M1271F05Soln_000

# M1271F05Soln_000 - Math 1271 Solutions for Fall 2005 Final...

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Math 1271 Solutions for Fall 2005 Final Exam 1) Since the equation x 2 + y 2 = e xy cannot be rearranged algebraically in order to write y as an explicit function of x , we must instead differentiate this relation implicitly with respect to x , in order to find an expression for y ’ : d dx ( x 2 + y 2 ) = d dx ( e xy ) 2 x + 2 y dy dx = d du ( e u ) du dx setting u = xy 2 x + 2 y dy dx = e u d dx ( xy ) 2 x + 2 y dy dx = e u ( x ' y + x y ' ) 2 x + 2 y dy dx = e xy (1 y + x dy dx ) 2 y dy dx x e xy dy dx = y e xy 2 x dy dx = y e xy 2 x 2 y x e xy . (A) 2) The two parts of the Fundamental Theorem of (Integral) Calculus can be put together to state that, for a function h ( x ) continuous on [ a , b ] , the definite integral , with a < x < b , equals f ( x ) f ( a ) , where f ’( x ) = h ( x ) . If we modify the upper limit of the integration to make it a function u ( x ) in the same interval, we then have The derivative of this integral is then f ( a ) is a constant The integrand given in the problem is continuous everywhere (since it is defined everywhere), so the Fundamental Theorem and its consequences can be applied here. For the integral function defined by sin( e t ) t 2 + 1 x 2 10 dt , we must first make an alteration to accommodate the conditions for the Theorem: we move the constant limit of integration to become the lower limit by “reversing the direction” of integration: h ( t ) dt a x " h ( t ) dt a u ( x ) " = f ( u ( x ) ) # f ( a ) . d dx h ( t ) dt a u ( x ) " = d dx f ( u ( x ) ) # f ( a ) [ ] = d dx f ( u ( x ) ) # d dx f ( a ) = d du f ( u ) " du dx # 0 = df dx u ( x ) " du dx or h ( u ( x ) ) " u ' ( x ) .

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sin( e t ) t 2 + 1 x 2 10 dt = sin( e t ) t 2 + 1 10 x 2 dt . With h ( x ) = sin( e x ) x 2 + 1 and u ( x ) = x 2 , the derivative of our integral function is d dx sin( e t ) t 2 + 1 x 2 10 dt = h ( u ( x ) ) u ' ( x ) = sin( e x 2 ) ( x 2 ) 2 + 1 ( x 2 )' = 2 x sin( e x 2 ) x 4 + 1 . (C) 3) There are many possible ways to choose bounds for the value of a definite integral; in view of the choices given, which involve integer results, we will want to look at something fairly simple to calculate. With the interval being [ 1 , 2 ] , the smallest value the integrand itself, x 4 + 9 sin 2 x , could possibly have would be if the second term within the radical were zero, reducing the function to x 4 = x 2 , which is equal to 1 at x = 1 (in fact, the “sine-squared” term is not zero there). With the width of the interval of integration being Δ x = 2 1 = 1 , we can definitely say that x 4 + 9 sin 2 x 1 2 dx 1 ⋅ Δ x = 1 . At the other extreme, the largest value the integrand could possibly have would be when sin 2 x is as large as possible, making the function equal to x 4 + 9 1 = x 4 + 9 ; this takes on its largest value on the interval at x = 2 , where x 4 + 9 = 4 + 9 = 16 + 9 = 5 . This result provides an upper bound for our integral of x 4 + 9 sin 2 x 1 2 dx 5 ⋅ Δ x = 5 .
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## This note was uploaded on 02/07/2012 for the course MATH 1271 taught by Professor Ming during the Fall '08 term at Minnesota.

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M1271F05Soln_000 - Math 1271 Solutions for Fall 2005 Final...

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