View of /trunk/NationalProblemLibrary/ma123DB/set4/s7_8_20.pg

    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 
   22 loadMacros(
   23 "PGbasicmacros.pl",
   24 "PGanswermacros.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