MATH116 Lecture 15: M116F15_P11sol

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cyanimpala400

9 Dec 2015

School

University of Waterloo

Department

Mathematics

Course

MATH116

Professor

Fiona Dunbar

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Practice solutions 11: find the area enclosed between the curves y = x and y = 5x x2. Solution: first we nd the point(s) of intersection: It helps to graph each function to see which is larger where: We can see that the area enclosed between them is on the interval [0, 4], and y = 5x x2 is larger. Thus the area enclosed between them is (cid:19) (0 0) (cid:90) 4 (5x x2) x dx = 3: consider the function f (x) = sin(2x). (a) find the average value of f on the interval [0, . 4 ]. (b) find the average value of f on the interval [0, 4 ]. 8 (cos(8 ) cos(0)) = 0: find the volumes of the following solids generated by rotating the region bounded by the given curves about the speci ed line. Solution: (a) the region is sketched as follows: and the solid is sketched as follows:

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Please help answer circled questions U substitution-exponential and logarithmic form.

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[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) >

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MATH116 Lecture 15: M116F15_P11sol

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