mid2pract-F09-Sol

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

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
MAT126 Fall 2009 Practice Midterm II The actual midterm will consist of six problems 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 11

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online