week11 - More on Integration 1. Simplify the integrand...

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

View Full Document Right Arrow Icon
More on Integration 1. Simplify the integrand before integrating. ex. Z 9 x 2 + 5 3 x dx = Z ± 3 x + 5 3 x - 1 ² dx = 3 x 2 2 + 5 3 ln | x | + C ex . Z 6 ( e 4 - 3 x ) 2 dx = Z 6 e 2(4 - 3 x ) dx = Z 6 e 8 - 6 x dx = Z 6 e u du - 6 , where u = 8 - 6 x ; du = - 6 dx ; dx = du - 6 = - 6 6 Z e u du = - e u + C = - e 2(4 - 3 x ) + C or = - ( e 4 - 3 x ) 2 + C ex . Z x 4 x 2 + 1 dx = Z x ( x 2 + 1) - 1 / 4 dx = Z u - 1 / 4 du 2 , where u = x 2 + 1; du = 2 xdx ; xdx = du 2 = 1 2 u - 1 / 4+1 3 / 4 + C = 2 3 ( x 2 + 1) 3 / 4 + C 2. Integrals involving ln x ex. Z 4 x ln(2 x 2 ) dx = 4 Z 1 x ln(2 x 2 ) dx = I Let u = ln(2 x 2 ). du = 1 2 x 2 · d dx (2 x 2 ) = 1 2 x 2 (4 x ) dx = 2 x dx . Thus, 1 x dx = du 2 I = 4 Z 1 u · du 2 = 4 2 Z 1 u du = 2 ln | u | + C = 2 ln | ln(2 x 2 ) | + C
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. Integrals involving b v ex . Z b v dv = Z ( e ln b ) v dv, since b = e ln b = Z e v ln b dv = Z e u du ln b , where u = v ln b ; du = (ln b ) dv, thus dv = du ln b = 1 ln b Z e u du, since (ln b ) is a constant = 1 ln b e u + C = 1
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/12/2011 for the course ECON 101 taught by Professor Bi during the Spring '11 term at York University.

Page1 / 5

week11 - More on Integration 1. Simplify the integrand...

This preview shows document pages 1 - 3. Sign up to view the full document.

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