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    1 ##KEYWORDS('integrals', 'improper')
2 ##DESCRIPTION
3 ## Determine if an improper integral converges and evaluate it.
4 ##ENDDESCRIPTION
5
6 ## Shotwell cleaned
7
8 ## DBsubject('Calculus')
9 ## DBchapter('Techniques of Integration')
10 ## DBsection('Improper Integrals')
11 ## Date('6/3/2002')
12 ## Author('')
13 ## Institution('')
14 ## TitleText1('Calculus Early Transcendentals')
15 ## EditionText1('4')
16 ## AuthorText1('Stewart')
17 ## Section1('7.8')
18 ## Problem1('20')
19
20 DOCUMENT();        # This should be the first executable line in the problem.
21
23 "PGbasicmacros.pl",
25 "PGauxiliaryFunctions.pl"
26 );
27
28 TEXT(beginproblem());
29 $showPartialCorrectAnswers = 1; 30 31$l = random(.1,.9,.1);
32 $m = random(-1,1,2); 33$n = 1+($l*$m);
34 $a=random(2,8,1); 35$b=random(2,5,1);
36 $soln = "(1+$a*$b)*e^(-$a*$b)/$b^2";
37
38 BEGIN_TEXT
39 Determine whether the integral is divergent or convergent.
40 If it is convergent, evaluate it. If not, enter $BITALIC div$EITALIC
41 $BR $\int_{a}^{\infty} x e^{-{b}x} dx$ 42 Answer: \{ans_rule( 30) \} 43 END_TEXT 44 ANS(num_cmp($soln, strings=>['div']));
45
46 $soln_ab =$a * $b; 47 48 &SOLUTION(EV3(<<'EOT')); 49 50$SOL $BR$BR
51
52 This integral requires integration by parts.  For a refresher, recall that
53 the rule is:
54   $55 \int u \; dv = u v - \int v \; du 56$
57 In this case, we will use the following substitutions:
58 $$u = x$$, $$du = dx$$, $$dv = e^{-b x} dx$$,
59 and $$v = \frac{-1}{b} e^{-b x}$$.
60
61 62 \begin{align*} 63 \int_{a}^{\infty} x e^{-b x} \; dx 64 &= \int_{a}^{\infty} u dv \\\\ 65 &= \left u v \right|_{a}^{\infty} - \int_{a}^{\infty} v \; du \\\\ 66 &= \left \frac{-x}{b} e^{-b x} \right|_{a}^{\infty} 67 + \frac{1}{b} \int_{a}^{\infty} e^{-b x} \; dx \\\\ 68 &= \left \frac{-x}{b} e^{-b x} \right|_{a}^{\infty} 69 - \left \frac{1}{b^2} e^{-b x} \right|_{a}^{\infty} \\\\ 70 &= \left \left( \frac{-x}{b} e^{-b x} 71 - \left \frac{1}{b^2} e^{-b x} \right) \right|_{a}^{\infty} \\\\ 72 &= \left \left( \frac{-b x - 1}{b^2} \right) 73 e^{-b x} \right|_{a}^{\infty} \\\\ 74 &= \lim_{x\to\infty} \left( \frac{-b x - 1}{b^2} \right) e^{-b x} 75 - \left( \frac{-b \cdot a - 1}{b^2} \right) e^{-b \cdot a} \\\\ 76 &= \frac{soln_ab + 1}{b^2} e^{-soln_ab} = soln 77 \end{align*} 78
79
80 (The limit term above disappeared because the $$e^{b x}$$ term in its
81 denominator is far more significant than any polynomial, like the simple
82 linear $$-b x - 1$$ in the numerator.  Thus the entire expression went to 0.)
83
84 EOT
85
86 ENDDOCUMENT();        # This should be the last executable line in the problem.
87