sol for exam 2 take home

U dv here treat x as a constant by integration by

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: e indefinite integral tx e−t dt . u dv Here, treat x as a constant. By “integration by parts”, we have tx e−t dt u tx · dv = − e−t u − e−t v x tx−1 dt v − du . In other words, tx e−t dt − e−t + x tx−1 dt = = tx · − e−t +C − tx e−t + C . The corresponding definite integral: tx e−t dt t=0 +∞ +∞ +∞ + t=0 − e−t 10 x tx−1 dt = − tx e−t t=0 . That is, +∞ +∞ t=0 tx e−t dt − x tx−1 e−t dt t=0 +∞ − tx e−t = = − t=0 tx e−t lim − t − +∞ → − 0x e−0 0 = 0 0. Here, the reason why the limit lim tx e−t t − +∞ → equals 0 is as follows: First, when x is a positive integer, then tx is a polynomial in t, and hence this is “Review of Lectures – XIV”, Formula A. Second, when x is not a positive integer, then choose any positive integer m which is bigger than x. Then 0 < tx e−t < tm e−t holds for an arbitrary real number t with t > 1...
View Full Document

This note was uploaded on 01/12/2014 for the course MATH 116 taught by Professor Scholle,minho during the Fall '08 term at Kansas.

Ask a homework question - tutors are online