Review Sheet 3 - Math 10b 1 Z b n X f(x dx = lim n a...

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Math 10b Solutions to Final Review Sheet 1. Z b a f ( x ) dx = lim n →∞ n X i =1 f ( x i x , where Δ x = b - a n and x i is the right endpoint of the i th subinterval. So x i = a + i · Δ x . In this integral, a = 1, b = 3 and f ( x ) = 2 x 2 + 1. So Δ x = 2 n and x i = 1 + i · Δ x = 1 + 2 i n . So n i =1 f ( x i x = n i =1 2(1 + 2 i n ) 2 + 1 2 n = 2 n n i =1 2(1 + 2 i n ) 2 + 1 = 4 n n i =1 (1 + 2 i n ) 2 + 2 n n i =1 1 = 4 n n i =1 (1 + 4 i n + 4 i 2 n 2 ) + 2 n n i =1 1 = 4 n n i =1 1 + 4 n n i =1 4 i n + 4 n n i =1 4 i 2 n 2 + 2 n n i =1 1 = 4 n n i =1 1 + 16 n 2 n i =1 i + 16 n 3 n i =1 i 2 + 2 n n i =1 1 = 4 n · n + 16 n 2 · n ( n +1) 2 + 16 n 3 · n ( n +1)(2 n +1) 6 + 2 n · n = 4 + 8 n 2 + n n 2 + 8 3 2 n 3 + 3 n 2 + n n 3 + 2 . Therefore lim n →∞ n i =1 f ( x i x = lim n →∞ 4 + 8 n 2 + n n 2 + 8 3 2 n 3 + 3 n 2 + n n 3 + 2 = 4 + 8 + 16 3 + 2 = 58 3 . So Z 3 1 (2 x 2 + 1) dx = 58 3 . 2. To express the limit as a definite integral, we must put it in the form Z b a f ( x ) dx , so we must find a , b and f ( x ). We can choose a = 0 (other choices are possible, but a = 0 is the simplest). Clearly Δ x = 2 n . Since Δ x = b - a n and a = 0, b must equal 2. Since x i = a + i Δ x and a = 0, then x i = 0+ 2 i n = 2 i n . Then f ( x i ) = e x i + x i , so f ( x ) = e x + x . So the integral is Z 2 0 ( e x + x ) dx . If we evaluate this integral, we get Z 2 0 ( e x + x ) dx = e x + x 2 2 # 2 0 = e 2 + 1 . 3. We must find Z 6 0 h 10 + 10 cos π 12 t i dt , which equals Z 6 0 10 dt + Z 6 0 10 cos π 12 t dt . Note that Z 6 0 10 dt = 10(6 - 0) = 60. For Z 6 0 10 cos π 12 t dt , use substitution with u = π 12 t and du = π 12 dt . Changing limits of integration: t = 0 u = 0 and t = 6 u = π 2 . So we get 12 π Z π 2 0 10 cos u du = 120 π sin u # π 2 0 = 120 π 1 - 0 = 120 π . So 60 + 120 π gallons of water flow into the reservoir during the first six hours after the pipe was opened. 4. d dx Z x 2 0 arctan t dt ! = q arctan( x 2 ) · 2 x , using the first Fundamental Theorem of Calculus and the chain rule.
5. Note that lim x 0 Z x 0 ln(1 + t ) dt = 0. So lim x 0 Z x 0 ln(1 + t ) dt x 2 is an indeterminate form of type 0 0 . So apply l’Hˆopital’s rule (twice), using the Fundamental Theorem of Calculus for the numerator, getting lim x 0 Z x 0 ln(1 + t ) dt x 2 = lim x 0 ln(1 + x ) 2 x = lim x 0 1 1 + x 2 = 1 2 . 6. (a) Z 4 1 4 x 3 / 2 + 9 x - 1 x dx . Using algebra to simplify gives us Z 4 1 4 x + 9 x 1 / 2 - x - 1 / 2 dx = 2 x 2 + 6 x 3 / 2 - 2 x 1 / 2 i 4 1 = 2(16) + 6(8) - 2(2) - 2(1) + 6(1) - 2(1) = 70. (b) Z e x sec 2 ( e x ) dx . Use substitution, with u = e x and du = e x dx . This gives Z sec 2 u du , which equals tan u + C . Resubstituting gives tan( e x ) + C . (c) Z 10 2 x 3 - 3 x 2 - 2 x dx . Use partial fractions. 10 x (2 x + 1)( x - 2) = A x + B 2 x + 1 + C x - 2 10 = A (2 x +1)( x - 2)+ Bx ( x - 2)+ Cx (2 x +1) . If we let x = 2, we get C = 1; if we let x = - 1 2 , we get B = 8; and if we let x = 0, we get A = - 5. So Z 10 2 x 3 - 3 x 2 - 2 x dx = - Z 5 x dx + Z 8 2 x + 1 dx + Z 1 x - 2 dx = - 5 ln | x | + 4 ln | 2 x + 1 | + ln | x - 2 | + C. (d) Z 2 1 dx (1 - x ) 2 / 3 . The integrand has a vertical asymptote at x = 1 so Z 2 1 dx (1 - x ) 2 / 3 = lim t 1 + Z 2 t dx (1 - x ) 2 / 3 .

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