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Unformatted text preview: Math 1910: Prelim I Solutions (October 2, 2008) Problem 1. (a) ( 12pts) There exists a straight line passing through the origin that divides the region under the parabola y = x x 2 and above the xaxis into two regions of equal area. What is the slope of that line? Solution First we compute the area of the region bounded by the parabola y = x x 2 and the xaxis. It is clear that area = integraldisplay 1 ( x x 2 ) d x = 1 6 . Now suppose that the equation for this line is y = kx , which intersects with the parabola at (0 , 0) and (1 k, k (1 k )). By the assumption, we have the following equation integraldisplay 1 k ( x x 2 kx ) d x = 1 12 . Direct computation yields 1 12 = integraldisplay 1 k ((1 k ) x x 2 ) d x = 1 2 (1 k )(1 k ) 2 1 3 (1 k ) 3 = 1 6 (1 k ) 3 , which implies that k = 1 3 radicalBig 1 2 . (b) ( 8pts) [Unrelated to (a).] Suppose v ( t ) is the speed of a car at the time t and V ( t ) is the average speed of the same car from time 0 to time t . Assuming that v ( t ) is continuous, prove the following: If V ( t ) has a maximum or a minimum at t > 0, then V ( t ) = v ( t ). Solution From the definition of average function, V ( t ) = integraltext t v ( s ) ds t . [Note: You can also show that this implies V (0) = v (0) using LHopitals rule.] Using the product rule and the Fundamental Theorem of Calculus, V ( t ) = ( 1 t ) parenleftbiggintegraldisplay t v ( s ) ds parenrightbigg +( 1 t ) v ( t ) = ( 1 t 2 ) parenleftbiggintegraldisplay t v ( s ) ds parenrightbigg +( 1 t ) v ( t ) = v ( t ) V ( t ) t . If V attains a minimum or a maximum at t (0 , T ), then V ( t ) = 0; thus, v ( t ) V ( t ) = 0. (Altrernatively, you could prove this without any calculus by simply realizing that V ( t ) increases whenever v > V and V decreases whenever v < V .) 1 Problem 2. Evaluate the following definite integrals: (a) ( 6pts) integraldisplay 2 2 (cos 4 x cos 3 x 5) sin x dx = Solution: Method 1: Notice that the limits are symmetrical about x=0. So test if integrand is an even or odd function: (cos 4 ( x ) cos 3 ( x ) 5) sin ( x ) = (cos 4 x cos 3 x 5) ( sin x ) = [(cos 4 x cos 3 x 5) sin x ] . The integrand is odd. Integrating an odd function from x = a to x = a gives zero, so integraldisplay 2 2 ( cos 4 x cos 3 x 5 ) sin x dx = 0 Method 2: Use substitution u = cos x . Then du dx = sin x , and integral is evaluated from cos ( 2) to cos 2. But cos ( 2) = cos (2), so the upper and lower limits are the same, giving integraldisplay cos(2) cos( 2) ( u 4 u 3 5 ) du = 0 = 0 Method 3: Consider the indefinite integral, and use substitution u = cos x ....
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This note was uploaded on 02/25/2009 for the course MATH 1910 taught by Professor Berman during the Fall '07 term at Cornell University (Engineering School).
 Fall '07
 BERMAN
 Math

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