Math 142, Exam 1 Information.

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1 Mth 14, Exm 1 Informtion. 9/14/10, LC 41, 9:30-10:45. Exm 1 will be bsed on: Sections The corresponding ssigned homework problems (see boyln/sccourses/14f10/14.html) At minimum, you need to understnd how to do the homework problems. Lecture notes: 8/19-9/8. Topic List (not necessrily comprehensive): You will need to know: theorems, results, nd definitions from clss. 7.1: Integrtion by prts. To integrte f(x) dx using the prts method, identify u nd dv so tht f(x) dx = u dv. To obtin v from dv, integrte: v = dv. The prts formul is then given by u dv = uv v du. You my hve to pply prts two or more times. Exmple: x cos x dx. Repeted itertion of prts my produce the originl integrl, I. Treting I s vrible, this gives liner eqution in the vrible I. To evlute the originl integrl solve the liner eqution for I. Exmple: e x cos x dx. Further exmples: Suppose tht f(x) = x j (trig. function) (ex. f(x) = x sin 4x), x j (exponentil function) (ex. f(x) = x 3 e x ), or x j (logrithm function) (ex. f(x) = x ln x). Then it is resonble to try to use prts to integrte f(x) dx.

2 7.: Trigonometric integrls. I = sin m x cos n x dx. It is helpful to recll tht sin x + cos x = 1, u = sin x = du = cos x dx, u = cos x = du = sin x dx. (i) Suppose tht m = j + 1 is odd. Let u = cos x; then du = sin x dx. We hve I = (sin x) j cos n x(sin x dx) = (1 cos x) j cos n x(sin x dx) = (1 u ) j u n du. Point: Since we hve n odd power of sin, we cn rewrite the integrl s (function of cos x)(sin x dx). (ii) Suppose tht n = k + 1 is odd. Let u = sin x; then du = cos x dx. We hve I = sin m x(cos x) k (cos x dx) = sin m x(1 sin x) k (cos x dx) = u m (1 u ) k du. Point: Since we hve n odd power of cos, we cn rewrite the integrl s (function of sin x)(cos x dx). (iii) Suppose tht neither m nor n re odd (both re even). In this cse pply suitble trigonometric identity such s one of the following: cos x = 1 sin x = cos x 1 = sin x = sin x = sin x cos x = 1 4 sin x = sin x cos x. 1 cos x, cos x = 1 + cos x ; Exmple: ( ) ( ) cos t sin t cos 4 t dt = sin t cos t cos t dt = 4 sin t dt = 1 ( ) sin t + sin t cos t dt 4 = 1 ( ) 1 cos 4t dt + 1 sin t cos t dt. 4 8

3 I = tn m x sec n x dx. It is helpful to recll tht tn x + 1 = sec x, u = tn x = du = sec x dx, u = sec x = du = sec x tn x dx. (i) Suppose tht n = k > 0 is even. Let u = tn x; then du = sec x dx. We hve I = tn m x(sec x) k 1 (sec x dx) = tn m (tn x + 1) k 1 (sec x dx) = u m (u + 1) k 1 du. Point: Since we hve n even power of sec, we cn rewrite the integrl s (function of tn x)(sec x dx). (ii) Suppose tht n 1 nd m = j is odd. Let u = sec x; then du = sec x tn x dx. We hve I = (tn x) j sec n 1 x(sec x tn x dx) = (sec x 1) j sec n 1 x(sec x tn x dx) = (u 1) j u n 1 du. Point: Since we hve n odd power of tn nd t lest one power of sec, we cn rewrite the integrl s (function of sec x)(sec x tn x dx). (iii) If you cnnot rewrite the integrl s either (function of tn x)(sec x dx) or (function of sec x)(sec x tn x dx), then you must use other methods nd fcts. For exmple, the formuls tn x dx = ln sec x + C, sec x dx = ln sec x + tn x + C re useful. For second exmple, one uses prts to compute sec 3 x dx. If the integrnd does not contin the nturl pirings (sin x, cos x), (tn x, sec x), or (cot x, csc x), it my be possible to use n identity to rewrite the integrnd in terms of these pirings. Exmple: tn x cos x dx = sin x cos x dx. I = sin x cos bx dx, sin x sin bx dx, cos x cos bx dx. One cn dd nd subtrct to obtin sin(x ± y) = sin x cos y ± sin y cos x, cos(x ± y) = cos x cos y sin x sin y sin A cos B = 1 (sin(a B) + sin(a + B)), sin A sin B = 1 (cos(a B) cos(a + B)), cos A cos B = 1 (cos(a B) + cos(a + B)). 3

4 7.3: Trigonometric substitution. Compute I = f(x) dx. In ech of the following cses (with > 0), we mke substitution involving the vrible θ. For this purpose, we drw right tringle with bse ngle thet. Suppose tht f(x) contins x. (i) The right tringle hs hypotenuse ; side opposite θ equl to x; side djcent to θ equl to x. (ii) The substitution is sin θ = x/, with π/ θ π/. Then sin θ = x, so dx = cos θ dθ. Moreover, cos θ = x, so cos θ = x. Suppose tht f(x) contins + x. (i) The right tringle hs hypotenuse + x ; side opposite θ equl to x; side djcent to θ equl to. (ii) The substitution is tn θ = x/, with π/ < θ < π/. Then tn θ = x, so dx = sec θ dθ. Moreover, sec θ = +x, so sec θ = + x. Suppose tht f(x) contins x. (i) The right tringle hs hypotenuse x; side opposite θ equl to x ; side djcent to θ equl to. (ii) The substitution is sec θ = x/ with 0 θ < π/ or π θ < 3π/. Then sec θ = x, so dx = sec θ tn θ dθ. Moreover, tn θ = x, so tn θ = x. More generlly, suppose tht f(x) contins rx + sx + t. To use the bove substitutions, complete the squre to convert rx + sx + t to the form x ± or x. 4

5 7.4: Prtil frctions. Q(x) fctors s product of distinct, liner fctors: Q(x) = ( 1 x + b 1 ) ( j x + b j ). (i) Write P (x) Q(x) = A 1 1 x + b A j. j x + b j (ii) Cler denomintors, nd solve for A 1,... A j. This cn be done severl wys: Compre coefficients of powers of x, giving liner system of equtions. Substitute specific vlues for x; specificlly, choose vlues of x which hve Q(x) = 0. Q(x) fctors s product of liner fctors, some of which re repeted. (i) Corresponding to repeted fctor (x + b) r of Q(x), we hve the terms A 1 x + b + A (x + b) + A r (x + b) r in the prtil frction decomposition of P (x)/q(x). (ii) Cler denomintors nd solve for the constnts in the decomposition s before. Q(x) contins irreducible qudrtic fctors, none of which is repeted. (i) Suppose tht Q(x) hs fctor x + bx + c with b 4c < 0 (so tht it is irreducible). Corresponding to this fctor, we hve the term Ax + B x + bx + c in the prtil decomposition of P (x)/q(x). (ii) Cler denomintors nd solve for constnts s before. (iii) The integrl my be useful. dx x + = 1 ( x ) tn 1 + C Q(x) contins repeted irreducible qudrtic fctor. (i) Suppose tht Q(x) hs the fctor (x + bx + c) r with b 4c < 0 (so tht x + bx + c is irreducible). Corresponding to this fctor, we hve the term A 1 x + B 1 x + bx + c + A x + B (x + bx + c) + + A r x + B r (x + bx + c) r in the prtil frction decomposition of P (x)/q(x). (ii) We cler denomintors nd solve for the constnts s before. Rtionlizing substitutions. If n integrnd contins n g(x), it my be resonble to try the substitution u = n g(x) (if ll else fils). 5

6 7.5: Integrtion strtegy. Other useful integrls (in ddition to those bove): x dx = x ln + C, dx ( x ) x = sin 1 + C. Do not forget the following: Simplify the integrnd if possible. Look for obvious u-substitutions. 6