Exam2Solutions

# Exam2Solutions - MATH 2300 Fall 2008 Exam 2 Solutions...

• Notes
• 3

This preview shows pages 1–3. Sign up to view the full content.

MATH 2300 - Fall 2008 Exam 2 Solutions Written By Patrick Newberry 1. Z 3 x 3 1 - x 2 dx x = sin θ dx = cos θ dθ Z 3 x 3 1 - x 2 dx = Z 3 sin 3 θ p 1 - sin 2 θ cos θ dθ = Z 3 sin 3 θ cos θ cos θ = Z 3 sin 3 θ dθ = 3 Z ( 1 - cos 2 θ ) sin θ dθ = 3 - cos θ + 1 3 cos 3 θ + C = - 3 1 - x 2 + ( 1 - x 2 ) 3 2 + C = ( - x 2 - 2 ) 1 - x 2 + C. 2. Z x - 10 2 x 2 - 5 x - 3 dx x - 10 (2 x + 1)( x - 3) = A 2 x + 1 + B x - 3 x - 10 = A ( x - 3) + B (2 x + 1) x = 3 : - 7 = 7 B - 1 = B x = 0 : - 10 = - 3 A - 1 - 9 = - 3 A 3 = A Z x - 10 2 x 2 - 5 x - 3 dx = Z 3 2 x + 1 + - 1 x - 3 dx = 3 2 ln | 2 x + 1 | - ln | x - 3 | + C.

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

3. Z 2 - 2 1 ( x - 1) 2 dx Z 2 1 1 ( x - 1) 2 dx = lim a 1 + Z 2 a 1 ( x - 1) 2 dx = lim a 1 + - 1 x - 1 2 a = lim a 1 + - 1 + 1 a - 1 = - 1 + = . So, Z 2 1 1 ( x - 1) 2 dx diverges. Thus, Z 2 - 2 1 ( x - 1) 2 dx = Z 1 - 2 1 ( x - 1) 2 dx + Z 2 1 1 ( x - 1) 2 dx also diverges. 4. (a) y 0 - x = - 5 y y 0 + 5 y = x μ ( x ) = e R 5 dx = e 5 x y ( x ) = e - 5 x Z xe 5 x dx = e - 5 x 1 5 xe 5 x - Z 1 5 e 5 x dx = e - 5 x 1 5 xe 5 x - 1 25 e 5 x + C = 1 5 x - 1 25 + Ce - 5 x 1 = y 1 5 = 1 25 - 1 25 + Ce - 1 e = C y ( x ) = 1 5 x - 1 25 + ee - 5 x = - 1 25 ( - 5 x + 1) + e - 5 x +1 . (b) dy dx = x 2 1 + y 2 Z 1 + y 2 dy = Z x 2 dx y + 1 3 y 3 = 1 3 x 3 + C y 3 + 3 y = x 3 + C.
5. (a) y 00 + y 0 - 6 y = 0 m 2 + m - 6 = 0 ( m + 3)( m - 2) = 0 y ( x ) = c 1 e - 3 x + c 2 e 2 x . (b) y 00 - 6 y 0 + 13 y = 0 m 2 - 6 m + 13 = 0 m = 6 ± 36 - 52 2 = 6 ± - 16 2 = 6 ± 4 i 2 = 3 ± 2 i y ( x ) = e 3 x (
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , ··· = ± 1 2 n-1 ² ∞ n =1 . lim n →∞ a n = lim n →∞ 1 2 n-1 = lim n →∞ ³ 1 2 ´ n-1 = 0 since µ µ µ µ 1 2 µ µ µ µ < 1 . (b) lim n →∞ 3 n 2 + 1 6 n 2 + 5 = lim n →∞ 3 + 1 n 2 6 + 5 n 2 = 3 + 0 6 + 0 = 1 2 . 7. a n = 2 n +1 ( n + 1)! a n +1 a n = 2 n +2 ( n + 2)! · ( n + 1)! 2 n +1 = 2 n + 2 < 2 0 + 2 = 1 for all n ≥ 1 . Thus, ± 2 n +1 ( n + 1)! ² ∞ n =1 is strictly decreasing, and so, not strictly increasing....
View Full Document

• Spring '08
• FRUGONI,ER

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern