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MAT1332sampleS - MAT1332 additional practice problems The...

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MAT1332 additional practice problems The practice problems here are in addition to the material given in the suggested exercises, the assignments and the tests. The problems here do NOT represent a sample exam. 1. Find the volume of the solid obtained by rotating the area bounded by y = 4 x - x 2 , y = 3, x = 1 and x = 3 about the x -axis. Sol. See the following Figure 1, 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 y=3 y=4x-x 2 Figure 1: The illustrative graph of Question 1. denote A 1 ( x ) = 4 x - x 2 and A 2 ( x ) = 3, we know that The volume of the solid is given by V = V 1 - V 2 = Z 3 1 π ( A 1 ( x )) 2 dx - Z 3 1 π ( A 2 ( x )) 2 dx = Z 3 1 π (4 x - x 2 ) 2 dx - Z 3 1 π · 3 2 dx = Z 3 1 π (16 x 2 - 8 x 3 + x 4 ) dx - Z 3 1 9 πdx = π ( 16 3 x 3 - 2 x 4 + 1 5 x 5 ) fl fl fl 3 1 - 9 πx fl fl fl 3 1 = 136 15 π. 1
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2. Consider the system x 0 = 3 x + 2 y y 0 = 11 x - 3 y (a) Show that (0 , 0) is the only equilibrium. (b) Write down the Jacobian matrix. (c) Show that the eigenvalues are λ = ± 5. (d) For each eigenvalue, find the corresponding eigenvectors. Sol. (a) Set x 0 = 0 and y 0 = 0, we have 3 x + 2 y = 0 11 x - 3 y = 0 Solving the above linear system for x and y , we obtain the unique solution x = 0 and y = 0. (b) The Jacobian matrix is given by J = 3 2 11 - 3 . (c) Note that tr( J ) = 0, and det( J ) = - 25, so the eigenvalues are given by λ = tr( J ) 2 ± r (tr( J )) 2 4 - det( J ) = ± 25 = ± 5 . (d) For λ 1 = 5, we have 3 - 5 2 | 0 11 - 3 - 5 | 0 ( 3 + 5) R 1 --------→ - 22 2( 3 + 5) | 0 11 - 3 - 5 | 0 1 2 R 1 --→ - 11 3 + 5 | 0 11 - 3 - 5 | 0 R 1 + R 2 -----→ - 11 3 + 5 | 0 0 0 | 0 Set x 2 as the free variable, i.e., x 2 = t , from the first row, we have - 11 x 1 + ( 3 + 5) x 2 = 0 to solve for x 1 , then x 1 = 3+5 11 t . Thus, the corresponding eigenvector is v 1 = 3+5 11 1 t, ( t 6 = 0) . For λ 2 = - 5, we have 3 + 5 2 | 0 11 - 3 + 5 | 0 ( 3 - 5) R 1 --------→ - 22 2( 3 - 5) | 0 11 - 3 + 5 | 0 1 2 R 1 --→ - 11 3 - 5 | 0 11 - 3 + 5 | 0 R 1 + R 2 -----→ - 11 3 - 5 | 0 0 0 | 0 Set x 2 as the free variable, i.e., x 2 = t , from the first row, we have - 11 x 1 + ( 3 - 5) x 2 = 0 to solve for x 1 , then x 1 = 3 - 5 11 t . Thus, the corresponding eigenvector is v 2 = 3 - 5 11 1 t, ( t 6 = 0) . 2
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3. Evaluate R 4 1 e x dx , Sol. Let u = x , then du = 1 2 1 x dx , i.e. dx = 2 udu . The integration limits are changed from x = 1 and x = 4 to u = 1 and u = 2. So by integration by parts, we have Z 4 1 e x dx = Z 2 1 e u · 2 udu = 2 Z 2 1 ude u = 2 ue u fl fl fl 2 1 - 2 Z 2 1 e u du = 2 ue u fl fl fl 2 1 - 2 e u fl fl fl 2 1 = 2 e 2 . 4. Evaluate R 2 1 ln( x 2 e x ) dx , Sol. Z 2 1 ln( x 2 e x ) dx = Z 2 1 (ln x 2 + ln e x ) dx = Z 2 1 (2 ln x + x ) dx = 2 Z 2 1 ln xdx + Z 2 1 xdx = 2 x ln x fl fl fl 2 1 - 2 Z 2 1 x · 1 x dx + Z 2 1 xdx = 2 x ln x fl fl fl 2 1 - 2 Z 2 1 dx + Z 2 1 xdx = 2 x ln x fl fl fl 2 1 - 2 x fl fl fl 2 1 + 1 2 x 2 fl fl fl 2 1 = 4 ln 2 - 1 2 . 5. Evaluate R 2 x - 1 ( x +4)( x +1) dx , Sol. 2 x - 1 ( x + 4)( x + 1) = A x + 4 + B x + 1 = A ( x + 1) + B ( x + 4) ( x + 4)( x + 1) = ( A + B ) x + ( A + 4 B ) ( x + 4)( x + 1) Setting A + B = 2 and A + 4 B = - 1, we have A = 3 and B = - 1. Hence Z 2 x - 1 ( x + 4)( x + 1) dx = Z 3 x + 4 + - 1 x + 1 dx = 3 ln | x + 4 | - ln | x + 1 | + C.
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