sol_assign1 - Solutions to Assignment 1 - MAT1332A - Fall...

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Solutions to Assignment 1 - MAT1332A - Fall 2009 (1) Evaluate π Z π/ 2 θ 3 cos θ 2 . Solution: First, let’s make a change of variable. Let u = θ 2 . So du = 2 θdθ . Then π Z π/ 2 θ 3 cos θ 2 = 1 2 π Z π/ 2 u cos udu. Now we use integration by parts (let f 0 ( u ) = cos u and g ( u ) = u ; then f ( u ) = sin u and g 0 ( u ) = 1): 1 2 π Z π/ 2 u cos udu = 1 2 ( u sin u + cos u ) ± ± ± ± π π/ 2 = 1 2 ² π sin π + cos π - π 2 sin π 2 - cos π 2 ³ = 1 2 ² π · 0 + ( - 1) - π 2 · 1 - 0 ³ = - π - 2 4 (2) Find the area bounded by the curves y = x 3 and y = 3 x . Solution: These two curves intersect at (0 , 0), ( - 1 , - 1), and (1 , 1). Exactly half of the area is in the first quadrant, while the other half is in the third quadrant. Therefore, the total area is the double of the area in the first quadrant. For 0 x 1, we have that 3 x x 3 . So the area in the first quadrant is: 1 Z 0 ( 3 x - x 3 ) dx = 1 Z 0 ( x 1 / 3 - x 3 ) dx = x 4 3 4 3 - x 4 4 ! ± ± ± ± 1
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sol_assign1 - Solutions to Assignment 1 - MAT1332A - Fall...

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