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Unformatted text preview: MAT 126: Calculus B, Fall 2009 Midterm I Solutions 1. Evaluate the integral by interpreting it as an area: Z 3 1 f ( x ) dx, where f ( x ) = ( 2 x if 1 ≤ x ≤ 2 x 2 if 2 ≤ x ≤ 3 Solution. The graph of f ( x ) is two lines intersecting at the point (2 , 0) . These two lines give the shape of triangles of base length 1 and height 1 . Hence, the value of the integral in terms of area is sum of the areas of the two triangles. Z 3 1 f ( x ) dx = A 4 1 + A 4 2 = 1 2 (1)(1) + 1 2 (1)(1) = 1 2 + 1 2 = 1 2. Evaluate lim n→∞ 1 n n X j =1 1 1 + ( j n ) 2 by interpreting the limit as an integral, and using the Evaluation Theorem. Solution. Here we are meant to apply the definition of the integral, which says, that an integral is a limit of a Riemann Sum. We are indeed given a Riemann Sum. What we should first notice is the 1 n term. This term is Δ x. Recall, Δ x = b a n . The important thing to know is, what are a and b ? Here is how we know what is should be. Observe in the sum that the term j n appears. Though this may seem a little different, it is in fact just the jth partition point...
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This note was uploaded on 02/22/2010 for the course MAT 126 taught by Professor Sutherland during the Fall '07 term at SUNY Stony Brook.
 Fall '07
 sutherland
 Calculus

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