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Unformatted text preview: Math 21B UC Davis, Winter 2010 Dan Romik Solutions to Practice Exam 1 1 Evaluate 20 k =1 (2 k 2). Solution. Using the properties of notation sums and the formula for the sum of the first n integers, we get: 20 X k =1 (2 k 2) = 20 X k =1 2 k 20 X k =1 2 = 2 20 X k =1 k 20 X k =1 2 = 2 20(20 + 1) 2 2 20 = 420 40 = 380 . 2 If f ( x ) is continuous and 25 R f ( x ) dx = 8, what is 5 R f ( x 2 ) xdx ? Solution. We perform the substitution u = x 2 . Using the substitution rule for definite integrals, this gives that Z 5 f ( x 2 ) xdx = 1 2 Z 5 f ( x 2 )(2 x ) dx = 1 2 Z 25 f ( u ) du = 1 2 Z 25 f ( x ) dx = 1 2 8 = 4 . 3 Find the area of the region bounded between the curves y = 3 x and y = x 2 . Solution. First, it is a good idea to sketch a diagram showing the graphs of the two curves and the region bounded between them. Figure 1 shows the result. Now we find the xordinates of the points of intersection of the two curves. This leads to the equation x 2 = 3 x , which has the two solutions...
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This note was uploaded on 11/13/2011 for the course MATH 21B taught by Professor Vershynin during the Winter '08 term at UC Davis.
 Winter '08
 Vershynin
 Differential Equations, Equations, Integers

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