Fix bug 4693

Author: gajennings

OpenProblemLibrary/Indiana/Indiana_setIntegrals4FTC/ur_in_4_11.pg

@@ -15,13 +15,15 @@ DOCUMENT();        # This should be the first executable line in the problem.
 
 loadMacros(
   "PGstandard.pl",
-  "PGchoicemacros.pl",
+  "MathObjects.pl",
   "PGcourse.pl"
 );
 
 TEXT(beginproblem());
 $showPartialCorrectAnswers = 1;
 
+Context("Numeric");
+
 $b = non_zero_random(1,2,1);
 $c = random(2,6,1);
 while ($b==$c) {$c = random(2,6,1)}
@@ -30,39 +32,36 @@ $m = random(2,7,1);
 $n = random(-3,-1,1);
 while ($m+$n == 1) {$n = random(-3,-1,1)}
 
-TEXT(EV2(<<EOT));
+BEGIN_TEXT
 Find a function \( f \) and a number \( a \) such that
 \[ $b + \int_{a}^{x} \frac {f(t)} {t^{$m}} dt = $c x^{$n} \]
 $PAR
-EOT
-
-HINT(EV2(<<EOT));
-There are two ways to solve this problem.
-$PAR
-The first (and better) way is to differentiate both sides of the above equation using the Fundamental Theorem of Calculus (Part I) on the integral.  This will give an equation that can be solved for f(x).  The value of the number a can be determined by no
-
-rmal integration (i.e. the Evaluation Theorem) of the original equation.
-$PAR
-The second way (and a bit harder) is to simply guess the form of f(x) and do normal integration to see if you are right.  The value of the number a is determined as in the preceding paragraph above.
-$PAR
-EOT
-
-TEXT(EV2(<<EOT));
 \( f(x) = \) \{ans_rule( 20)\}
 $PAR
 \( a = \) \{ans_rule( 20)\}
-EOT
+END_TEXT
 
 $d = $c*$n;
 $e = $m+$n-1;
+$f = $n-1;
+
+$ans1=Compute("$d*x^$e");
+$ans2=Compute(($b/$c)**(1/$n));
 
-$ans1="$d*x**$e" ;
-$ans2=($c/$b)**(-1/$n);
+ANS($ans1->cmp());
+ANS($ans2->cmp());
 
-ANS(fun_cmp($ans1));
-ANS(num_cmp($ans2));
 
-&SOLUTION(EV3(<<'EOT'));
+BEGIN_HINT
+There are two ways to solve this problem.
+$PAR
+The first (and better) way is to differentiate both sides of the above equation using the Fundamental Theorem of Calculus (Part I) on the integral.  This will give an equation that can be solved for f(x).  The value of the number a can be determined by integrating the original equation (i.e. the Evaluation Theorem).
+$PAR
+The second way (and a bit harder) is to simply guess the form of f(x) and integrate to see if you are right.  The value of the number a is determined as in the preceding paragraph above.
+$PAR
+END_HINT
+
+BEGIN_SOLUTION
 $SOL $BR
 As the hint suggests, the best way to solve this problem is to first determine the form
 of the function \( f(x) \) by using the first portion of the Fundamental Theorem of 
@@ -73,28 +72,26 @@ below:
 \]
 Next, we differentiate both sides to get:
 \[
-  \frac{f(x)}{x^{$m}} = \{$c*$n\} x^{ \{$n-1\} }
+  \frac{f(x)}{x^{$m}} = $d x^{ $f }
 \]
 Therefore, the function desired is:
 \[
-  f(x) = \{ $c*$n \} x^{ \{$n-1+$m\} }
+  f(x) = $d x^{ $e }
 \]
 Now, to determine the value of \( a \), we need to integrate the original expression
 using our new information about \( f(t) \).  That is, we now have:
 \[
   \begin{aligned}
-    $b + \int_{a}^{x} \frac{ $d t^$e}{t^3} dt &= $c x^{$n} \\
-    \int_{a}^{x} $d t^{ \{$e-3\} } dt &= $c x^{$n} - $b \\
-    \frac{$d}{ \{$e-2 \} }\left( x^{ \{$e-2\} } - a^{ \{$e-2\} } \right) &= $c x^{$n} - $b \\
-    $c x^{$n} - $c a^{$n} &= $c x^{$n} - $b \\
+    $b + \int_{a}^{x} \frac{ $d t^$e}{t^$m} dt &= $c x^{$n} \\
+    \int_{a}^{x} $d t^{ \{$f\} } dt &= $c x^{$n} - $b \\
+    $c \left( x^{ $n } - a^{ $n } \right) &= $c x^{$n} - $b \\
     $c a^{$n} &= $b \\
     a^{$n} &= \frac{$b}{$c} \\
-    \frac{1}{a^{\{($n==-1)?'':-$n\}}} &= \frac{$b}{$c} \\
-    a &= \sqrt[\{($n==-1)?'':-$n\}]{\frac{$c}{$b}}
+    a &=  $ans2
   \end{aligned}
 \]
+END_SOLUTION
 
-EOT
 ENDDOCUMENT();        # This should be the last executable line in the problem.