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OpenProblemLibrary/Rochester/setIntegrals0Theory/sc5_2_30.pg

Lines changed: 22 additions & 36 deletions
Original file line numberDiff line numberDiff line change
@@ -18,54 +18,40 @@
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## Problem1('61')
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## KEYWORDS('integral')
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DOCUMENT(); # This should be the first executable line in the problem.
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DOCUMENT(); # This should be the first executable line in the problem.
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loadMacros(
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"PGstandard.pl",
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"PGchoicemacros.pl",
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"PGcourse.pl"
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);
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loadMacros('PGstandard.pl', 'PGML.pl', 'PGcourse.pl');
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TEXT(beginproblem());
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$showPartialCorrectAnswers = 1;
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$a= random(1, 10, 1);
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$add1= random(1, 10, 1);
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$add2= random(1, 10, 1);
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$b=$a+$add1;
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$c=$b+$add2;
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$a = random(1, 10, 1);
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$add1 = random(1, 10, 1);
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$add2 = random(1, 10, 1);
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$b = $a + $add1;
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$c = $b + $add2;
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TEXT(EV2(<<EOT));
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\[ \int_{$a}^{$c} f(x) \,dx - \int_{$a}^{$b} f(x) \,dx = \int_{a}^{b} f(x) \,dx\]
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$BR where \( a= \)
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\{ans_rule( 10)\}
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and \( b= \) \{ans_rule( 10)\}.
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EOT
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BEGIN_PGML
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[``` \int_{[$a]}^{[$c]} f(x) \,dx - \int_{[$a]}^{[$b]} f(x) \,dx = \int_{a}^{b} f(x) \,dx```]
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$lowerbound=$b;
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$upperbound=$c;
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ANS(num_cmp($lowerbound), num_cmp($upperbound));
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where [` a= `] [_]{$b}
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and [` b= `] [_]{$c}.
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END_PGML
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BEGIN_PGML_SOLUTION
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&SOLUTION(EV3(<<'EOT'));
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$SOL $BR
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First recall the that
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\[ \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx = \int_{a}^{c} f(x) \,dx \]
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First recall that
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[``` \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx = \int_{a}^{c} f(x) \,dx ```]
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and therefore we can rearrange it to look like this:
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\[ \int_{a}^{c} f(x) \,dx - \int_{a}^{b} f(x) \,dx = \int_{b}^{c} f(x) \,dx \]
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$BR
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[``` \int_{a}^{c} f(x) \,dx - \int_{a}^{b} f(x) \,dx = \int_{b}^{c} f(x) \,dx ```]
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Applying that in this specific context, we find:
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\[
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\int_{$a}^{$c} f(x) \,dx - \int_{$a}^{$b} f(x) \,dx = \int_{$b}^{$c} f(x) \,dx
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\]
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Thus we find \(a=$b\) and \(b=$c\).
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EOT
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Applying that in this specific context, we find:
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[```
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\int_{[$a]}^{[$c]} f(x) \,dx - \int_{[$a]}^{[$b]} f(x) \,dx = \int_{[$b]}^{[$c]} f(x) \,dx
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```]
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Thus we find [` a=[$b]`] and [`b=[$c]`].
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END_PGML_SOLUTION
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ENDDOCUMENT(); # This should be the last executable line in the problem.
71-

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