Homework_12_Revised_Solutions - Math 1110 Name: Homework 12...

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Unformatted text preview: Math 1110 Name: Homework 12 Due 12/1 or 12/2 in class Please print out these pages. Write your answers to the “text exercises” on separate paper and staple it to these pages. You should include computational details. These problems will be assessed for completeness. Always write neatly and legibly. Please answer the “presentation problems” in the spaces provided. Include full explanations and write your answers in complete, mathematically and grammatically correct sentences. Your answers will be assessed for style and accuracy, and you will be given written feedback on these problems. GRADES Text exercises / 20 Pres probs / 20 Staple /1 Text exercises. Please do the following problems from the book. §5.5 #2, 10, 11, 17, 18, 34, 35, 38, 56, 63, 68, 72, 75, 79 §5.6 #2, 3, 8, 13, 20, 31, 50, 53, 58, 68, 73, 84, 89, 93, 99, 104, 107, 111, 112 Question 1. Please solve the following two problems. ￿ ￿x￿ ￿x￿ (a) tan3 sec2 dx. 4 4 ￿￿ ￿x￿ sec2 x dx 4 Let u = tan , so that du = . 4 4 ￿ ￿ ￿x￿ ￿x￿ ￿x￿ Then tan3 sec2 dx = 4u3 du = u4 + C = tan4 + C, where C is an arbitrary 4 4 4 constant. (b) Find the area of the region enclosed between the line y = −3 and the curve y = 2x − x2. First we find the points where the line and the curve intersect. −3 = 2x − x2 x2 − 2x − 3 = 0 (x − 3)(x + 1) = 0 So the points of intersection occur at x = 3 and x = −1. Now we need to determine whether the line or the curve has greater values on the entire interval (−1, 3). To do this, we evaluate both expressions at point in the interval; for simplicity we pick x = 0. For the line we have a y-value of −3, and for the curve we have a y-value of 0. We conclude that the curve takes on greater values on the entire interval (−1, 3). ￿3 ￿3 ￿3 ￿ 2 13 2 2 2 To find the area, we compute 2x − x − (−3) dx = 2x − x + 3 dx = x − x + 3x￿ = 10 ￿ 3 3 −1 −1 −1 Math 1110 (Fall 2011) HW5 Presentation Problems 2 Question 2. Let f be a function that is differentiable on [a, b]. In Chapter 2 of our text, we defined the average rate of change of f over [a, b] to be f(b) − f(a) b−a and the instantaneous rate of change of f at x to be f ￿ (x). In Chapter 5, we defined the average value of a function. For the new definition of average value to be consistent with the old one, we should have f(b) − f(a) = average value of f ￿ on [a, b]. b−a Is this the case? Give reasons for your answer. ￿b 1 Using the definition in Chapter 5, the average value of on [a, b] is given by f ￿ (x) dx. b−a a f(b) − f(a) Using the Fundamental Theorem of Calculus we see that this is equal to , since f is an b−a ￿ on [a, b]. antiderivative of f f￿ Homework 12 Book Problem Answers: Section 5.5: Section 5.6: ...
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This note was uploaded on 03/01/2012 for the course MATH 1110 at Cornell University (Engineering School).

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