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11Breview8sol

# 11Breview8sol - econ 11b ucsc ams 11b Review Questions 8...

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econ 11 b ucsc ams 11 b Review Questions 8 Solutions 1. Compute the following integrals a. Z 5 x 7 - 2 x dx = 5 · Z x 7 - 2 x dx = 10( - 6 x - 14)(7 - 2 x ) 3 / 2 60 + C = - (3 x + 7)(7 - 2 x ) 3 / 2 3 + C . Use the formula Z u a + bu du = 2(3 bu - 2 a )( a + bu ) 3 / 2 15 b 2 + C , with a = 7 and b = - 2. b. Z 7 t 2 + 3 t - 1 2 + 5 t dt = 7 · Z t 2 dt 2 + 5 t + 3 · Z t dt 2 + 5 t - Z dt 2 + 5 t = 7 t 2 10 - 2 t 25 + 4 125 ln | 2 + 5 t | +3 t 5 - 2 25 ln | 2 + 5 t | - 1 5 ln | 2 + 5 t | + C = 7 10 t 2 + 1 25 t - 27 125 ln | 2 + 5 t | + C . Use the formulas (i) Z du a + bu = 1 b ln | a + bu | + C, (ii) Z u du a + bu = u b - a b 2 ln | a + bu | + C, and (iii) Z u 2 du a + bu = u 2 2 b - au b 2 + a 2 b 3 ln | a + bu | + C , ( all with a = 2 and b = 5). c. Z 500 t 2 e - 0 . 04 t dt = 500 t 2 e - 0 . 04 t - 0 . 04 + 2 0 . 04 Z te - 0 . 04 t dt = 500 t 2 e - 0 . 04 t - 0 . 04 + 2 0 . 04 e - 0 . 04 t 0 . 0016 ( - 0 . 04 t - 1) + C = - 12500 e - 0 . 04 t ( t 2 + 50 t + 1250 ) + C . Use the formulas (i) Z u n e au du = u n e au a - n a Z u n - 1 e au du , (with n = 2 and a = - 0 . 04), and (ii) Z ue au du = e au a 2 ( au - 1) + C , (with a = - 0 . 04). d. Z 3 e 2 x 4 + e x dx = 3 Z e x 4 + e x · e x dx = 3 Z u 4 + u du , (substituting u = e x and du = e x dx ), = 3 2( u - 8) 4 + u 3 + C = 2( e x - 8) 4 + e x + C . Use the formula Z u du a + bu = 2( bu - 2 a ) a + bu 3 b 2 + C , (with a = 4 and b = 1).

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e. Z 300 1 + 0 . 25 e - 0 . 1 t dt = 300 - 0 . 01 t - ln | 1 + 0 . 25 e - 0 . 01 t | - 0 . 01 + C = 30000 ( 0 . 01 t + ln | 1 + 0 . 25 e - 0 . 01 t | ) + C . Use the formula Z du a + be ku = ku - ln a + be ku ak + C , (with a = 1, b = 0 . 25 and k = - 0 . 1). f. Z 4(ln x ) 2 3 x 2 + 7 ln x dx = 4 3 Z (ln x ) 2 x 2 + 7 ln x dx = 4 3 Z u 2 du 2 + 7 u , (substituting u = ln x , du = 1 x dx ), = 4 3 2(147 u 2 - 56 u + 32) 2 + 7 u 5145 + C = 8(147(ln x ) 2 - 56 ln x + 32) 2 + 7 ln x 15435 + C . Use the formula Z u 2 du a + bu = 2(3 b 2 u 2 - 4 abu + 8 a 2 ) a + bu 15 b 3 + C , (with a = 2 and b = 7). 2. Let y = f ( x ) satisfy (i) dy dx = 3 xy 2 and (ii) y (1) = 2. Find the function f ( x ). Separate: dy y 2 = 3 x dx . Integrate: Z dy y 2 = Z 3 x dx = - 1 y = 3 x 2 2 + C . (This is the implicit solution.) Solve for y : y = 2 C - 3 x 2 . Solve for C : y (1) = 2 = 2 = 2 C - 12 = C - 12 = 1 = C = 13.
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