11Breview5sol - ucsc econ/ams 11b Review Questions 5...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ucsc econ/ams 11b Review Questions 5 Solutions 1. Compute the following integrals a. Substitute u = x 2 + 1, du = 2 xdx , then 3 xdx = (3 / 2) du and Z 3 xdx 3 √ x 2 + 1 = 3 2 Z u- 1 / 3 du = 3 2 · u 2 / 3 2 / 3 + C = 9 4 ( x 2 + 1) 2 / 3 + C . b. Substitute u = x 3 +3 x 2- 1, du = (3 x 2 +6 x ) dx , then ( x 2 +2 x ) dx = (1 / 3) du and Z ( x 2 +2 x )( x 3 +3 x 2- 1) 3 dx = 1 3 Z u 3 du = 1 12 u 4 + C = 1 12 ( x 3 +3 x 2- 1) 4 + C . c. Substitute v = ln x , dv = 1 x dx , then Z dx x ln x = Z 1 ln x · 1 x dx = Z dv v = ln | v | + C = ln | ln x | + C . d. Substitution does not work here (why?), but the integrand is a polynomial: Z ( x 2 + x )( x 3 + x 2- 1) 2 dx = Z ( x 2 + x )( x 6 + 2 x 5 + x 4- 2 x 3- 2 x 2 + 1) dx = Z x 8 +2 x 7 +3 x 6- x 5- 4 x 4- 2 x 3 + x 2 + xdx = x 9 9 + x 8 4 + 3 x 7 7- x 6 6- 4 x 5 5- x 4 2 + x 3 3 + x 2 2 + C . e. Substitute u = √ x 2 + 1, giving du = x √ x 2 + 1 dx , (check!!), so Z 3 x · e √ x 2 +1 √ x 2 + 1 dx = 3 Z e u du = 3 e u + C = 3 e √ x 2 +1 + C. Sometimes the ‘ugliest’ integrals are very easy. f. Substitute u =- . 05 t , du =- . 05 dt , then dt =- 20 du and Z 1000 e- . 05 t dt =- 20000 Z e u du =- 20000 e u + C =- 20000 e- . 05 t + C ....
View Full Document

This note was uploaded on 09/16/2011 for the course ECON 11B taught by Professor Binici during the Spring '08 term at University of California, Santa Cruz.

Page1 / 4

11Breview5sol - ucsc econ/ams 11b Review Questions 5...

This preview shows document pages 1 - 2. Sign up to view the full document.

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