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Unformatted text preview: Differentiation of integrals and Leibniz rule We are interested in solving problems of the type d dx Z v ( x ) u ( x ) f ( x , t ) dt Notice in addition to the limits depending on x , the function f ( x , t ) also depends on x First lets take the case where the limits are constants ( u ( x ) = b and v ( x ) = a ) In most instances we can take the derivative inside the integral (as a partial derivative) d dx Z b a f ( x , t ) dt = Z b a f ( x , t ) x dt Patrick K. Schelling Introduction to Theoretical Methods Example Find R t n e kt 2 dt for n odd First solve for n = 1, I = Z te kt 2 dt Use u = kt 2 and du = 2 ktdt , so I = Z te kt 2 dt = 1 2 k Z e u du = 1 2 k Next notice dI dk = R  t 3 e kt 2 dt = 1 2 k 2 We can take successive derivatives with respect to k , so we find Z t 2 n +1 e kt 2 dt = n ! 2 k n +1 Patrick K. Schelling Introduction to Theoretical Methods Another example... the even n case The even n case can be done of R t n e kt 2 dt Take for I in this case (writing in terms of x instead of t ) I = Z  e kx 2 dx We can write for I 2 , I 2 = Z  Z  e k ( x 2 + y 2 ) dxdy Make a change of variables to polar coordinates, x = r cos , y = r sin , and x 2 + y 2 = r 2 The dxdy in the integrand becomes rdrd (we will explore this more carefully in the next chapter) Patrick K. Schelling Introduction to Theoretical Methods Example: continued Then in polar coordinates we solve the I 2 integral, I 2 = Z 2 Z e kr 2 rdrd = k So we find I = p k To find R x 2 e kx 2 dx , we take 1 2 dI dk = ( 1 2 )( 1 2 ) 1 / 2 k 3 / 2 For R x 4 e kx 2 dx = 1 2 d 2 I dk 2 = ( 1 2 )( 1 2 )( 3 2 )...
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 Spring '03
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