epe3k

# epe3k - etc 1 6 lim t →∞-xe-3 x 3-e-3 x 9 | t = 1 9...

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

Exam III, 100 minutes, (Key) April 1, 2003 The average was about 64/100. 1) This was HW problem Ch 6.4-23 and was also on Exam II. L = 3 1 [1 + ( f ) 2 ] 1 / 2 dx = 3 1 [1 + ( x 2 / 2 - x - 2 / 2) 2 ] 1 / 2 dx = 3 1 [ x 4 / 4 + 1 / 2 + x - 4 / 4] 1 / 2 dx = 3 1 x 2 / 2 + x - 2 / 2 dx = x 3 / 6 - x - 1 / 2 | 3 1 = 27 / 6 - 1 / 6 - 1 / 6 + 1 / 2 = 14 / 3 Mistakes - mostly algebra and/or f . 2) There are two good trig subns, and two good answers: A) u = sec t leads to (sec 8 t ) / 8 - (sec 6 t ) / 3 + (sec 4 t ) / 4 + C B) u = tan t leads to (tan 6 t ) / 6 + (tan 8 t ) / 8 + C 3) Set x = 3 sec u , so dx = 3 sec u tan u du . Get 3 tan 2 u du = 3 sec 2 u - 1 du = 3 tan u - 3 u + C = x 2 - 9 - 3 sec - 1 ( x/ 3) + C 4) The partial fraction method gives 1 /x +4 / ( x - 4) = ln | x | +4 ln | x - 4 | . Other methods are possible, but are generally harder. 5) Split the numerator into (2 x +4)+1 and solve the 2 integrals (by a u sub, and by completing-the-square) and get ln | x 2 + 4 x + 5 | + tan - 1 ( x + 2) + C . Another way (slightly harder, I think) is to c-the-square first and set x + 2 = tan u and go thru

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.

Unformatted text preview: etc. 1 6) lim t →∞-xe-3 x / 3-e-3 x / 9 | t = 1 / 9 Using ”lim” correctly and the IBP work were worth about 3-4 points each. 7a) (Advertised) See lecture notes or page 698. 7b) (Advertised) See lecture notes or page 686. 7c) See lecture notes. 8a) This was done in class and labelled ”the trick”. Do IBP twice to get: Answer = e x sin x-e x cos x-Answer. Then solve for Answer. 8b) See Ex.6 on page 499. 9) FTTFT TFTFT Bonus) Split the fractions up so that every numerator is 1 and regroup them into geo-series: [1 / 2 + 1 / 4 + 1 / 8 + . . . ] + [1 / 4 + 1 / 8 + 1 / 16 + . . . ] + [1 / 8 + 1 / 16 + . . . ] + . . . = 1 + 1 / 2 + 1 / 4 + 1 / 8 + . . . = 2 2...
View Full 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