APPM 1360 Midterm: appm1360spring2016exam1_sol

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turquoisegnu128

31 Jan 2019

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Spring 2016: (30 points, 10 points each) evaluate the following integrals. d (a) z (b) z (c) z sin . 1 dt cos2 + cos 2 t2 8t + 20 (x2 1)3/2 dx. Solution: (a) let u = cos , du = sin d . d =z cos2 + cos 2 sin . A(u 1) + b(u + 2) = 1 (a + b)u + ( a + 2b) = 1. Z sin cos2 + cos 2. A + b = 0, a + 2b = 1 a = d =z (cid:18) 3(u 1)(cid:19) du = ln|cos 1| + c = ln|cos + 2| . + c cos + 2 ln(cid:12)(cid:12)(cid:12)(cid:12) cos 1(cid:12)(cid:12)(cid:12)(cid:12) tan 1(cid:16) x. 1 a a(cid:17) + c. (b) complete the square in the denominator, then use the formulaz dx x2 + a2 = Let u = t 4, du = dt. 2 (cid:19) + c (c) this problem is equivalent to webassign hw 6. 2 #3.

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tealskunk114

[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

APPM 1360 Midterm: appm1360spring2016exam1_sol

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