math10b1sol

math10b1sol - Math 10B Test 1 Solutions 100 pts October 21,...

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Math 10B Test 1 Solutions 100 pts October 21, 2009 Professor Evans Directions: Answers alone are not sufcient. Show all work. Each problem is 20 points. (1) Let f ( x ) denote the Function e x 2 . Recall that e = 2 . 71828 ... . Jane partitioned the interval [0 , 1] into 175 equal parts and then computed a corresponding leFt Riemann sum For the Function f ( x ) above. CareFully explain why the value oF Jane’s Riemann sum must be within .01 oF the exact value oF R 1 0 f ( x ) dx . Solution (1) Since f is increasing, the exact area R 1 0 f ( x ) dx lies between the leFt Riemann sum LRS (underestimate) and the right Riemann sum RRS (overestimate). Thus it sufces to show that RRS minus LRS is less than .01. This is true, because the RRS minus LRS equals ( f (1) - f (0))( b - a ) /n = ( e 1 - e 0 )(1 / 175) = ( e - 1) / 175, which is less than .01 (as can be seen by cross multiplying). (2) ±ind the exact area oF the (finite) region bounded between the parabola y = x 2 and the line y = 3 x . Hint
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This note was uploaded on 09/13/2010 for the course MATH math 10b taught by Professor Reed during the Summer '10 term at UCSD.

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math10b1sol - Math 10B Test 1 Solutions 100 pts October 21,...

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