Assignment 3 Solutions

# Assignment 3 Solutions - Professor Jason Levy University of...

This preview shows pages 1–2. Sign up to view the full content.

Professor Jason Levy, University of Ottawa, MAT 1332C, Winter 2011 Assignment 3 Solutions 1. For each of the following improper integrals, determine whether it converges, and evaluate it if if does. (a) Z 0 t 2 4 + t 6 dt (b) Z 1 e - t sin(2 t ) dt Solution: (a) We express the improper integral as a limit, and use two substitu- tions, u = t 3 ( du = 3 t 2 dt ), and then v = u/ 2 ( dv = du/ 2): Z 0 t 2 4 + t 6 dt = lim T →∞ Z T 0 t 2 4 + t 6 dt = lim T →∞ Z T 3 0 du/ 3 4 + u 2 = lim T →∞ 1 3 * 4 Z T 3 0 du 1 + ( u/ 2) 2 = lim T →∞ 1 12 Z T 3 / 2 0 2 dv 1 + ( v ) 2 = lim T →∞ 1 6 arctan( v ) T 3 / 2 0 = lim T →∞ 1 6 (arctan( T 3 / 2) - 0) . As T → ∞ , T 3 / 2 also → ∞ , and arctan( T 3 / 2) π/ 2. So the final answer is 1 6 ( π/ 2) = π 12 . (b) Let’s first evaluate the indefinite integral, as this is a little tricky: Write I for R e - t sin(2 t ) dt . Then I = Z e - t sin(2 t ) dt = - e - t sin(2 t ) + 2 Z e - t cos(2 t ) dt by parts, with u = sin(2 t ) , dv = e - t dt = - e - t sin(2 t ) - 2 e - t cos(2 t ) - 4 Z e - t sin(2 t ) dt by parts, with u = cos(2 t ) , dv = e - t dt = - e - t sin(2 t ) - 2 e - t cos(2 t ) - 4 I, Notes: (1) Really the equation should be I = - e - t sin(2 t ) - 2 e - t cos(2 t ) - 4 I + C , C any constant. The + C is because of the inherent ambiguity in an indefinite integral: it isn’t a particular function, rather it is a function plus an arbitrary constant. You don’t really need to worry about this, as long as your final answer

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ 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