// Numbas version: exam_results_page_options {"name": "Integration by substitution", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"extensions": [], "advice": "\n \n \n
This exercise is best solved by using substitution.
Note that if we let $u=\\simplify[std]{{a} * (x ^ 2) + {b}}$ then $du=\\simplify[std]{({2*a} * x)*dx }$
Hence we can replace $xdx$ by $\\frac{1}{\\var{2*a}}du$.
Hence the integral becomes:
\n \n \n \n\\[\\begin{eqnarray*} I&=&\\simplify[std]{Int((1/{2*a})u^{m},u)}\\\\\n \n &=&\\simplify[std]{(1/{2*a})u^{m+1}/{m+1}+C}\\\\\n \n &=& \\simplify[std]{({a} * (x ^ 2) + {b})^{m+1}/{2*a*(m+1)}+C}\n \n \\end{eqnarray*}\\]
\n \n \n \nA Useful Result
This example can be generalised.
Suppose \\[I = \\int\\; f'(x)g(f(x))\\;dx\\]
The using the substitution $u=f(x)$ we find that $du=f'(x)\\;dx$ and so using the same method as above:
\\[I = \\int g(u)\\;du \\]
And if we can find this simpler integral in terms of $u$ we can replace $u$ by $f(x)$ and get the result in terms of $x$.
Find $\\displaystyle \\int x(a x ^ 2 + b)^{m}\\;dx$
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", "strings": ["."], "partialCredit": 0, "showStrings": false}, "useCustomName": false, "vsetRangePoints": 5, "extendBaseMarkingAlgorithm": true, "scripts": {}, "answerSimplification": "std", "showPreview": true, "customName": "", "customMarkingAlgorithm": "", "marks": 3, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true}], "showFeedbackIcon": true, "prompt": "\\[I=\\simplify[std]{Int( x*({a} * x ^ 2 + {b})^{m},x)}\\]
\n$I=\\;$[[0]]
\nInput numbers in your answer as integers or fractions and not as decimals.
\nClick on Show steps to get further help. You will lose 1 mark if you do so.
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\nDon't forget $C$!
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