MATH 1190 Exam 4 (Version 2) Solutions December 1, 2006 S. F. Ellermeyer Name

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1 MATH 90 Exam 4 (Version ) Solutions December, 006 S. F. Ellermeyer Name Instructions. Your work on this exam will be graded according to two criteria: mathematical correctness and clarity of presentation. In other words, you must know what you are doing (mathematically) and you must also express yourself clearly. In particular, write answers to questions using correct notation and using complete sentences where appropriate. Also, you must supply su cient detail in your solutions (relevant calculations, written explanations of why you are doing these calculations, etc.). It is not su cient to just write down an answer with no explanation of how you arrived at that answer. As a rule of thumb, the harder that I have to work to interpret what you are trying to say, the less credit you will get. You may use your calculator on this exam but you may not use any books or notes. This exam contains 7 questions but you only need to do 5 of them. You can choose any 5. Even if you work on more than 5 of them, I will only grade 5 of them. You must choose which 5 you want me to grade. Below, circle the numbers of the 5 questions that you want me to grade For the function f (x) = x + x on the interval [a; b] = [ ; ], (a) Find the slope, m, of the secant line joining the points (a; f (a)) and (b; f (b)). (b) Find a point c in the interval (a; b) such that f 0 (c) = m. (c) Graph f together with its tangent line at the point (c; f (c)). Also include, in this same picture, the secant line joining the points (a; f (a)) and (b; f (b)). Solution: a. Note that f ( ) = ( ) + ( ) = 4 and that f () = () + = 4. The slope of the secant line joining the points ( ; 4) and (; 4) is m = 4 4 ( ) = 0. b. Since f 0 (x) = x +, we see that c = = is the point in the interval ( ; ) at which f 0 (c) = m = 0. Note that f (c) = f ( =) = ( =) + ( =) = 9=4. The equation of the tangent line to the graph of f at the point ( =; 9=4) is y = 9=4. Also, the equation of the secant line joining the points ( ; 4) and (; 4) is y = 4.

2 c.. In parts a, b, and c below, nd the family of antiderivatives of the given function f. State your results using complete sentences of the form The family of antiderivatives of the function f (x) = consists of all functions of the form F (x) =. (a) f (x) = 9 (b) f (x) = e x 4 (c) f (x) = sin (x) cos (x) Answers: a. The family of antiderivatives of the function f (x) = 9 consists of all functions of the form F (x) = 9x + C (where C can be any constant). b. The family of antiderivatives of the function f (x) = e x 4 consists of all functions of the form F (x) = e x 4 + C (where C can be any constant). (d) The family of antiderivatives of the function f (x) = sin (x) cos (x) consists of all functions of the form F (x) = sin (x) + C (where C can be any constant).. Speedometer readings for a motorcycle at second intervals are given in the table below. The speed of the motorcycle is constantly decreasing throughout the entire 60 second time interval. Find lower and upper estimates for the total distance that the motorcycle travelled during this 60 second time interval. (You must show your calculations in detail and include sentences that explain what you are doing and what your conclusions are.) t (seconds) v (feet/sec) Solution: Since the motorcycle s speed is always decreasing, we can get an upper estimate of the total distance travelled by using the speed at the beginning of each

3 second interval (and assuming this to be the constant speed throughout the second interval). Thus a lower estimate of the distance travelled is (0 ft = s) ( s) + (8 ft = s) ( s) + (5 ft = s) ( s) + ( ft = s) ( s) + ( ft = s) ( s) = ; 5 ft and (by similar reasoning) a lower estimate of the distance travelled is (8 ft = s) ( s) + (5 ft = s) ( s) + ( ft = s) ( s) + ( ft = s) ( s) + (8 ft = s) ( s) = ; 68 ft. 4. Use the de nition of the de nite integral Z b a f (x) dx = lim n! R n to show that Z x dx = 7. You must include all details of your work. Note: Even though it is easy to use the Evaluation Theorem (Fundamental Theorem of Calculus, Part II) on this problem, that is not the point of the problem. You must use the de nition of the integral. Solution: We have (for any integer n ) (x) n = n = n and for any integer i (with i n) f (x i ) = x i = f (x i ) (x) n = x i = + i n = + i n + i = + i n n + i n + i n + i n n = n + i n + i n

4 and R n = nx f (x i ) (x) n i= nx = n + i n + i n i= = nx + nx i + nx i n n n i= i= i= = n (n) + n (n + ) + n (n + ) (n + ) n n 6 n + = + + n + n + n 6 n n = n 6 n n Using the fact that lim n! n Z x dx = lim = 0, we obtain n! R n = lim n! n 6 n n = + ( + 0) + ( + 0) ( + 0) 6 = The Evaluation Theorem (Fundamental Theorem of Calculus, Part II) tells us that if the function f is continuous on the interval [a; b] and if F is an antiderivative of f on [a; b], then Z b a f (x) dx = F (b) F (a). For example, since f (x) = x is continuous on [; ] and F (x) = x is an antiderivative of f on [; ], then we conclude that Z x dx = = 7. (a) Find an antiderivative of the function f (x) = x + x + 4x + and then use the Evaluation Theorem to show that Z x + x + 4x + dx = 44. 4

5 (b) The graph of f (x) = x + x + 4x + on the interval [ ; ] is shown below. Explain why it makes sense, in terms of the graph that is shown, that the above de nite integral is equal to a positive number. It would be helpful to shade certain parts of the given picture and to refer to these shadings in your explanation. (Write in complete sentences.) Graph of f (x) = x + x + 4x + Solution: An antiderivative of f (x) = x + x + 4x + is Thus Z x + x + 4x + dx F (x) = 4 x4 + x + x + x. = F () F ( ) = 4 ()4 + () + () + () 4 ( )4 + ( ) + ( ) + ( ) = 44. Referring to the picture below: Since the given integral gives us the area of the black region minus the area of the red region and since the black region is bigger than the red region, it makes sense that the value of the integral is positive. 5

6 6. Let F be the function F (x) = Z 0 The derivative of F is (circle the correct choice): (a) F 0 (x) = tan (x) (b) F 0 (x) = tan (0) (c) F 0 (x) = sec (x) (d) F 0 (x) = sec (0) tan (x) sec (x) (e) None of the above choices is correct. x tan (t) dt. You must explain how you arrive at your answer. The grading of this problem will be as follows Work Points Awarded correct choice not circled 0 correct choice circled but explanation not correct correct choice circled and explanation correct 0 Provide your explanation here: The Fundamental Theorem of Calculus (Part I) tells us that if f is continuous on some interval [a; b], then Z d x f (t) dt = f (x). dx a Since Z 0 tan (t) dt = x Z x 0 tan (t) dt, we can combine the FTC and the Constant Multiple Rule to obtain Z d 0 tan (t) dt = d Z x tan (t) dt = d Z x tan (t) dt = tan (x). dx dx dx x 7. The graph of a function, f, on the interval [ 4; 4] is shown below

7 Graph of f Let g be the function de ned, for all x in the interval [ 4; 4], by g (x) = (a) Complete the following table for g: Z x 4 f (t) dt. x g (x) (b) Determine all intervals on which g is increasing and all intervals on which g is decreasing. (Hint: There is some interval on which g is neither increasing nor decreasing; that is, g remains constant on some interval.) Explain your reasoning (referring to the graph of f in your explanation and writing in complete sentences which do not contain the word it ). (c) Determine all intervals on which g is concave up and all intervals on which g is concave down. (Hint: There are some intervals on which g has no concavity; that is, g is linear on some intervals.) Explain your reasoning (referring to the graph of f in your explanation and writing in complete sentences which do not contain the word it ). (d) Based on your ndings in parts a, b, and c above, draw the graph of g. (e) The Fundamental Theorem of Calculus (Part I) tells us that g 0 (x) = f (x). Does the graph that you drew in part d above support this fact? Explain. 7

8 Solution: Since g 0 (x) = f (x) and f (x) > 0 on the intervals ( 4; :5) and (:5; 4), we conclude that g is increasing on the intervals ( 4; :5) and (:5; 4). Since f (x) < 0 on the intervals ( :5; ) and (; :5), we conclude that g is decreasing on the intervals ( :5; ) and (; :5). Since f (x) = 0 on the interval (; ), we conclude that g remains constant throughout the interval (; ). Since g 00 (x) = f 0 (x) and f 0 (x) < 0 on the intervals ( 4; ) and (; ), we conclude that g is concave down on these intervals. Since f 0 (x) > 0 on the intervals (0; ) and (; 4), we conclude that g is concave up on these intervals. Since f 0 (x) = 0 on the intervals ( ; 0) and (; ), g has no concavity on these intervals. g is linear throughout the intervals ( ; 0) and (; ). The graph of g is shown below

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