# y=5 y=1+x 2 AP Calculus Chapter 5 Testbank Part I. Multiple-Choice Questions

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1 AP Calculus Chapter 5 Testbank Part I. Multiple-Choice Questions. Which of the following integrals correctly corresponds to the area of the shaded region in the figure to the right? (A) (B) (C) (D) (E) ( 4) d (4 ) d ( 4) d ( + 4) d (4 ) d y= y y= d t 4 t d dt = (A) t 4 t (B) 4 (C) 4 4 (D) (E) 4

2 . π 4 sin d + π 4 cos d = (A) (B) (C) (D) (E) 4. If f() d = A and 5 f() d = B, then 5 f() d = (A) A + B (B) A B (C) (D) B A (E) 5. The average value of the function f() = ( ) on the interval from = to = 5 is (A) 6 (B) 6 (C) 64 (D) 66 (E) The average value of the function f() = ln on the interval [, 4] is (A).4 (B).4 (C).59 (D).48 (E) d d cos t dt = (A) sin (B) sin (C) cos (D) sin (E) cos

3 8. If the definite integral the Trapezoid Rule with n = 4, the error is (A) (B) 7 (C) ( + ) d is approimated by using (D) 65 6 (E) ln d = ln (A) + C ln (B) + C (C) ln + C (D) ln + C (E) ln + C. d = (A) C (B) C (C) 5 + C (D) + C (E) 5 + C. The average value of f() = e 4 on the interval [ 4, 4 ] is (A).7 (B).545 (C).9 (D).8 (E) 4.6

4 . Find the distance traveled (to three decimal places) in the first four seconds, for a particle whose velocity is given by v(t) = 7e t, where t stands for time. (A).976 (B) 6.4 (C) 6.59 (D).7 (E) 7.. Approimate to three decimal places. sin d using the Trapezoid Rule with n = 4 (A).77 (B).7 (C).555 (D).9 (E).9 4. Use the Trapezoid Rule with n = 4 to approimate the area between the curve y = and the -ais over the interval [, 4]. (A) 5.66 (B) (C) 6.6 (D).56 (E) The graph of the function f shown below consists of a semicircle and a straight line segment. Then y=f() 4 y 4 f() d = (A) π + (B) (π ) (C) π + (D) (π + ) (E) π +

5 6. d 5 cos t dt = d (A) 5 cos 5 cos (B) 5 sin 5 sin (C) cos 5 cos (D) sin 5 sin (E) 5 cos 5 sin 7. d d 5 cos t dt = (A) cos 5 5 cos (B) 5 cos cos 5 (C) cos 5 cos 5 (D) 5 cos 5 cos (E) 5 cos 5 + cos 8. Given that the function f is continuous on the interval [, ), and that f(t) dt =, then f (t) dt = (A) (B) /4 (C) 4 ln (D) 4 (ln ) (E) Can t be determined 9. Given that the function f is continuous on the interval [, ), and that f(t) dt =, then f (t) dt = (A) (B) 6 (C) 4 (D) (E)

6 Part II. Free-Response Questions. On the graph below, shade in the appropriate area indicated by the integral f()d. In the graph to the right, indicate the rectangles that would used in computing a RRAM (right rectangular approimation) to f()d using five rectangles, each of width. (Don t try to compute anything; just draw the relevant picture.) Does it appear the result gives an overestimate or an underestimate of the true area?

7 4 y y=f() The graph of the function f, consisting of three line segments, is given above. Let g() = f(t) dt. (a) Compute g(4) and g( ). g(4) = g( ) = f(t) dt = 4 f(t) dt = 6. f(t) dt = 5 ; (b) Find the instantaneous rate of change of g, with respect to, at =. By the Fundamental Theorem of Calculus, we have g () = f() = 4. (c) Find the minimum value of g on the closed interval [, 4]. Justify your answer. We have g () = f() = when = or when =. Comparing with endpoints we have g( ) = 6 (from part (a)), g() = f(t) dt =, and g(4) = 5 (from part (a)). Therefore, the minimum value of g on the closed interval [, 4] is 6 (which happens at = ).

8 . Let f be the function depicted in the graph to the right. Arrange the following numbers in increasing order: y y=f() f() d, f() d, f () d. f () d < f() d < f() d. 4. Assume that the average value of a function f over the interval [, ] is.5. Compute f()d. f()d is equal to 4 times its average value over this interval. Therefore, f()d = 4.5 =. 5. Epress the area of a circle of radius r as a definite integral. r r This is clearly r d = 4 r d. r 6. Find an antiderivative for each function below: (a) f() = + f() d = + + C. (b) g() = e g() d = + e + C. (c) v(t) = gt + v (here, v is just a constant) v(t) dt = gt + v t + C

9 (d) h() = /, where < (This is not hard, you just need to be careful!) h() d = ln( ) + C (This says that, in d general, = ln + C.) (e) k() = ( + )/ k() d = ln + + C. (f) θ() = /( + ) θ() d = ln + + C. 7. For each definite integral below, sketch the area represented. (Don t try to compute anything, just draw the relevant picture.) y (a) (b) e π π ln d cos d 4 Equation : y=ln y π π/ π/ π Equation : y=cos y (c) (d) t( t) dt Equation : y=( ) 4 f() d, where { if f() = if >. 4 y Equation : y=² Equation : y= Equation : y=² (( ))/ (( )) Equation 4: y=( ) (( )( ))/ (( )( ))

10 8. Compute the definite integrals (a) (9 )d = (9 ) d = 9 = 7 9 = 8. π (b) sin d = cos π = (c) 4 e d = e 4 = e 4. d (d) = sin = π/ d (e) + = tan = π/4 (f) d + = ln( + ) = ln 9. Solve for a in each of the below: a (a) ( )d = ( ) = a = a a a = a d (b) + = π 4 = a tan = tan a a = (c) a d = = ln a a = ln a a = e.. Suppose that f is an even function and that where a >. Compute a a f()d. Since f is an even function, we have a a f()d = a a f()d = =. f()d =,. Suppose that f is an odd function and that a >. Compute a a f()d. This is =. To prove this, note that a f() d =

11 a a a a f() d u= = f()d = f()d = a a a a f( u) du = f(u) du. That is to say, f() d, and so f()d+ a a a f()d = f() d+ f() d =.. Solve the differential equations: (a) y =, y() =. y = ( ) d = + C. Also, = y() = + C C =. Therefore, y =. (b) y = e, y() =. y = ( e ) d = + e + C. Also, = y() = + e + C C =. Therefore, y = + e. (c) ( + )y d =, y() =. y = = ln + + C. Also, + = y() = ln + C C =. Therefore, y = ln +. (d) y = y, y() =. Write this as dy = y, which separates as d dy y = d ln y = + C. Therefore, y = e +C ; it s better to write this as y = Ke, where K is a (possibly negative) arbitrary constant. From = y() = Ke = K, get K = and so y = e. (e) y = y, y() =. Argue almost eactly as in (d) to arrive at the solution y = e.. Suppose that you know that f () =. What can you say about f()? Write down the most general form that f() could have. Clearly the most general solution must be of the form f() = A + B, where A and B are arbitrary constants. 4. Suppose that the velocity of a particle is given by v(t) = + sin t, t, where v is given in units of cm/sec. How far has this particle traveled after sec?

12 We know that the TOTAL DISTANCE the particle travels is given by the integral s(t) dt, where s(t) is the speed of the particle. Net, since s(t) = v(t), we have that the total distance traveled is v(t) dt = +sin t dt = (+sin t) dt = t cos t cos.84 cm. = 5. Assume that a water pump is pumping water into a large tank at a variable rate: after t hours, water is being pumped at a rate of v(t) = 5t + t gallons/hour. (a) Write down the definite integral that epresses how much water has been pumped into the vessel after 4 hours. (Don t 4 5t dt compute this.) This is clearly + t. (b) Compute how much water is in the tank after 4 hours, assuming that the tank was empty to begin with and that F (t) = 5 (t ln( + t)) is an antiderivative for v(t). The amount pumped into the tank over the first 4 hours is 4 5t dt + t = 5(t ln(+t)) 4 = 5(4 ln 5) 9.6 gallons.

13 6. The rate at which people enter an amusement park on a given day is modeled by the function E defined by E(t) = 56 t 4t + 5. the rate at which people leave the same amusement park on the same day is modeled by the function L defined by L(t) = 989 t 8t + 7. Both E(t) and L(t) are measured in people per hour and time t is measured in hours after midnight. These functions are valid for 9 t, the hours during which the park is open. At time t = 9, there are no people in the park. (a) How many people have entered the park by 5: P.M. (t = 7)? Round your answer to the nearest whole number. This is the integral of the rate of people entering the park: dt, 746 people, where the approimation was done via numerical integration on a TI-84 calcula- t 4t + 5 tor. (b) The price of admission to the park is \$5 until 5: P.M. (t = 7). After 5: P.M., the price of admission to the park is \$. How many dollars are collected from admission to the park on the given day? Round your answer to the nearest whole number. The total in receipts would be the sum \$5 7 7 E(t) dt + \$ 56 dt = 5 t 4t \$ = \$5, \$, 56 = \$7, E(t) dt 56 dt t 4t + 5 E(t) dt =

14 (c) Let H(t) = t 9 (E() L()) d for 9 t. The value of H(7) to the nearest whole number is 75. Find the value of H (7) and eplain the meaning of H(7) and H (7) in the contet of the amusement park. The meaning of H(7) is the total number of people having arrived in the park between 9 a.m. and 5 p.m. less the total number of people having left the park during this same time. In other words, H(7) is the total number of people in the park at 5 p.m. H (7) = E(7) L(7) which is the net rate of people (in people per hour) entering the park at 5 p.m. (d) At what time, t, for 9 t, does the model predict that the number of people in the park is a maimum? As H(t) gives the total number of people in the park at time t, the maimum number of people in the park will happen when H (t) =, i.e., when E() = L(). Using a graphics calculator, one infers that this happens when t 6.7, i.e. at roughly 4: p.m.

15 (, ) Graph of f (, ) 7. The graph of the function f shown above consists of two line segments. Let g be the funtion given by g() = (a) Find g( ), g ( ), and g ( ). g( ) =.5. g ( ) = f( ) = ; g ( ) = f ( ) =. f(t) dt. f(t) dt = (b) Over which interval(s) within (, ) is g increasing? Eplain your reasoning. g is increasing where g () >, i.e., where f() >. This latter condition is satisfied on the interval < <. (c) Over which interval(s) within (, ) is the graph of g concave down? Eplain your reasoning. The graph of g is concave down where g () <, i.e., where f () <. This happens on the interval < <. (d) On the aes provided, sketch the graph of g on the closed interval [, ]. f(t) dt =

16 t (minutes) R(t) (gallons per minute) The rate of fuel consumption, in gallons per minute, recorded during an airplane flight is given by a twice-differentiable and strictly increasing function R of t. A table of selected values of R(t), for the time interval t 9 minutes, is shown above. (a) Use data from the table to find an approimation for R (45). Show the computations that lead to your answer. Indicate units of measure. R R(5) R(4) (45) = =.5 (gallons/min ). (b) The rate of fuel consumption is increasing fastest at time t = 45. What is the value of R (45)? Eplain your reasoning. For the rate of fuel consumption to be increasing at its fastest, we must have that R (t) is a maimum. This happens (since R is twice-differentiable) when R (t) = ; therefore, we infer that R (45) =. (c) Approimate the value of 9 R(t) dt using a left Riemann sum with the five subintervals indicated by the data in the table. Is this numerical approimation less than the value of 9 9 R(t) dt? Eplain your reasoning. We have R(t) dt =, 7 gallons. Since R is a strictly-increasing function of t, we conclude that the left Riemann sum must give an underestimate of the total fuel consumption over the first 9 minutes.

17 b (d) For < b 9 minutes, eplain the meaning of in terms of fuel consumption for the plane. meaning of b b R(t) dt Eplain the R(t) dt in terms of fuel consumption for this plane. Indicate units of measure in both answers. b R(t) dt is the total fuel consumption (in gallons) over the first b minutes, where < b 9. The integral R(t) dt is the average fuel consumption (in gallons per minute) over the first b b minutes. b

18 y y=8- y=f() R S 9. Let f be the function given by f() = 4, and let L be the line y = 8, where L is tangent to the graph of f. Let R be the region bounded by the graph of f and the -ais, and let S be the region bounded by the graph of f, the line l, and the -ais, as shown above. (a) Show that L is tangent to the graph of y = f() at the point =. We have that f () = 8 = = 4 7 =, and so the line has the correct slope. Also, f() = 4 = 9 = 8, and so the line and the graph of y = f() both intersect where =. These two facts imply that the given straight line is tangent to the graph of y = f() at the indicated point. (b) Find the area of S. Area (S) = 6 = 7 = 7 4 (8 ) d (4 ) d ( 4 ) ( ) ( ) 4 = 95.

19 4. Traffic flow is defined as the rate at which cars pass through an intersection, measured in cars per minute. The traffic flow at a particular intersection is modeled by the function F defined by F (t) = sin ( ) t for t, where F (t) is measured in cars per minute and t is measured in minutes. (a) To the nearest whole number, how many cars pass through the intersection over the -minute period? The total number of cars passing through the given intersection over the -minute period is the integral: F (t) dt = ( sin ( )) t dt, 474 cars. (b) Is the traffic flow increasing or decreasing at t = 7? Give a reason for your answer. The traffic flow is increasing or decreasing at t = 7 according as to whether F (7) > or F (7) <. However, ( ) 7 F (7) = cos.87 <, and so the traffic flow is decreasing at t = 7. (c) What is the average value of the traffic flow over the time interval t 5? Indicate units of measure. The average traffic flow over the time interval t 5 is 5 5 ( sin ( t )) dt 8.9 cars/min (d) What is the average rate of change of the traffic flow over the time interval t 5? Indicate units of measure. The average rate of change of the traffic flow over the time interval t 5 is the difference quotient F (5) F () 5 = 5 ( 4 sin ( ) 5 4 sin ( )).5 cars/min

20 4 y y=f() Let f be the function defined on the closed interval [, 7]. The graph of f, consisting of four line segments, is shown above. Let g be the function given by g() = (a) Find g(), g (), and g (). g() = f() =, g () = f () =. f(t) dt. f(t) dt =, g () = (b) Find the average rate of change of g on the interval g() g(). This is = ( 4) = 7. (c) For how many values c, where < c <, is g (c) equal to the average rate found in part (b)? Eplain your reasoning. As g (c) = f(c), we are seeking the number of solutions of f(c) = 7 on the interval < c <. An inspection of the graph reveals that there are two such solutions. (d) Find the -coordinate of each point of inflection of the graph of g on the interval < < 7. Justify your answer. There are no solutions of g () = f () = on the interval < < 7. However, there are changes in the concavity of the graph of y = g() at = and at = 5 since the sign of f () changes sign at these two values of.

21 y y=f() R - y=g() 4. Let f and g be the functions given by f() = ( ) and g() = ( ) for. The graphs of f and g are shown in the figure above. Find the area of the region R enclosed by the graphs of f and g. This area is given by the integral (f() g()) d = = (( ) ( ) ) d ( 6 ) 5 5/ + / = = 7 5.

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