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Unformatted text preview: MAT 1332 W2011 TEST1 Max = 30 Solutions 1. Area calculation . (Total=3) Marking: 1 point for sketch, 1 point for correctly setting up the integral including correct limits of integration, 0.5 point for correct ant derivative. 0.5 points for correct end result (can be in exact form). 1. Let R be the region bounded by the curves y = 0 . 5 x ( x + 4) and y = x ( x + 4). (a) Sketch (b) We have 0 . 5 x ( x + 4) = x ( x + 4) ⇐⇒ x ( x + 4) = 0 ⇐⇒ x = 0 or x = 4. Moreover the curve y = 0 . 5 x ( x + 4) lies below the curve y = x ( x + 4) for 4 < x < 0. We conclude that the area in question is A = Z 4 x ( x + 4) . 5 x ( x + 4) dx = 1 . 5 Z 4 x ( x + 4) dx = 1 . 5[ 1 3 x 3 + 2 x 2 ] 4 = 3 2 [ 64 3 2 * 16] = 32 + 48 = 16 . 2. Let R be the region bounded by the curves y = 0 . 5 x ( x + 1) and y = 3 x ( x + 1). (a) Sketch (b) We have 0 . 5 x ( x + 1) = 3 x ( x + 1) ⇐⇒ x ( x + 1) = 0 ⇐⇒ x = 0 or x = 1. Moreover the curve y = 0 . 5 x ( x + 1) lies below the curve y = 3 x ( x + 1) for 1 < x < 0. We conclude that the area in question is A = Z 1 3 x ( x + 4) . 5 x ( x + 4) dx = 3 . 5 Z 1 x ( x + 1) dx = 3 . 5[ 1 3 x 3 + 1 2 x 2 ] 1 = 7 2 [ 1 3 1 2 ] = 7 2 [ 1 6 ] = 7 12 . 3. Let R be the region bounded by the curves y = 0 . 5 x ( x 1) and y = 2 x ( x 1). (a) Sketch (b) We have 0 . 5 x ( x 1) = 3 x ( x 1) ⇐⇒ x ( x 1) = 0 ⇐⇒ x = 0 or x = 1. Moreover the curve y = 0 . 5 x ( x 1) lies below the curve y = 2 x ( x 1) for < x < 1. We conclude that the area in question is A = Z 1 2 x ( x 1) . 5 x ( x 1) dx = 2 . 5 Z 1 x ( x 1) dx = 2 . 5[ 1 3 x 3 1 2 x 2 ] 1 = 5 2 [ 1 3 1 2 ] = 5 2 [ 1 6 ] = 5 12 . 4. Let R be the region bounded by the curves y = . 5 x ( x 2) and y = 2 x ( x 2). (a) Sketch (b) We have . 5 x ( x 2) = 2 x ( x 2) ⇐⇒ x ( x 2) = 0 ⇐⇒ x = 0 or x = 2. Moreover the curve y = 2 x ( x 2) lies below the curve y = . 5 x ( x 2) for < x < 2. We conclude that the area in question is A = Z 2 . 5 x ( x 2) 2 x ( x 2) dx = 2 . 5 Z 1 x ( x 2) dx = 2 . 5[ 1 3 x 3 x 2 ] 2 = 5 2 [ 8 3 4] = 5 2 [ 4 3 ] = 10 3 . 5. Let R be the region bounded by the curves y = 2 x ( x + 1) and y = 3 x ( x + 1). (a) Sketch (b) We have 2 x ( x + 1) = 3 x ( x + 1) ⇐⇒ x ( x + 1) = 0 ⇐⇒ x = 0 or x = 1. Moreover the curve y = 3 x ( x + 1) lies below the curve y = 2 x ( x + 1) for 1 < x < 0. We conclude that the area in question is A = Z 1  2 x ( x + 1) 3 x ( x + 1) dx = 5 Z 1 x ( x + 1) dx = 5[ 1 3 x 3 + 1 2 x 2 ] 1 = 5[ 1 3 1 2 ] = 5[ 1 6 ] = 5 6 . 2. Volume calculation (Total=3) Let R be the region bounded by the graph of y = e bx + c , the xaxis, the yaxis and the vertical lines x = u and x = l . We have V = π Z u l ( e bx + c ) 2 dx = π Z u l ( e 2 bx +2 c ) dx = π Z 2 bu +2 c 2 bl +2 c 1 2 b e z dz where we use the substitution z = 2 bx + 2 c , dz/dx = 2 b hence dx = dz/ (2 b ). Therefore the volume of the solid is π Z 2 bu +2 c 2 bl +2 c 1 2 b e z dz = π 2 b [ e...
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 Winter '07
 MUNTEANU
 Derivative, Limits, lim, dx

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