x dx: Z 2 2 sin x(1 − cos x) dx = = Z Z −(1 − u2 )2 du −(1 − 2u2 + u4 ) du 1 2 = −u + u3 − u5 + C 3 5 2 1 = − cos x + cos3 x − cos5 x + C. 3 5 203 204 Chapter 10 Techniques of Integration EXAMPLE 10.1.2 Z Evaluate function: sin6 x dx. Use sin2 x = (1 − cos(2x))/2 to rewrite the (1 − cos 2x)3 dx 8 Z 1 = 1 − 3 cos 2x + 3 cos2 2x − cos3 2x dx. 8 Now we have four integrals to evaluate: Z 1 dx = x Z 6 sin x dx = Z 2 3 (sin x) dx = and Z Z 3 −3 cos 2x dx = − sin 2x 2 are easy. The cos3 2x integral is like the previous example: Z Z 3 − cos 2x dx = − cos 2x cos2 2x dx Z = − cos 2x(1 − sin2 2x) dx Z 1 = − (1 − u2 ) du 2   u3 1 u− =− 2 3   1 sin3 2x =− sin 2x − . 2 3 And finally we use another trigonometric identity, cos2 x = (1 + cos(2x))/2:   Z Z sin 4x 1 + cos 4x 3 2 x+ . 3 cos 2x dx = 3 dx = 2 2 4 So at long last we get     Z sin3 2x 3 sin 4x 3 1 x 6 sin 2x − + x+ + C. sin 2x − sin x dx = − 8 16 16 3 16 4 EXAMPLE 10.1.3 Evaluate Z sin2 x cos2 x dx. Use the formulas sin2 x = (1−cos(2x))/2 and cos2 x = (1 + cos(2x))/2 to get: Z Z 1 − cos(2x) 1 + cos(2x) 2 2 sin x cos x dx = · dx. 2 2 The remainder is left as an exercise. 10.2 Trigonometric Substitutions 205 Exercises 10.1. Find Zthe antiderivatives. 2 sin x dx ⇒ 1. 2. Z sin3 x dx ⇒ 4. Z cos2 x sin3 x dx ⇒ 6. Z sin2 x cos2 x dx ⇒ sin x(cos x)3/2 dx ⇒ tan3 x sec x dx ⇒ 3. Z 5. Z 7. Z cos x sin x dx ⇒ 8. Z 9. Z sec2 x csc2 x dx ⇒ 10. Z 10.2 4 sin x dx ⇒ 3 cos x dx ⇒ 3 2 Trigonometri Substitutions So far we have seen that it sometimes helps to replace a subexpression of a function by a single variable. Occasionally it can help to replace the original variable by something more complicated. This seems like a “reverse” substitution, but it is really no different in principle than ordinary substitution. EXAMPLE 10.2.1 Evaluate Z p 1 − x2 dx. Let x = sin u so dx = cos u du. Then Z p Z p Z √ 2 2 1 − x dx = 1 − sin u cos u du = cos2 u cos u du. √ We would like to replace cos2 u by cos u, but this is valid only if cos u is positive, since √ cos2 u is positive. Consider again the substitution x = sin u. We could just as well think of this as u = arcsin x. If we do, then by the definition of the arcsine, −π/2 ≤ u ≤ π/2, so cos u ≥ 0. Then we continue: Z Z Z √ u sin 2u 1 + cos 2u 2 2 du = + +C cos u cos u du = cos u du = 2 2 4 = arcsin x sin(2 arcsin x) + + C. 2 4 This is a perfectly good answer, though the term sin(2 arcsin x) is a bit unpleasant. It is possible to simplify this. Using the identity sin 2x = 2 sin x cos x, we can write sin 2u = q p p 2 2 sin u cos u = 2 sin(arcsin x) 1 − sin u = 2x 1 − sin2 (arcsin x) = 2x 1 − x2 . Then the full antiderivative is √ √ arcsin x x 1 − x2 arcsin x 2x 1 − x2 + = + + C. 2 4 2 2 206 Chapter 10 Techniques of Integration This type of substitution is usually indicated when the function you wish to integrate contains a polynomial expression that might allow you to use the fundamental identity sin2 x + cos2 x = 1 in one of three forms: cos2 x = 1 − sin2 x sec2 x = 1 + tan2 x tan2 x = sec2 x − 1. If your function contains 1 − x2 , as in the example above, try x = sin u; if it contains 1 + x2 try x = tan u; and if it contains x2 − 1, try x = sec u. Sometimes you will need to try something a bit different to handle constants other than one. EXAMPLE 10.2.2 Evaluate Z p 4 − 9x2 dx. We start by rewriting this so that it looks more like the previous example: Z p Z p Z p 2 2 4 − 9x dx = 4(1 − (3x/2) ) dx = 2 1 − (3x/2)2 dx. Now let 3x/2 = sin u so (3/2) dx = cos u du or dx = (2/3) cos u du. Then Z p Z Z p 4 2 2 cos2 u du 2 1 − (3x/2) dx = 2 1 − sin u (2/3) cos u du = 3 4u 4 sin 2u + +C = 6 12 2 arcsin(3x/2) 2 sin u cos u + +C = 3 3 2 arcsin(3x/2) 2 sin(arcsin(3x/2)) cos(arcsin(3x/2)) = + +C 3 3 p 2 arcsin(3x/2) 2(3x/2) 1 − (3x/2)2 = + +C 3 3 √ 2 arcsin(3x/2) x 4 − 9x2 = + + C, 3 2 using some of the work from example 10.2.1. EXAMPLE 10.2.3 Z p 1+ Evaluate x2 Z p 1 + x2 dx. Let x = tan u, dx = sec2 u du, so Z p Z √ 2 2 dx = 1 + tan u sec u du = sec2 u sec2 u du. √ Since u = arctan(x), −π/2 ≤ u ≤ π/2 and sec u ≥ 0, so sec2 u = sec u. Then Z √ Z 2 sec2 u sec u du = sec3 u du. In problems of this type, two integrals come up frequently: Z 3 sec u du and Both have relatively nice expressions but they are a bit tricky to discover. Z sec u du. 10.2 First we do R Trigonometric Substitutions sec u du, which we will need to compute Z sec u du = = Z Z sec u Z 207 sec3 u du: sec u + tan u du sec u + tan u sec2 u + sec u tan u du. sec u + tan u Now let w = sec u + tan u, dw = sec u tan u + sec2 u du, exactly the numerator of the function we are integrating. Thus Z Z Z 1 sec2 u + sec u tan u du = dw = ln |w| + C sec u du = sec u + tan u w = ln | sec u + tan u| + C. Now for Z sec3 u du: sec3 u sec3 u sec3 u (tan2 u + 1) sec u sec u = + = + 2 2 2 2 3 2 3 sec u sec u tan u sec u sec u + sec u tan2 u sec u = + + = + . 2 2 2 2 2 3 We already know how to integrate sec u, so we just need the first quotient. This is “simply” a matter of recognizing the product rule in action: Z sec3 u + sec u tan2 u du = sec u tan u. So putting these together we get Z sec u tan u ln | sec u + tan u| + + C, sec3 u du = 2 2 and reverting to the original variable x: Z p sec u tan u ln | sec u + tan u| + +C 1 + x2 dx = 2 2 sec(arctan x) tan(arctan x) ln | sec(arctan x) + tan(arctan x)| + +C 2 2 √ √ ln | 1 + x2 + x| x 1 + x2 + + C, = 2 2 q p using tan(arctan x) = x and sec(arctan x) = 1 + tan2 (arctan x) = 1 + x2 . = 208 Chapter 10 Techniques of Integration Exercises 10.2. Find Zthe antiderivatives. csc x dx ⇒ 1. 3. 5. 7. 9. 11. 2. csc3 x dx ⇒ Z p 4. 9 + 4x2 dx ⇒ Z p 6. x2 1 − x2 dx ⇒ Z p x2 + 2x dx ⇒ 8. Z p x2 − 1 dx ⇒ Z p x 1 − x2 dx ⇒ Z 1 √ dx ⇒ 1 + x2 Z 1 dx ⇒ 2 x (1 + x2 ) √ Z x √ dx ⇒ 1−x 10.3 Z x2 √ dx ⇒ 4 − x2 Z x3 √ dx ⇒ 12. 4x2 − 1 10. Z Integration by Parts We have already seen that recognizing the product rule can be useful, when we noticed that Z sec3 u + sec u tan2 u du = sec u tan u. As with substitution, we do not have to rely on insight or cleverness to discover such antiderivatives; there is a technique that will often help to uncover the product rule. Start with the product rule: d f (x)g(x) = f ′ (x)g(x) + f (x)g ′ (x). dx We can rewrite this as f (x)g(x) dx + Z f (x)g ′ (x) dx, f (x)g (x) dx = f (x)g(x) − Z f ′ (x)g(x) dx. f (x)g(x) = and then Z Z ′ ′ This may not seem particularly useful at first glance, but it turns out that in many cases we have an integral of the form Z f (x)g ′ (x) dx but that Z f ′ (x)g(x) dx is easier. This technique for turning one integral into another is called integration by parts, and is usually written in more compact form. If we let u = f (x) and v = g(x) then 10.3 Integration by Parts 209 du = f ′ (x) dx and dv = g ′ (x) dx and Z u dv = uv − Z v du. To use this technique we need to identify likely candidates for u = f (x) and dv = g ′ (x) dx. EXAMPLE 10.3.1 Evaluate let dv = x dx so v = x2 /2 and Z x2 ln x x ln x dx = − 2 EXAMPLE 10.3.2 Z Z Evaluate dv = sin x dx so v = − cos x and Z x sin x dx = −x cos x − Z x ln x dx. Let u = ln x so du = 1/x dx. Then we must x2 1 x2 ln x dx = − 2 x 2 Z EXAMPLE 10.3.3 Evaluate x x2 ln x x2 dx = − + C. 2 2 4 x sin x dx. Let u = x so du = dx. Then we must let − cos x dx = −x cos x + Z Z Z cos x dx = −x cos x + sin x + C. sec3 x dx. Of course we already know the answer to this, but we needed to be clever to discover it. Here we’ll use the new technique to discover the antiderivative. Let u = sec x and dv = sec2 x dx. Then du = sec x tan x dx and v = tan x and Z Z 3 sec x dx = sec x tan x − tan2 x sec x dx Z = sec x tan x − (sec2 x − 1) sec x dx Z Z 3 = sec x tan x − sec x dx + sec x dx. 210 Chapter 10 Techniques of Integration At first this looks useless—we’re right back to Z Z 3 sec x dx + 2 Z Z Z sec3 x dx. But looking more closely: 3 sec x dx = sec x tan x − 3 sec x dx = sec x tan x + 3 sec x dx = sec x tan x + Z Z 3 sec x dx + Z sec x dx sec x dx Z sec x dx Z Z sec x tan x 1 3 sec x dx = sec x dx + 2 2 = sec x tan x ln | sec x + tan x| + + C. 2 2 Z Evaluate x2 sin x dx. Let u = x2 , dv = sin x dx; then du = 2x dx Z Z 2 2 and v = − cos x. Now x sin x dx = −x cos x + 2x cos x dx. This is better than the EXAMPLE 10.3.4 original integral, but we need to do integration by parts again. Let u = 2x, dv = cos x dx; then du = 2 and v = sin x, and Z 2 2 x sin x dx = −x cos x + Z 2x cos x dx Z 2 = −x cos x + 2x sin x − 2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C. Such repeated use of integration by parts is fairly common, but it can be a bit tedious to accomplish, and it is easy to make errors, especially sign errors involving the subtraction in the formula. There is a nice tabular method to accomplish the calculation that minimizes the chance for error and speeds up the whole process. We illustrate with the previous example. Here is the table: sign u dv u dv x2 sin x x2 sin x − 2x − cos x −2x − cos x − 0 0 cos x 2 − sin x cos x or 2 − sin x 10.3 Integration by Parts 211 To form the first table, we start with u at the top of the second column and repeatedly compute the derivative; starting with dv at the top of the third column, we repeatedly compute the antiderivative. In the first column, we place a “−” in every second row. To form the second table we combine the first and second columns by ignoring the boundary; if you do this by hand, you may simply start with two columns and add a “−” to every second row. To compute with this second table we begin at the top. Multiply the first entry in column u by the second entry in column dv to get −x2 cos x, and add this to the integral of the product of the second entry in column u and second entry in column dv. This gives: 2 −x cos x + Z 2x cos x dx, or exactly the result of the first application of integration by parts. Since this integral is not yet easy, we return to the table. Now we multiply twice on the diagonal, (x2 )(− cos x) and (−2x)(− sin x) and then once straight across, (2)(− sin x), and combine these as 2 −x cos x + 2x sin x − Z 2 sin x dx, giving the same result as the second application of integration by parts. While this integral is easy, we may return yet once more to the table. Now multiply three times on the diagonal to get (x2 )(− cos x), (−2x)(− sin x), and (2)(cos x), and once straight across, (0)(cos x). We combine these as before to get 2 −x cos x + 2x sin x + 2 cos x + Z 0 dx = −x2 cos x + 2x sin x + 2 cos x + C. Typically we would fill in the table one line at a time, until the “straight across” multiplication gives an easy integral. If we can see that the u column will eventually become zero, we can instead fill in the whole table; computing the products as indicated will then give the entire integral, including the “+C ”, as above. Exercises 10.3. Find Zthe antiderivatives. 1. 3. Z 5. Z x cos x dx ⇒ 2. Z x2 cos x dx ⇒ xex dx ⇒ 4. Z xex dx ⇒ 6. Z ln x dx ⇒ 2 sin x dx ⇒ 2 212 Chapter 10 Techniques of Integration 7. Z x arctan x dx ⇒ 9. Z x3 cos x dx ⇒ 11. Z x sin x cos x dx ⇒ 13. Z √ sin( x) dx ⇒ 10.4 8. Z x3 sin x dx ⇒ 10. Z x sin2 x dx ⇒ 12. Z √ arctan( x) dx ⇒ 14. Z sec2 x csc2 x dx ⇒ Rational Fun tions A rational function is a fraction with polynomials in the numerator and denominator. For example, x3 1 x2 + 1 , , , x2 + x − 6 (x − 3)2 x2 − 1 are all rational functions of x. There is a general technique called “partial fractions” that, in principle, allows us to integrate any rational function. The algebraic steps in the technique are rather cumbersome if the polynomial in the denominator has degree more than 2, and the technique requires that we factor the denominator, something that is not always possible. However, in practice one does not often run across rational functions with high degree polynomials in the denominator for which one has to find the antiderivative function. So we shall explain how to find the antiderivative of a rational function only when the denominator is a quadratic polynomial ax2 + bx + c. We should mention a special type of rational function that we already know how to integrate: If the denominator has the form (ax + b)n , the substitution u = ax + b will always work. The denominator becomes un , and each x in the numerator is replaced by (u − b)/a, and dx = du/a. While it may be tedious to complete the integration if the numerator has high degree, it is merely a matter of algebra. 10.4 Find EXAMPLE 10.4.1 Z 3 1 x dx = (3 − 2x)5 −2 = 1 16 1 = 16 1 = 16 =− Z Z    Z u−3 −2 u5 213 Rational Functions x3 dx. Using the substitution u = 3 − 2x we get (3 − 2x)5 3 1 du = 16 Z u3 − 9u2 + 27u − 27 du u5 u−2 − 9u−3 + 27u−4 − 27u−5 du u−1 9u−2 27u−3 27u−4 − + − −1 −2 −3 −4  +C (3 − 2x)−1 9(3 − 2x)−2 27(3 − 2x)−3 27(3 − 2x)−4 − + − −1 −2 −3 −4 1 9 9 27 + − + +C 2 3 16(3 − 2x) 32(3 − 2x) 16(3 − 2x) 64(3 − 2x)4  +C We now proceed to the case in which the denominator is a quadratic polynomial. We can always factor out the coefficient of x2 and put it outside the integral, so we can assume that the denominator has the form x2 + bx + c. There are three possible cases, depending on how the quadratic factors: either x2 + bx + c = (x − r)(x − s), x2 + bx + c = (x − r)2 , or it doesn’t factor. We can use the quadratic formula to decide which of these we have, and to factor the quadratic if it is possible. EXAMPLE 10.4.2 Determine whether x2 + x + 1 factors, and factor it if possible. The quadratic formula tells us that x2 + x + 1 = 0 when x= −1 ± √ 2 1−4 . Since there is no square root of −3, this quadratic does not factor. EXAMPLE 10.4.3 Determine whether x2 − x − 1 factors, and factor it if possible. The quadratic formula tells us that x2 − x − 1 = 0 when x= Therefore x2 − x − 1 = 1± √ 1+4 1± 5 = . 2 2 √ √ ! √ ! 1− 5 1+ 5 x− . x− 2 2 214 Chapter 10 Techniques of Integration If x2 + bx + c = (x − r)2 then we have the special case we have already seen, that can be handled with a substitution. The other two cases require different approaches. If x2 + bx + c = (x − r)(x − s), we have an integral of the form Z p(x) dx (x − r)(x − s) where p(x) is a polynomial. The first step is to make sure that p(x) has degree less than 2. x3 dx in terms of an integral with a numer(x − 2)(x + 3) ator that has degree less than 2. To do this we use long division of polynomials to discover that EXAMPLE 10.4.4 so Rewrite Z x3 7x − 6 7x − 6 x3 = 2 =x−1+ 2 =x−1+ , (x − 2)(x + 3) x +x−6 x +x−6 (x − 2)(x + 3) Z x3 dx = (x − 2)(x + 3) Z x − 1 dx + Z 7x − 6 dx. (x − 2)(x + 3) The first integral is easy, so only the second requires some work. Now consider the following simple algebra of fractions: A B A(x − s) + B(x − r) (A + B)x − As − Br + = = . x−r x−s (x − r)(x − s) (x − r)(x − s) That is, adding two fractions with constant numerator and denominators (x−r) and (x−s) produces a fraction with denominator (x − r)(x − s) and a polynomial of degree less than 2 for the numerator. We want to reverse this process: starting with a single fraction, we want to write it as a sum of two simpler fractions. An example should make it clear how to proceed. x3 7x − 6 dx. We start by writing (x − 2)(x + 3) (x − 2)(x + 3) as the sum of two fractions. We want to end up with EXAMPLE 10.4.5 Evaluate Z 7x − 6 A B = + . (x − 2)(x + 3) x−2 x+3 If we go ahead and add the fractions on the right hand side we get 7x − 6 (A + B)x + 3A − 2B = . (x − 2)(x + 3) (x − 2)(x + 3) So all we need to do is find A and B so that 7x − 6 = (A + B)x + 3A − 2B, which is to say, we need 7 = A + B and −6 = 3A − 2B. This is a problem you’ve seen before: solve a 10.4 Rational Functions 215 system of two equations in two unknowns. There are many ways to proceed; here’s one: If 7 = A+B then B = 7−A and so −6 = 3A−2B = 3A−2(7−A) = 3A−14+2A = 5A−14. This is easy to solve for A: A = 8/5, and then B = 7 − A = 7 − 8/5 = 27/5. Thus Z Z 7x − 6 27 1 8 27 8 1 dx = + dx = ln |x − 2| + ln |x + 3| + C. (x − 2)(x + 3) 5x−2 5 x+3 5 5 The answer to the original problem is now Z Z Z x3 7x − 6 dx = x − 1 dx + dx (x − 2)(x + 3) (x − 2)(x + 3) 8 27 x2 − x + ln |x − 2| + ln |x + 3| + C. 2 5 5 = Now suppose that x2 + bx + c doesn’t factor. Again we can use long division to ensure that the numerator has degree less than 2, then we complete the square. x+1 dx. The quadratic denominator does not + 4x + 8 factor. We could complete the square and use a trigonometric substitution, but it is simpler to rearrange the integrand: Z Z Z x+2 1 x+1 dx = dx − dx. 2 2 2 x + 4x + 8 x + 4x + 8 x + 4x + 8 EXAMPLE 10.4.6 Evaluate Z x2 The first integral is an easy substitution problem, using u = x2 + 4x + 8: Z Z x+2 1 du 1 dx = = ln |x2 + 4x + 8|. 2 x + 4x + 8 2 u 2 For the second integral we complete the square: ! 2  x + 2 +1 , x2 + 4x + 8 = (x + 2)2 + 4 = 4 2 making the integral 1 4 Using u = Z 1
x+2 2 2 +1 dx. x+2 we get 2   Z Z 1 1 1 2 1 x+2 . dx = du = arctan
x+2 2 4 4 u2 + 1 2 2 + 1 2 The final answer is now   Z 1 1 x+2 x+1 2 + C. dx = ln |x + 4x + 8| − arctan x2 + 4x + 8 2 2 2 216 Chapter 10 Techniques of Integration Exercises 10.4. Find the antiderivatives. Z 1 dx ⇒ 1. 4 − x2 Z 1 3. dx ⇒ x2 + 10x + 25 Z x4 5. dx ⇒ 4 + x2 Z x3 7. dx ⇒ 4 + x2 Z 1 dx ⇒ 9. 2 2x − x − 3 10.5 x4 dx ⇒ 4 − x2 Z x2 4. dx ⇒ 4 − x2 Z 1 6. dx ⇒ 2 x + 10x + 29 Z 1 8. dx ⇒ x2 + 10x + 21 Z 1 10. dx ⇒ 2 x + 3x 2. Z Numeri al Integration We have now seen some of the most generally useful methods for discovering antiderivatives, and there are others. Unfortunately, some functions have no simple antiderivatives; in such cases if the value of a definite integral is needed it will have to be approximated. We will see two methods that work reasonably well and yet are fairly simple; in some cases more sophisticated techniques will be needed. Of course, we already know one way to approximate an integral: if we think of the integral as computing an area, we can add up the areas of some rectangles. While this is quite simple, it is usually the case that a large number of rectangles is needed to get acceptable accuracy. A similar approach is much better: we approximate the area under a curve over a small interval as the area of a trapezoid. In figure 10.5.1 we see an area under a curve approximated by rectangles and by trapezoids; it is apparent that the trapezoids give a substantially better approximation on each subinterval. .... ........ .............. ..... ..... .... ..... . . .... . .... ... . ... . .. ... . ... . . ... . . . ... . . ... . . . ... . ... . ... ... ... ... ... ... . . .... . .... ... .... ... .... ... . . ..... .. ..... .... ..... ..... ...... ..................... Figure 10.5.1 .......... ....... ........... ..... ... .............. ... .... ........ . . .. .. ..... .. ... .... ... ... ... ... ... ...... . ...... . . . . ....... . ... . . ...... . .... . ...... . .... . ...... . .... ..... ... ...... ..... .. .. ....... ...... ..... . . ....... ... . ...... .. ... ....... ... .. .. ..... ....... . . ........ . ... ... ........ ........ ..... ....... ........... ...... .............. Approximating an area with rectangles and with trapezoids. (AP) As with rectangles, we divide the interval into n equal subintervals of length ∆x. A f (xi ) + f (xi+1 ) typical trapezoid is pictured in figure 10.5.2; it has area ∆x. If we add up 2 10.5 Numerical Integration 217 the areas of all trapezoids we get f (x1 ) + f (x2 ) f (xn−1 ) + f (xn ) f (x0 ) + f (x1 ) ∆x + ∆x + · · · + ∆x = 2 2 2   f (x0 ) f (xn ) ∆x. + f (x1 ) + f (x2 ) + · · · + f (xn−1 ) + 2 2 This is usually known as the Trapezoid Rule. For a modest number of subintervals this is not too difficult to do with a calculator; a computer can easily do many subintervals. (xi , f (xi )) ................... ... ..... ... .... ... ..... ... .... ... ... ....... ...... ...... ...... .... xi Figure 10.5.2 (xi+1 , f (xi+1 )) xi+1 A single trapezoid. In practice, an approximation is useful only if we know how accurate it is; for example, we might need a particular value accurate to three decimal places. When we compute a particular approximation to an integral, the error is the difference between the approximation and the true value of the integral. For any approximation technique, we need an error estimate, a value that is guaranteed to be larger than the actual error. If A is an approximation and E is the associated error estimate, then we know that the true value of the integral is between A − E and A + E. In the case of our approximation of the integral, we want E = E(∆x) to be a function of ∆x that gets small rapidly as ∆x gets small. Fortunately, for many functions, there is such an error estimate associated with the trapezoid approximation. THEOREM 10.5.1 Suppose f has a second derivative f ′′ everywhere on the interval [a, b], and |f ′′ (x)| ≤ M for all x in the interval. With ∆x = (b − a)/n, an error estimate for the trapezoid approximation is E(∆x) = Let’s see how we can use this. b−a (b − a)3 M (∆x)2 = M. 12 12n2 218 Chapter 10 Techniques of Integration EXAMPLE 10.5.2 Approximate Z 1 2 e−x dx to two decimal places. The second deriva- 0 −x2 2 −x2 2 tive of f = e is (4x −2)e , and it is not hard to see that on [0, 1], |(4x2 −2)e−x | ≤ 2. We begin by estimating the number of subintervals we are likely to need. To get two decimal places of accuracy, we will certainly need E(∆x) < 0.005 or 1 1 (2) 2 < 0.005 12 n 1 (200) < n2 6 r 100