mid2pract-F09-Sol

# mid2pract-F09-Sol - MAT126 Fall 2009 Practice Midterm II...

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MAT126 Fall 2009 Practice Midterm II The actual midterm will consist of six problems 1

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Problem 1 If the function g ( x ) is given by g ( x ) = x 2 Z 2 x t ln tdt, compute the derivative g 0 ( x ) a) by using the Fundamental Theorem of Calculus to diﬀerentiate the inte- gral b) by using the Evaluation Theorem to ﬁrst evaluate g ( x ) explicitly, and then diﬀerentiating. Solution: a) We note that g ( x ) has the form similar to the fundamental theorem of calculus, but the limits of integration are not as required to apply the theorem. To apply the fundamental theorem of calculus, we denote F ( x ) = Z x 0 t ln tdt, so that then g ( x ) = F ( x 2 ) - F (2 x ) (recall that the integral of a function from 2 x to x 2 is equal to the diﬀerence of the integrals from 0 to x 2 and from 0 to 2 x ). By the fundamental theorem of calculus we have F 0 ( x ) = x ln x , and thus by applying the chain rule we have g 0 ( x ) = F 0 ( x 2 ) · 2 x - F 0 (2 x ) · 2 = x 2 ln( x 2 ) · 2 x - 2 x ln(2 x ) · 2 = 4 x 3 ln( x ) - 4 x ln(2 x ) , where we have used ln( x 2 ) = 2 ln x for the last identity. b) We are going to compute the integral x 2 Z 2 x t ln tdt by parts. Since we want to get the ln t diﬀerentiated, we will use u = ln t , which forces us to have dv = tdt , so that du = dt t and v = t 2 2 . We thus obtain g ( x ) = x 2 Z 2 x t ln tdt = ± t 2 ln t 2 ² x 2 2 x - x 2 Z 2 x t 2 dt 2 t 2
= ± t 2 ln t 2 ² x 2 2 x - x 2 Z 2 x tdt 2 = ± t 2 ln t 2 - t 2 4 ² x 2 2 x = x 4 ln( x 2 ) - 4 x 2 ln(2 x ) 2 - x 4 - 4 x 2 4 = x 4 ln( x ) - 2 x 2 ln(2 x ) + x 2 - x 4 4 , where we have again used ln( x 2 ) = 2 ln x . We now use the chain and product rules to compute the derivative g 0 ( x ) = 4 x 3 ln( x ) + x 4 x - 4 x ln(2 x ) - 2 x 2 · 2 2 x + 2 x - x 3 = 4 x 3 ln( x ) + x 3 - 4 x ln(2 x ) - 2 x + 2 x - x 3 = 4 x 3 ln( x ) - 4 x ln(2 x ) matching the answer in part a). 3

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Problem 2 Evaluate the following deﬁnite integrals: 1. 2 Z 0 x 2 p 4 - x 2 dx 2. e π Z 1 cos(ln x ) sin 2 (ln x ) x dx 3. 2 Z 1 sin(1 /x ) x 2 dx Solution: 1. In this case a trigonometric substitution is called for. We substitute x = 2 sin θ , in which case p 4 - x 2 = p 4 - 4 sin 2 θ = 4 cos 2 θ = 2 cos θ and dx = d (sin θ ) = cos θdθ so we get (don’t forget to change the limits of integration, using
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## This note was uploaded on 10/21/2010 for the course MAT 126 taught by Professor Sutherland during the Spring '07 term at SUNY Stony Brook.

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mid2pract-F09-Sol - MAT126 Fall 2009 Practice Midterm II...

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