SampleMT1-4 - Math 1B, Practice material for the First...

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

View Full Document Right Arrow Icon
Math 1B, Practice material for the First Midterm Examination Practice exam 1 1.( 15 points ) Evaluate the integral ± ( x x + 1 ) 2 dx 2.( 20 pnts ) Evaluate the integral ± ² 1 + x 1 - x dx 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
3.( 25 pnts ) Determine whether each improper integral is convergent or di- vergent. Evaluate the integrals which are convergent. (a) ± 2 dx ( x 6 - x 2 ) 1 6 (b) ± 2 1 dx x (ln x ) 2 3 4.( 20 pnts ) We approximate ± 2 1 10(57 x + 1 x ) dx with the Trapezoidal rule, and want to make sure the error is at most 0 . 0001. Which of the following is true? ( Explain your answer) 2
Background image of page 2
(a) We can take any N 50 (b) We can take any 100 N 400 c) We can take any N 50 d) We can take any N 5 5.( 20 pnts ) Consider the curve given by the equation y = 1 2 (sin x - ± x 0 dx cos y ) = 1 2 (sin x - ln(sec x - tan x )) (1) Find the arclength of this curve for 0 x π/ 4. 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
1. Using the decomposition into partial fractions we have: x 2 ( x + 1) 2 = 1 - 2 x + 1 + 1 ( x + 1) 2 Thus: ± x 2 ( x + 1) 2 dx = ± dx - ± 2 x + 1 dx + ± 1 ( x + 1) 2 dx = x - 2 ln | x = 1 |- 1 ( x + 1) 2 + C 2. ² x - 1 x + 1 = ³ (1 - x ) 2 1 - x 2 After the substitution x = sin θ , taking into account 1 - sin 2 θ = cos 2 θ we have ² x - 1 x + 1 = 1 - sin θ cos θ Thus, for the integral we have: ± ² x - 1 x + 1 dx = ± 1 - sin θ cos θ cos θdθ = ± - ± sin θdθ = θ + cos θ + C 3(a). x 6 - x 2 < x 6 for x > 2. Therefore 1 ( x 6 - x 2 ) 1 / 6 > 1 ( x 6 ) 1 / 6 = 1 x The integral ± 2 dx x diverges, therefore by the comparison test the integral in question also diverges. 3(b). ± dx x (ln x ) 2 / 3 = ± d (ln x ) (ln x ) 2 / 3 = ± dt t 2 / 3 where t = ln x . Thus, the question becomes: Is the integral ± ln 2 0 dt t 2 / 3 convergent? This is a “ p -integral” for p = 2 / 3 between 0 and ln2. It is conver- gent. 4. We want the error of the approximation to be not greater then 0
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 06/26/2010 for the course MATH 58455 taught by Professor Daniel during the Spring '09 term at University of California, Berkeley.

Page1 / 15

SampleMT1-4 - Math 1B, Practice material for the First...

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