Akos_M101_10prac1_sollutionspdf

# Akos_M101_10prac1_sollutionspdf - SOLUTIONS MATH 101...

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Unformatted text preview: SOLUTIONS MATH 101 - PRACTICE MIDTERM I Problem I. (12 pts) Short answer questions. Here you only need to give the correct answer no detailed explanations/ calculations are needed to support your answer. (a) (2 pts) Z π- π sin 3 x cos 2 x dx = 0 SOLUTION: Use symmetry: sin(- x ) =- sin x, cos(- x ) = cos x , thus f (- x ) =- f ( x ), where f ( x ) = sin 3 x cos 2 x is the integrand. Since the range of integration [- π,π ] is symmetric to the origin the integral is 0. (b) (2 pts) Express the limit as a definite integral lim n →∞ 1 n n X i =1 n 2 n 2 + i 2 = Z 1 1 1 + x 2 dx SOLUTION: Write: n 2 n 2 + i 2 = 1 1+( i/n ) 2 , set: x i = i/n and 4 x = 1 /n . Then the sum becomes: n X i =1 1 1 + x 2 i 4 x (c) (3 pts) Let R be the region between the curve y = sin 2 x and the x-axis, for 0 ≤ x ≤ π 2 . Find the area of the region. Simplify your answer completely! SOLUTION: If 0 ≤ x ≤ π/ 2 then 0 ≤ 2 x ≤ π , thus sin 2 x ≥ 0. The area is: Z π/ 2 sin 2 x dx = 1 2 Z π sin u du =- 1 2 cos u | π = 1...
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## This note was uploaded on 10/26/2010 for the course MATH MATH100 taught by Professor Akos during the Spring '10 term at UBC.

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Akos_M101_10prac1_sollutionspdf - SOLUTIONS MATH 101...

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