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Unformatted text preview: 18.014 Problem Set 8 Solutions Total: 24 points Problem 1: Compute 1 xf (2 x ) dx given that f is continuous for all x , and f (0) = 1, f (0) = 3, f (1) = 5, f (1) = 2, f (2) = 7, f (2) = 4. Solution (4 points) Applying integration by parts (theorem 5.5), we have 1 xf (2 x ) 1 1 1 1 1 1 1 1 xf (2 x ) dx = f (2 x ) dx = f (2) f (2) + f (0) = . 2 2 2 4 4 2 We can use this theorem because x is differentiable with constant derivative 1 that is continuous and never changes sign, and f (2 x ) is continuous by hypothesis. Problem 2: Use the definition a x = e x log a to derive the following properties of general exponentials: (b) ( ab ) x = a x b x . (c) a x a y = a x + y . (d) ( a x ) y = ( a y ) x = a xy (e) Suppose a > 0, a = 1. Then y = a x if and only if x = log a y . Solution (4 points) (b) By the definition of the exponential function, part (ii) of theorem 3 of course notes M, part (i) of theorem 2 of course notes M, and the definition of the exponential function, we have ( ab ) x = e x log( ab ) = e x log( a )+ x log( b ) = e x log( a ) e x log( b ) = a x b x . (c) By the definition of the exponential function, part (i) of theorem 2 of course notes M, and the definition of the exponential function, we have x y x log( a ) y log( a ) ( x + y ) log( a ) x + y a a = e e = e = a . (d) After twice using the definition of the exponential function, using that the ex ponential function and the logarithmic function are inverses, and again using the definition of the exponential function, we obtain ( a x ) y = e y log( a x ) = e y log( e x log a ) = e yx log( a ) = a xy ....
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This note was uploaded on 01/18/2012 for the course MATH 18.014 taught by Professor Christinebreiner during the Fall '10 term at MIT.
 Fall '10
 ChristineBreiner
 Calculus, Integration By Parts

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