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Unformatted text preview: University of Ottawa MAT 1332B Midterm Exam Feb. 13, 2008. Duration: 80 minutes. Instructor: Frithjof Lutscher Solutions Question 1. [3 points] Consider the two functions f ( x ) = 1 1 + 2 x , g ( x ) = 1 x. 1. (1 point) Show that the functions intersect at the two points x 1 = 0 and x 2 = 1 / 2 . 2. (2 points) Find the area enclosed by the two functions between the two points of inter section. Solution: For the intersection points, we solve f ( x ) = g ( x ) , which gives the quadratic equation (1 x )(1 2 x ) = 1 , or 2 x 2 x = 0 or x (2 x 1) = 0 . Hence, the intersections occur at x 1 = 0 and x 2 = 1 / 2 . Since f (1 / 4) = 2 / 3 < 3 / 4 = g (1 / 4) , we know that g ( x ) > f ( x ) for x 1 < x < x 2 . Therefore the area enclosed by the two curves is given by Z x 2 x 1 ( g ( x ) f ( x )) dx = Z 1 / 2 1 x 1 1 + 2 x dx = x x 2 2 1 2 ln(  1 + 2 x  )  1 / 2 = 1 2 1 8 1 2 ln(2) 0 = 3 8 1 2 ln(2) . Question 2. [5 points] Suppose that the mass of a worm grows according to the equation dW dt = 4 e t/ 2 , W (0) = 10 . The units are in milligrams. 1. (2 point) How much is the mass gain between t = 0 and t = 2? 2. (2 points) Is the mass gain until t = finite or infinite? 3. (1 point) Will the mass of the worm ever reach 20 mg? 1 Solution: The mass gain is given by the definite integral Z 2 dW dt dt = Z 2 4 e t/ 2 dt = 8( e t/ 2 )  2 = 8(1 e 1 ) ....
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