f(x) g(x) = [f (x)g(x) dx + f(x)g (x)dx

Transcription

1 Chapter 7 is concerned with all the integrals that can t be evaluated with simple antidifferentiation. Chart of Integrals on Page Integration by Parts Like with the Chain Rule substitutions with u and du, we re going to look at integration by considering derivatives. Let s look at the Product Rule: d dx [f(x)g(x)] = f (x)g(x) + f(x)g (x) d dx [f(x)g(x)]dx = (f (x)g(x) + f(x)g (x)) dx f(x) g(x) = [f (x)g(x) dx + f(x)g (x)dx f(x)g (x)dx = f(x)g(x) - [f (x)g(x) dx or Integration by Parts: u dv = uv v du Ex: Find x cos x dx Let u = x du = dx dv = cos x dx v = sin x x cox x dx = x sin x - sin x dx = x sin x + cos x + C

2 How can we check our answer? In general, we want to choose the expression that will simplify when differentiated to be u. Ex: Find ln x dx Let: u = ln x du = 1 x dv = dx v = x Sometimes you have to use the technique more than once. Ex: Find t 2 e t dt

3 Sometimes it s not obvious that you re making any headway. Ex: Find e x sin x dx (Hint: Use Integration by Parts twice, and then fold the equation.)

4 Ex: What is a Reduction Formula? Ex: (lnx) n dx = x(lnx) n n (lnx) n 1 dx

5 7.2 Trigonometric Integrals First, some review: Trigonometricpalooza cos 2 x = 1 sin 2 x sin 2 x = 1 cos 2 x cos 2 x = ½ (1 + cos 2x) sin 2 x = ½ (1 - cos 2x) sin A cos B = ½ [sin (A-B) + sin(a+b)] sin A sin B = ½ [cos (A-B) - cos(a+b)] cos A cos B = ½ [cos (A-B) + cos(a+b)] d dx d dx sinx = cosx cosx = sinx d dx tanx = sec2 x d dx secx = secx tanx tanx dx = ln sec x + C secx dx = ln sec x + tanx + C

6 Ex: Find sinx cosx dx Ex: Find sin 2 x cosx dx

7 Ex: Find cos 3 x dx Ex: Find cos 3 x sin 2 x dx

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9 Ex: Find cos 2 x dx (Hint: Use the half-angle formula cos 2 x = ½ (1 + cos 2x) )

10 Ex: Find tan 3 x dx (Hint: Use tan 2 x = sec 2 x 1, plus the fact that tanxdx = ln sec x + C ) Ex: Evaluate sin4x cos 5x dx (Hint: Use one of the product rules: sin A cos B = ½[sin(A-B) + sin (A+B)]

11 7.3 Trigonometric Substitution (Note: Triangles are optional. Use them if they help, but skip them if they don t.) Ex: Find 9 x 2 x dx

12 Ex: Find 9 x 2 dx Oh no! We can t use a substitution to solve this. In our desperation, we try something ridiculous: Let x = 3 sin θ ( π θ π ) 2 2 then dx = and 9 x 2 = So the integral becomes 9 x 2 dx = 9 9sin 2 θ 3 cosθ dθ = 9cos 2 θ 3 cosθ dθ = 3 cos θ 3 cosθ dθ = 9 cos 2 θ dθ = 9 1 ( 1 + cos 2θ) dθ 2 = 9 2 [ θ + ½ sin 2θ] + C

13 = 9 [θ + ½ * 2 sin θ cos θ ] + C 2 Note that x = 3 sin θ, so sin θ = x 3 so θ = arcsin x 3 and cos θ = 1 sin 2 θ = 1 x2 9 = 9 2 [ arcsin x 3 + x 9 9 x2 ] + C = 9 2 arcsin x x 9 x2 + C But are we allowed to do this? Yes, as long as we limit θ so that x = 3 sin θ is one-to-one. This is called an inverse substitution.

14 7.4 Integration of Rational Functions by Partial Fractions This section deals with integrals of rational expressions. The first type we ll look at are those of the form R(x) = P(x)/Q(x) where deg (P) > deg (Q) (improper fraction) Ex: Find 3t 2 t+1 dt Our first step is to divide the expression, using techniques from Math 90.

15 So, 3t 2 5 dt = [3 ]dt t+1 t+1 = 3t 5 ln t+1 + C

16 The second type of rational equation is one in which the denominator can be completely factored: Ex: Find x2 + 2x 1 2x 3 + 3x 2 2x dx First, check to make sure that it s not improper. Then, try factoring the denominator: Next, write the rational expression as a sum of simpler fractions: x 2 + 2x 1 2x 3 + 3x 2 2x = A x + B + C 2x 1 x+2 The idea is to simplify the problem so that the integrations will be easier, but finding A, B and C is going to take some work.

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18 One way to find A, B and C is to clear fractions in this equation: x 2 + 2x 1 2x 3 + 3x 2 2x = A x + B 2x 1 + C x+2

19 The result is called the Partial Fraction Decomposition We then integrate each piece:

20 Ex: Find x4 2x 2 +4x+1 dx x 3 x 2 x+1 First, divide, since the degree on top is greater than on the bottom: Next, factor the denominator: We notice that the linear factor (x-1) appears twice, which means that we ll have the partial fraction decomposition: 4x = A + (x 1) 2 (x+1) (x 1) B (x 1) 2 + C (x+1)

21 Find A, B and C:

22 and then integrate:

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24 7.5 Strategy for Integration Four-Step Strategy for Tricky Integrals: 1) Simplify the integrand, if possible, by multiplying out, or rewriting formulas: Ex: x(1 x + x) dx Ex: tanθ sec 2 θ dθ

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26 2) Use a u substitution Ex: x 1 x 2 dx Ex: sin(4x) dx

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28 3) Consult one of the Four Horsemen of the Calcupocalypse a) Trigonometric Functions from 7.2 b) Rational Expression/Partial Fraction Decomposition 7.4 c) Integration by Parts 7.1 d) Radicals 7.3 (if they aren t easily dealt with using a u substitution) 4) The book helpfully suggests you try again at this point. Ex: Find e x dx There s a cool trick to this one let u = x Then u 2 = x, so 2u du = dx so the integral becomes e u 2u du = 2 e u u du which we can integrate by parts:

29 7.6 Integration Using Tables and Computer Algebra Systems This section has been somewhat obviated by the appearance of online integral calculators that even show their steps, but here is the procedure anyway: Ex: Use the Table of Integrals to find x2 5 4x 2 dx (See Reference Pages 6 10, at the back of the book.) On Reference Page 7, #34 reads u2 du a 2 u 2 = u 2 a2 u 2 + a2 2 sin 1 u a + C How do we use this formula in our case? u = 2x du = a 2 =

30 7.7 Approximate Integration Sometimes the exact value of an integral cannot be calculated, either because the antiderivative is difficult or impossible to find, or because a function is defined by data points and lacks a formula. As long as we can evaluate f(x) at various values of x on the interval [a, b], we can use Riemann sums to approximate b f(x)dx a Midpoint Rule You might recall from before that we looked at left endpoint approximations and right endpoint approximations, and then discussed the fact that using the midpoints in each subinterval seemed to give us the best results:

31 The Midpoint Rule b f(x)dx a Mn = x [f(x ) 1 + f(x ) f(x ) n where x = b a n and x i = ½ (xi 1 + xi) = midpoint of [ xi-1, xi] If f (x) K for a x b, then the error, EM, in using the

32 Midpoint Rule can be calculated as: EM K(b a)3 24n 2 Ex: Use the Midpoint Rule with n=5 to approximate dx x 2 (Note: 1 dx = ln 2 - ln 1 = ln 2 1 x = ) n = 5 x = = 0.2 Intervals are: [1, 1.2], [1.2, 1.4], [1.4, 1.6], [1.6, 1.8] and [1.8, 2] Midpoints are: 1.1, 1.3, 1.5, 1.7, 1.9

33 Estimate = 0.2 ( ) What s the actual error in using the Midpoint Rule in this case? Error What is the predicted upper bound on the error? f(x) = 1 x = x-1 f (x) = - x -2 f (x) = 2 x -3 on [1, 2], f (x) 2 (1) -3 = 2 EM 2 (2 1)3 = 2 = (5) The Trapezoidal Rule Another way to estimate integrals is to average the values obtained by looking at the left and right endpoints of each interval.

34 The Trapezoidal Rule b f(x)dx a Tn = x 2 [f (x 0 ) + 2f(x 1 ) + 2f(x 2 ) + +2f(x n 1 ) + f(x n ) where x = b a n and xi = a + i x If f (x) K for a x b, then the error, ET, in using the Trapezoidal Rule can be calculated as: ET K(b a)3 12n 2 Ex: How large should we take n in order to guarantee that the Trapezoidal Rule 2 approximation for dx is accurate to 1 x

35 within ? Remember that, from above, we calculated that f (x) < 2 on [1, 2].

36 Simpson s Rule The third technique for approximating integrals comes from the idea of using parabolas, instead of lines, to approximate the function values. (See page 512 for the derivation.) The formula is as follows: Simpson s Rule b f(x)dx a Sn = x 3 [f (x 0 ) + 4f(x 1 ) + 2f(x 2 ) + 4f(x 3 ) +2f(x n 2 ) + 4f(x n 1 ) + f(x n ) where x = b a n If f (4) (x) K for a x b, then the error, ES, in using Simpson s Rule can be calculated as: ES K(b a)5 180n 4

37 2 Ex: Use Simpson s Rule with n = 10 to approximate dx n = 10 x = b a n = 1 10 = 0.1 S (f(1) + 4f(1.1) + 2f(1.2) + 4f(1.3) + + 2f(1.8) + 4f(1.9) + f(2)) 1 x

38 4.4 Indeterminate Forms and L Hospital s Rule Ex: Calculate lim ex x x 2 Argh. Since lim e x = and lim x 2 =, x x this is called an indeterminate form of type. Normally, we d try to divide by x or look at the graph, since we weren t able to evaluate this limit until now. *Cue Exciting Music * L Hospital s Rule Suppose f and g are differentiable and g (x) 0 on an open interval l that contains a (except possibly at a.) Suppose that or that lim f(x) = 0 and lim g(x) = 0 x a x a lim f(x) = ± and lim g(x) = ± x a x a (In other words, we have an indeterminate form of type 0 0 or.) Then lim x a f(x) g(x) = lim x a f (x) g (x) if the limit on the right side exists (or is or. )

39 So lim = lim x 2 x e x x e x 2x = lim x e x 2 = Ex: Find lim x 1 ln x x 1

40 Big Warning L Hopital doesn t always work: lim x π sin x 1 cos x =? L Hopital would suggest this: BUT, while sin π = 0, 1 cos π = 1 (-1) = 2, so the limit equals 0 2 = 0, not 0 0, so you can t use L Hopital. Ex: lim xln x x 0 + (Hint: x ln x = ln x 1 x )

41 7.8 Improper Integrals There are two different categories of integrals which are considered improper. Type 1 are integrals where one or both limits are infinite. Ex: dx 1 x 2 The only way to evaluate this integral is to use a limit: dx 1 x 2 t dx t 1 x 2 = lim = lim[ 1 ]t t x 1 = lim[ 1 + 1] t t = lim t [1 1 t ] = 1

42 Ex: = lim dx 1 x t dx t 1 x = lim t [ln x] t 1 = lim t [ln t ln 1] = lim t [ln t] = So, what happened? Why did one converge and the other diverge? In the case of y = 1/x, the values simply don t get small enough fast enough for the area to converge. 1 1 x p dx is convergent if p>1 and divergent if p<1

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44 Ex: x 3 e x4 dx In order to evaluate this integral, we need to pick some value c to break this integral into two pieces. We can pick any value, so let s pick c = 0. x 3 0 x 3 = lim e x4 0 dx = x 3 e x4 dx = lim [ e x4 t 4 ] 0 t 0 t t x3 = lim [ 1 1 ] = 1-0 t 4 4e t4 4 x 3 0 e x4 dx + x 3 0 e x4 dx e x4 dx = lim x 3 e x4 dx t 0 t e x4 dx = lim [ e x4 t 4 ] t 0 = lim t [ 1 4e t4 1 4 ] = 1 4 x 3 e x4 dx = = 0 Type 2 are integrals where the functions are discontinuous on the interval [a, b].

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46 Ex: Evaluate 3 dx 0 x 1 Why is this integral improper? 1/(x-1) is discontinuous at x = 1. 3 dx 0 x 1 1 dx 0 x 1 = lim 1 = dx 0 = lim t dx t 1 0 x 1 x dx 1 t dx t 1 0 x 1 x 1 = lim t 1 ln x 1 t 0 = lim ln t 1 - ln -1 t 1 = - (It diverges) 1 Ex: Evaluate ln x dx 0 Why is this integral improper? y = ln x has a vertical asymptote at x = 0, so it s discontinuous at x = 0. 1 ln x dx 0 = lim t t lnx dx Integrate by parts: u = ln x dv = dx du = dx x v = x lim t 0 1 t + lnx dx = lim t 0 + [ xlnx 1 t dx ] = lim [1 ln1 t ln t (1 t )] t 0 +

47 = lim [ t ln t + t 1] t 0 + lim [ t ln t] =? Use L Hospital s Rule: t 0 + -t ln t = lnt 1 t lim t 0 + lnt 1 t = lim t t 1 t 2 = lim t 0 + ( t) = 0 So, 1 ln x dx 0 = lim [1 ln1 t ln t (1 t )] = = -1 t 0 + Comparison Theorem Suppose that f and g are continuous functions with f(x) > g(x) > 0 for x > a. a) If f(x)dx a b) If g(x)dx a is convergent, then is divergent, then a a g(x)dx is convergent. f(x)dx is divergent. Ex: Does 1+e x 1 x dx converge or diverge? Since e x > 0, 1+e x > 1 x x and 1 1 x dx diverges because of the p rule, So

48 1+e x 1 x dx diverges.