on the interval (x + 1) 0! x < ", where x represents feet from the first fence post. How many square feet of fence had to be painted?
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1 Calculus II MAT 46 Improper Inegrals A mahemaician asked a fence painer o complee he unique ask of paining one side of a fence whose face could be described by he funcion y f (x on he inerval (x + x <, where x represens fee from he firs fence pos. How many square fee of fence had o be pained? Here s a graph showing he face of he fence: The graph shows ha he fence is alles when x and diminishes in heigh as x grows larger. I seems, however, ha because he fence has infinie lengh, we would be unable o deermine he area of he fence. Suppose for a minue, however, ha he area can be represened by one specific value. How could we find i? You migh have guessed ha he definie inegral (x + describes he area we seek. Because he definie inegral includes as a limi of inegraion, we call i an improper inegral. Any definie inegral wih one or boh limis involving infiniy is an improper inegral. Because we canno direcly evaluae an inegral ha includes ± as a limi of inegraion, we rely on he use of a limi o help complee he evaluaion. Here s how. A second ype of improper inegral, described laer in hese noes, is one for which he limis of inegraion include a verical asympoe of he funcion. The inegral x is an example of his second ype of improper inegral because x is a verical asympoe of he funcion.
2 Example (x + lim # (x + lim $ ( ' * # &' x + * lim $ # + $ + $.( ' - * &, / lim $ ( # & ' + * $ Because he limi exiss, we say ha he improper inegral (x + converges. For our example, here are square fee of area o be pained on he fence. Here s anoher example. Example : Evaluae x. This is an improper inegral because one of he limis of inegraion is. Following he process described in Example, we have x lim # x lim ln x # lim ln $ ln # $ ln Because a finie limi does no exis, we say ha he improper inegral x diverges. Replace wih a variable here we use and hen ake he limi of he resuling inegral. Find he desired aniderivaive and evaluae from o. Evaluae he limi. Every improper inegral of he ype explored in hese wo examples will eiher converge or diverge. If he associaed limi exiss (i.e., is a finie value, he improper inegral converges. If
3 he associaed limi does no exis (i.e., approaches ± or oherwise canno be evaluaed, he improper inegral diverges. Anoher Form of Improper Inegral Our mahemaician friend also asked he fence painer o pain one side of anoher fence. This fence also had a face ha could be described by a funcion, namely, y on he inerval, x < x where x represens fee from he firs fence pos. How many square fee of fence had o be pained on one side of his fence? Here s a graph showing he face of his fence: The graph shows ha he fence is alles when x and diminishes in heigh as x grows larger. Alhough his fence does no have infinie lengh, i does appear o have infinie heigh a x, so again we seem unable o deermine he area of he fence. If, however, we could compue he area, how could we find i? Righ, evaluae he definie inegral x. This inegral is also an improper inegral, because he funcion y / x has a verical asympoe a one of he limis of inegraion. Again we call on limis o help us deermine wheher he improper inegral converges or diverges and, if i converges, o deermine he value of he definie inegral.
4 Example x lim + x lim ln x + lim ln # ln + ln # #$ ( $ Because he limi does no exis, he improper inegral diverges. Example 4: Describe why x is an improper inegral and deermine wheher i converges or diverges. If i converges, deermine is value. This is an improper inegral of he second ype because he funcion y has a verical asympoe a one of he x limis of inegraion. We use limis o evaluae he inegral. x lim # + $ & x ' ( lim [ x ] # + lim # + Example : Randi evaluaed and deermined ha x ln x x ln ln ln ln. Deermine wha Randi did incorrecly and hen correcly evaluae he definie inegral.
5 I is correc ha ln x + C, bu Randi evaluaed x his incorrecly. Because he limis of inegraion include a value of x ha is a disconinuiy of he funcion, a x, we need o be more careful in is evaluaion. The definie inegral is an improper inegral because a disconinuiy in he x funcion occurs wihin he limis of inegraion. To correcly evaluae he improper inegral, we need o express he original inegral as he sum of wo inegrals so ha he value of x a which a disconinuiy occurs is eiher an upper or lower limi of inegraion: + x x x We evaluae he resuling wo inegrals, using limis as shown in previous examples, and deermine wheher he improper inegrals converge or diverge. If one or boh of he new improper inegrals diverges, hen he original inegral is said o diverge. If boh of he new improper inegrals converge, he original inegral converges and we express is value as a finie number. $ ' lim & x # x ( lim ln x # lim ln ln # ln ln * ln * Because one of he new improper inegrals diverges, he original improper inegral diverges. There is no need o evaluae he second of he wo new improper inegrals.
MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.
![MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.](/thumbs/86/93221008.jpg "MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.")
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