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exam1 - Version 085 EXAM 1 BERG(56525 This print-out should...

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Version 085 – EXAM 1 – BERG – (56525) 1 This print-out should have 19 questions. Multiple-choice 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 -1 0 1 2 3 4 5 6 2 4 6 2 4 6 Determine the value of the definite integral I = integraldisplay 6 0 f ( x ) dx . 1. I = 22 2. I = 26 3. I = 25 4. I = 23 5. I = 24 correct Explanation: Since the graph of f lies above the x -axis everywhere on [0 , 6], i.e. , f ( x ) > 0 for all 0 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 0 1 2 3 4 5 6 1 0 1 2 3 4 5 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 + 4 parenrightBig , A 5 = 1 2 parenleftBig 4 + 2 parenrightBig , A 6 = 1 2 parenleftBig 2 + 6 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 + 4 + 2 + 6 2 = 24 . Consequently, I = Area = 24 . 002 10.0 points Estimate the area, A , under the graph of f ( x ) = 4 x on [1 , 5] by dividing [1 , 5] into four equal subintervals and using right endpoints. 1. A 76 15
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Version 085 – EXAM 1 – BERG – (56525) 2 2. A 26 5 3. A 74 15 4. A 77 15 correct 5. A 5 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 2 + 4 3 + 1 + 4 5 = 77 15 . 003 10.0 points The graph of f is shown in the figure -1 0 1 2 3 4 5 6 7 8 9 10 11 2 4 6 8 10 2 4 6 2 If the function g is defined by g ( x ) = integraldisplay x 1 f ( t ) dt, for what value of x does g ( x ) have a maxi- mum? 1. x = 5 correct 2. not enough information given 3. x = 1 4. x = 7 5. x = 6 6. x = 2 . 5 Explanation: By the Fundamental theorem of calculus, if g ( x ) = integraldisplay x 1 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 + 1 5 7 for g . This shows that the maximum value of g occurs at x = 5 since the sign of g changes from positive to negative at x = 5. 004 10.0 points Determine F ( x ) when F ( x ) = integraldisplay x 5 8 cos t t dt . 1. F ( x ) = 8 cos x x 2. F ( x ) = 8 sin x x 3. F ( x ) = 4 sin( x ) x 4. F ( x ) = 4 cos( x ) x 5. F ( x ) = 8 sin( x ) x
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Version 085 – EXAM 1 – BERG – (56525) 3 6. F ( x ) = 4 cos( x ) x correct 7. F ( x ) = 8 sin x x 8. F ( x ) = 4 cos x x Explanation: By the Fundamental Theorem of Calculus and the Chain Rule, d dx parenleftBig integraldisplay g ( x ) a f ( t ) dt parenrightBig = f ( g ( x )) g ( x ) .
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