HW01-solutions.pdf

# HW01-solutions.pdf - young(toy68 HW01 villafuerte...

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young (toy68) – HW01 – villafuerte altu – (53615) 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 10.0points If f is a linear function whose graph has slope m and y -intercept b , evaluate the inte- gral I = integraldisplay 2 0 f ( x ) dx . 1. I = m 2. I = 2 m + b 3. I = 2 m correct 4. I = 1 2 m 5. I = m + b 6. I = 1 2 m + b Explanation: Since the graph of f has slope m and y - intercept b , f ( x ) = mx + b , so f ( x ) = m . But then by the Fundamental Theorem of Calculus, integraldisplay 2 0 f ( x ) dx = integraldisplay 2 0 m dx = bracketleftBig mx bracketrightBig 2 0 . Consequently, I = 2 m . Alternatively, we could use the Fundamen- tal Theorem of Calculus directly: integraldisplay 2 0 f ( x ) dx = f (2) - f (0) = 2 m . 002 10.0points The graph of f is shown in the figure 2 4 6 8 10 2 4 6 - 2 If g ( x ) = integraldisplay x 2 f ( t ) dt, for what value of x does g ( x ) have a maxi- mum? 1. x = 8 2. x = 7 3. x = 2 4. x = 3 . 5 5. not enough information given 6. x = 6 correct Explanation: By the Fundamental theorem of calculus, if g ( x ) = integraldisplay x 2 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 + - 2 6 8 for g . This shows that the maximum value of g occurs at x = 6

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young (toy68) – HW01 – villafuerte altu – (53615) 2 since the sign of g changes from positive to negative at x = 6. 003 10.0points The graph of f is shown in the figure 2 4 6 8 10 2 4 6 - 2 If g ( x ) = integraldisplay x 1 f ( t ) dt, for what value of x does g ( x ) have a maxi- mum? 1. x = 1 2. x = 5 correct 3. x = 2 . 5 4. x = 7 5. x = 6 6. not enough information given 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 10.0points If F ( x ) = d dx parenleftBig integraldisplay x 2 0 5 t 4 dt parenrightBig , determine the value of F (1). 1. F (1) = 2 2. F (1) = 20 3. F (1) = 10 correct 4. F (1) = 4 5. F (1) = 40 Explanation: By the Fundamental Theorem of Calculus, integraldisplay x 2 0 5 t 4 dt = bracketleftBig t 5 bracketrightBig x 2 0 = x 10 . In this case, F ( x ) = d dx parenleftBig x 10 parenrightBig = 10 x 9 Consequently, F (1) = 10 . 005 10.0points If f is a continuous function such that integraldisplay x 1 f ( t ) dt = 3 x 5 x 2 + 2 + 3 7 , find the value of f (0).
young (toy68) – HW01 – villafuerte altu – (53615) 3 1. f (0) = 9 2 2. f (0) = 0 3. f (0) = 3 4 4. f (0) = 3 2 correct 5. f (0) = 1 6. f (0) = 2 Explanation: By the Fundamental Theorem of Calculus, d dx parenleftBig integraldisplay x 1 f ( t ) dt parenrightBig = f ( x ) .

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• Fall '09
• GOGOLEV
• Derivative, lim, villafuerte altu

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