Be sure this exam has 8 pages including the cover The University of British Columbia MATH 103 Midterm Exam II Mar 14, 2012

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1 Be sure this exam has 8 pages including the cover The University of British Columbia MATH Midterm Exam II Mar 4, 22 Family Name Student Number Given Name Signature Section Number This exam consists of 4 questions. No notes nor calculators. Note the number of marks for each question. Use your time wisely. Problem max score score total 5 The following formula may be helpful: ln(t) dt = t ln(t) t + C.

2 Feb 8, 22 Math Midterm Page 2 of 8 (9 points). Answer the following multiple choice questions. Check your answer very carefully. Your answer will be marked right or wrong (work will not be considered for this problem). ( points) (a) The value of the integral xe.5x2 dx is A. e.5 B. 2e 2 C..5e D. E. 2 Answer: A ( points) (b) The value of the integral π x sin(2x)dx is A. π/2 B. 2 C. π D. 2 E. π/2 F. π Answer: E ( points) (c) The integral x2 x 2 dx is equal to A. π 2 B. π 2 C. π 2 Answer: C +cos(2t) 4 dt sin2 (t) cos(t)dt cos(4t) 8 dt D. π 2 sin2 (t)( cos(2t))dt (a) (b) (c)

3 Feb 8, 22 Math Midterm Page of 8 ( points) 2. Compute the following integrals and simplify your answer to the best possible. Show in detail how you arrive at your answer (work will be considered for this problem). (5 points) (a) The value of the integral 5 2x 4 dx is 2x 2 7x+6 (a) (5 points) (b) The value of the integral (ln x)2 dx is (b) Solution: Solution. Factoring the denominator we have 2x 2 7x + 6 = (2x )(x 2). We need to use Partial Fractions. Then we have that This gives the linear system 2x 4 (2x )(x 2) = A 2x + B x 2 2x 4 = A(x 2) + B(2x ) 2x 4 = (A + 2B)x 2A B Therefore A + 2B = 2 2A B = 4 A = 2( B) = B = 4 2[2( B)] = A = 2 B = 5 2x 4 5 2x 2 7x + 6 dx = 2 2x dx Using substitution, let u = 2x so that du = 2dx or dx = du. The limits of 2 integration become x = = u = and x = 5 = u = 7 then we have 5 Solution x dx = 2 u ( 2 ) du = ln u 7 = ln(7) ln() = ln We could have also used the fact that the numerator will simplify the denominator into the same integral we just solved. We have 5 ( ) 2 7 dx = ln. (2x ) 2x 4 5 (2x )(x 2) dx = 2(x 2) 5 (2x )(x 2) dx = ( 7 ). (b) The value of the integral (ln x)2 dx is

4 Feb 8, 22 Math Midterm Page 4 of 8 Solution: We need to use integration by parts. Using the fact that (ln x) 2 can be written as ()(ln x) 2, we let u = (ln x) 2 dv = du = 2 ln x ( x) dx v = x Then, using the formula b udv = a uv b a b vdu, we have a (ln x) 2 dx = x(ln x) 2 e 2 x ln x dx x = e(ln e) 2 ( ln()) 2 2 = e 2 ln x dx ln x dx Using ln e = and ln() =. We need to use integration by parts again. Using the fact that ln x can be written as () ln x, we let u = ln x dv = du = dx x v = x Then e [ ( ) ] ln x dx = e 2 x ln x e x dx x [ ] = e 2 e ln e ln() ()dx = e 2e + 2 x e = e + 2(e ) = e 2.

5 Feb 8, 22 Math Midterm Page 5 of 8 (8 points). The function f(x) = e (x2), for x, when rotated about the y-axis forms an upward-facing bowl. (2 points) (a) Make a sketch of this function and the resulting bowl. (6 points) (b) Find the volume V of the bowl. Solution: (a) (b) To calculate the volume of the resulting bowl, we stack disc along the y-axis. A disc has radius r(y) = ln(y) and width dy. We need to integrate along the y-axis from y = f() = to y = f() = e. Hence V = = π π(r(y)) 2 dy = π ( ln(y) ) 2 dy ln(y) dy = π (y ln(y) y) e = π(e ln(e) e) π( ln() ) = π. We use the hint provided on the front page to find the antiderivative of ln(y). Alternative solution: Shell method. It is also possible to calculate volume using the shell method. Here the shells have height h(x) = e e x2, radius x and width dx. We need to integrate along the x-axis

6 Feb 8, 22 Math Midterm Page 6 of 8 from x = to x =. Hence V = 2πxh(x) dx = π = 2πe x dx π = 2πe x2 2 πe x2 2x(e e x2 ) dx 2xe x2 dx = (πe ) π(e ) = π.

7 Feb 8, 22 Math Midterm Page 7 of 8 (8 points) 4. Consider a rare disease whose prevalence in the population is p, i.e., each person has a probability p of being infected. The population has N individuals. ( points) (a) Let I be the number of infected people. What is its mean, I? ( points) (b) What is the variance of the number of infected people? (2 points) (c) What is the probability that 2 people are infected? Solution: Infection is a Bernoulli process, hence I = Np, Var I = Np( p), Prob (I = 2) = C(N, 2)p 2 ( p) N 2 N(N ) = p 2 ( p) N 2. 2 Suppose that the N people are tested for the disease. Assume that the test has a false positive rate of r, i.e., someone who is not infected still has probability r of testing positive. If someone is infected, then the test always gives a positive result. (2 points) (d) What is the probability that a randomly chosen individual tests positive? Solution: Let P be the event that a person tests positive. We are given the probabilities Prob(P I) = and Prob(P not I) = r, hence we use the fact that I and (not I) are mutually exclusive to write Prob(P ) = Prob(P and I) + Prob(P and noti) = Prob(I)Prob(P I) + Prob( not I)Prob(P not I) = p + ( p)r ( points) (e) Let X be the number of positive results in the test. Find its mean X. Solution: Testing positive is a Bernoulli process, hence X = N Prob(P ) = N(p + ( p)r) ( points) (f) Evaluate your answers for I and X for p = /, r = /2, and N =. You may use the approximation 999/. Solution: I = =, ( X = ) 2 = 5. ( + ) 2 2 = 2 Note the huge disparity between these numbers. Only a small fraction /5 of the people who test positive actually have the disease! Note: one could use / = 999/ to obtain X 5, which is correct to the same level of approximation.

8 Feb 8, 22 Math Midterm Page 8 of 8 This is an extra page for answers.

Integration by Parts

Calculus 2 Lia Vas Integration by Parts Using integration by parts one transforms an integral of a product of two functions into a simpler integral. Divide the initial function into two parts called u