Math 2260 Exam #2 Solutions. Answer: The plan is to use integration by parts with u = 2x and dv = cos(3x) dx: dv = cos(3x) dx

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1 Math 6 Eam # Solutions. Evaluate the indefinite integral cos( d. Answer: The plan is to use integration by parts with u = and dv = cos( d: u = du = d dv = cos( d v = sin(. Then the above integral is equal to sin( sin( d = sin( + cos( + C. 9. Shown below is the graph of the function f( = of the shaded region? Simplify as much as possible. 5 between = and =. What is the area 4 5 Answer: Notice, first of all, that since the region lies entirely below the ais, the area of the region is times the integral of the function from to. In other words, the goal is to compute 5 d = 5 d after bringing the minus sign inside the integral. Now, this is a classic setup for using partial fractions. Notice that the denominator factors as so we write the fraction as the sum of fractions Clearing denominators yields the equation 5 = ( + (, 5 = ( + ( = A + + B. = A( + B( + = (A + B + (B A. Equating like terms then yields the system of equations = A + B = B A,

2 which we now solve for A and B. From the first equation, we know that A = B; substituting into the second equation yields = B ( B = B 6 + 6B = 7B 6. Therefore, after adding 6 to both sides, we see that 7 = 7B, and so B =. In turn, this means that A = B = ( = 4. Therefore, 5 d = ( 4 + d [ ln + ln ] = = ( ln 5 ln ( ln ln = ln 5 + ln. Now, we can simplify this using the rules of logarithms to Therefore, the area of the region is ln 75. ln 5 + ln = ln(5 + ln = ln( 5 = ln 75.. Does the improper integral converge or diverge? ( + d Answer: There are a variety of ways to approach this problem. Here are three: Direct Comparison: Notice that + > for any. Then so long as > (which is the only region we care about, ( + >, and so Moreover, Therefore, the improper integral ( + <. d b b b d d b ( [ ] b b b =. ( b + d converges. Since Comparison Test implies that the improper integral (+ < for all >, the Direct (+ d converges as well.

3 Limit Comparison: Again, we want to compare to, so compute the limit since lim lim (+ ( ( + = + + +/ = and =. Since < < and we already saw above that d converges, the Limit Comparison Test implies that (+ d also converges. Computing the Integral: We could also just compute the value of this definite integral: 4. Evaluate the definite integral d ( + b b b ( + d ( + d b ( [( ] + b b b = 8. d. ( (b Answer: Notice, first of all, that this is an improper integral since the denominator goes to zero as goes to. Therefore, b d d. b Now, trig substitution and u-substitution both look unpromising, so integration by parts seems like the best bet. Since I want to choose a u that gets nicer when it gets differentiated, u = is a much better choice than u =. Therefore, the good choices for u and dv are Hence, the above integral is equal to ( [ ] b b + lim b u = dv = d = ( / d du = d v = ( / =. d b b = 4. ( [ ] b ( [ b b + ] [ 4 ( / ] b [ 4 ( b/ 4 5. A research group publishes a paper claiming that the world wombat population is governed by the differential equation (4 t / dw = W, where W = W (t is the wombat population after t years. ]

4 (a If there are currently wombats in the world, what is the wombat population as a function of time according to this group s model? Answer: The goal is to solve the differential equation for W, so first we separate variables: dw W =. (4 t / Now, integrate both sides: ln W = dw W = (4 t / (4 t / (I don t have to worry about absolute value signs in the natural log because the wombat population can never be negative. Now, to evaluate the integral on the right hand side, the most promising approach is to use the trig substitution t = sin θ, so that = cos θ dθ and we can re-write as 4 cos θ dθ (4 4 sin cos θ dθ = θ / 4 / ( sin θ / 4 cos θ dθ = 8(cos θ / cos θ dθ = = = cos θ dθ cos θ sec θ dθ = tan θ + C. Now, we have to translate from θ back into the variable t. Since we made the substitution t = sin θ, we know that sin θ = t, and so we can realize θ as the indicated angle in the following triangle: t Θ Therefore, and so we conclude that tan θ = t 4 t, ln W = (4 t ln W = t + C. 4 t / 4

5 Now, we know that the wombat population at time t = is equal to, so substituting t = and W = into the above equation yields so we know that C = ln and we know that ln( = + C = C, 4 ln W = t + ln. 4 t Now, eponentiating both sides gives us the wombat population as a function of t: W = e t +ln 4 t W = e ln e t 4 t W = e t 4 t (b What does this model say is going to happen to the wombat population in 4 years? Answer: Nothing very sensible. As t goes to the model predicts that the wombat population is going to infinity. Past that point, the model predicts an imaginary number of wombats, since 4 t is an imaginary number when t >. While it is possible that this model is making a rather profound philosophical or psychological point ( any wombat you happen meet four years from now must be a product of a deluded imagination, it s more likely that it s just not a very good model. 5