MAT293Solutions_2008

# MAT293Solutions_2008 - Prepared by B. Fitzpatrick...

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Unformatted text preview: Prepared by B. Fitzpatrick Handwritten solutions by J.W. Davis 1.2 Differentiability of an Integral with respect to its Parameter Most of these problems require Leibniz's formula: Let f be a differentiable function of two variables, let a and b be differentiable functions of a single variable, and define the function F by F t = a t b t f t , x d t then d d t F t = f t , b t d d t b t K f t , a t d d t a t C a t b t v v t f t , x d x 1. Since the limits of integration are constants, the first two terms in Leibniz's formula are zero for this question. Of course, whether integrating first or differentiating first, the results are the same. We proceed by integrating first: i) F x = 3 x y 3 K y x d y F x = 81 4 x K 9 2 x d 2 d x 2 F x = K 9 x 3 ii) F x = 1 3 ! 1 2 ! sin y cos x d y F x = 1 2 cos x d d x F x = K 1 2 sin x iii) F x = a b e x ln x y d y F x = K e x a ln x a C e x a C e x b ln x b K e x b d d x F x = K e x a ln x a K e x a x C e x a C e x b ln x b C e x b x K e x b When differentiating first, since the limits of integration are constants, the differentiation operator will pass directly into the integral. All the results are the same as when integrating first. 2. While we can do the first two integrations before differentiating, the third part of this question is most simply done by differentiating first. Hence, we will need to use the full-blown version of Leibniz's formula for the third part. i) F t = t y 2 d y F t = 1 3 t 3 d d t F t = t 2 ii) F t = e t ln y d y F t = e t ln e t K e t Assuming t > 0, F t = e t t K 1 d d t F t = e t t For the above integration, one should be careful to define `0' as a limit approached from the right, to avoid the expression 0 \$ ln 0 , which is an infinity-zero type singularity. The limit can be expressed as a quotient and evaluates to zero using l'Hopital's rule. iii) F t = t t 2 arctan e y d y Applying Leibniz's formula gives d d t F t = 2 arctan e t 2 t K arctan e t C t t 2 v v t arctan e y d y The integrand on the right vanishes, giving d d t F t = 2 arctan e t 2 t K arctan e t 3. i) This derivative is easily found by integrating first: F x = x x 2 y d y F x = 1 2 x 4 d d x F x = 2 x 3 ii) For this question, we'll need to apply Leibniz's formula: F x = x 2 e x arcsin y d y the result is d d x F x = 2 e x arcsin x 2 x C x 2 v v x e x arcsin y d y d d x...
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## This note was uploaded on 10/18/2010 for the course ENGINEERIN MAT293 taught by Professor J.davis during the Fall '08 term at University of Toronto.

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MAT293Solutions_2008 - Prepared by B. Fitzpatrick...

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