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Unformatted text preview: Version 044 – EXAM 1 – Zheng – (56555) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points On the interval [0 , 6] the continuous func tion f has graph 2 4 6 2 4 6 Determine the value of the definite integral I = integraldisplay 6 f ( x ) dx . 1. I = 57 2 2. I = 51 2 3. I = 55 2 4. I = 53 2 5. I = 49 2 correct Explanation: Since the graph of f lies above the xaxis everywhere on [0 , 6], i.e. , f ( x ) > 0 for all ≤ x ≤ 6, the value of I is just the area of the region ‘trapped under’ the graph of f on [0 , 6]. Furthermore, this area can be calculated explicitly because f is piecewise linear. In fact I is the sum I = A 1 + A 2 + A 3 + A 4 + A 5 + A 6 of the areas of the ‘hashed’ trapezoidal regions in 2 4 6 2 4 6 1 1 2 3 4 5 6 2 4 6 2 4 6 But A 1 = 1 2 parenleftBig 2 + 6 parenrightBig , A 2 = 1 2 parenleftBig 6 + 5 parenrightBig , A 3 = 1 2 parenleftBig 5 + 3 parenrightBig , A 4 = 1 2 parenleftBig 3 + 5 parenrightBig , A 5 = 1 2 parenleftBig 5 + 2 parenrightBig , A 6 = 1 2 parenleftBig 2 + 5 parenrightBig . Thus the area of the region trapped under the graph of f on the interval [0 , 6] is given by Area = 2 2 + 6 + 5 + 3 + 5 + 2 + 5 2 = 49 2 . Consequently, I = Area = 49 2 . 002 10.0 points Estimate the area, A , under the graph of f ( x ) = 5 x on [1 , 5] by dividing [1 , 5] into four equal subintervals and using right endpoints. Version 044 – EXAM 1 – Zheng – (56555) 2 1. A ≈ 25 4 2. A ≈ 77 12 correct 3. A ≈ 37 6 4. A ≈ 13 2 5. A ≈ 19 3 Explanation: With four equal subintervals and right end points as sample points, A ≈ braceleftBig f (2) + f (3) + f (4) + f (5) bracerightBig 1 since x i = x ∗ i = i + 1. Consequently, A ≈ 5 2 + 5 3 + 5 4 + 1 = 77 12 . 003 10.0 points The graph of f is shown in the figure 2 4 6 8 10 2 4 6 2 If the function g is defined by g ( x ) = integraldisplay x 4 f ( t ) dt, for what value of x does g ( x ) have a maxi mum? 1. x = 8 correct 2. not enough information given 3. x = 10 4. x = 4 5. x = 9 6. x = 5 . 5 Explanation: By the Fundamental theorem of calculus, if g ( x ) = integraldisplay x 4 f ( t ) dt, then g ′ ( x ) = f ( x ). Thus the critical points of g occur at the zeros of f , i.e. , at the x intercepts of the graph of f . To determine which of these gives a local maximum of g we use the sign chart g ′ + − 4 8 10 for g ′ . This shows that the maximum value of g occurs at x = 8 since the sign of g ′ changes from positive to negative at x = 8. 004 10.0 points Determine F ′ ( x ) when F ( x ) = integraldisplay √ x 5 8 sin t t dt . 1. F ′ ( x ) = 4 sin x x 2. F ′ ( x ) = − 8 cos x √ x 3. F ′ ( x ) = − 4 sin( √ x ) √ x 4. F ′ ( x ) = 4 cos( √ x ) x Version 044 – EXAM 1 – Zheng – (56555) 3 5. F ′ ( x ) = 8 sin x √ x 6. F ′ ( x ) = 4 sin( √ x ) x correct 7. F ′ ( x ) = − 8 cos x x 8. F ′ ( x ) = − 8 cos( √ x )...
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 Spring '10
 ZHENG
 Math, Calculus, Fundamental Theorem Of Calculus, dx, D10

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