Do not input numbers as decimals, only as integers without the decimal point, or fractions
", "showStrings": false, "partialCredit": 0, "strings": ["."]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "answersimplification": "all", "marks": 2, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n\t\t\t$I=\\displaystyle \\int \\simplify[std]{({a}x+{b})*e^({c}x)} dx $
You are given that the answer is of the form \\[I=g(x)e^{\\var{c}x}+C\\] for a polynomial $g(x)$. You have to find $g(x)$.
$g(x)=\\;$[[0]]
\n\t\t\tInput all numbers as fractions or integers and not decimals.
\n\t\t\tYou can get help by clicking on Show steps. You will lose 1 mark if you do.
\n\t\t\t", "steps": [{"type": "information", "prompt": "\n\t\t\t\t\t \n\t\t\t\t\t \n\t\t\t\t\tThe formula for integrating by parts is
\\[ \\int u\\frac{dv}{dx} dx = uv - \\int v \\frac{du}{dx} dx. \\]
Do not input numbers as decimals, only as integers without the decimal point, or fractions
", "showStrings": false, "partialCredit": 0, "strings": ["."]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "answersimplification": "all", "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n\t\t\tUse the result from the first part to find:
\n\t\t\t$\\displaystyle I=\\int \\simplify[std]{({a}x+{b})^2*e^({c}x)} dx $
\n\t\t\tYou are given that the answer is of the form \\[I=h(x)e^{\\var{c}x}+C\\] for a polynomial $h(x)$. You have to find $h(x)$.
\n\t\t\t$h(x)=\\;$[[0]]
\n\t\t\tInput all numbers as fractions or integers and not decimals.
\n\t\t\t", "showCorrectAnswer": true, "marks": 0}], "statement": "\n\tFind the following indefinite integrals.
\n\tInput all numbers as fractions or integers and not decimals.
\n\t", "tags": ["Calculus", "MAS1601", "Steps", "algebraic manipulation", "checked2015", "exponential function", "integration", "integration by parts", "integration of exponential function"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n\t\t3/08/2012:
\n\t\tAdded tags.
\n\t\tAdded description.
\n\t\tChecked calculation. OK.
\n\t\tGot rid of redundant instructions about inputting constant of integration.
\n\t\tPenalised use of steps in first part, 1 mark. Added message to that effect in first part.
\n\t\tAdded message about not inputting decimals in appropriate places.
\n\t\tChanged marks reflecting the use of steps and degree of difficulty in second part.
\n\t\tImproved Advice display.
\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "Given $\\displaystyle \\int (ax+b)e^{cx}\\;dx =g(x)e^{cx}+C$, find $g(x)$. Find $h(x)$, $\\displaystyle \\int (ax+b)^2e^{cx}\\;dx =h(x)e^{cx}+C$.
"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n\ta)
\n\tThe formula for integrating by parts is
\\[ \\int u\\frac{dv}{dx} dx = uv - \\int v \\frac{du}{dx} dx. \\]
We choose $u = \\simplify[std]{{a}x+{b}}$ and $\\displaystyle\\frac{dv}{dx} = \\simplify[std]{e^({c}x)}$.
\n\tSo $\\displaystyle \\frac{du}{dx} = \\var{a}$ and $\\displaystyle v = \\simplify[std]{(1/{c})*e^({c}*x)}$.
\n\tHence,
\\[ \\begin{eqnarray} \\int \\simplify[std]{({a}*x+{b})*e^({c}*x)} dx &=& uv - \\int v \\frac{du}{dx} dx \\\\ &=& \\simplify[std]{({1}/{c})*({a}x+{b})*e^({c}x) - (1/{c})*Int(({a})*e^({c}x),x)} \\\\ &=& \\simplify[std]{(1/{c})*({a}x+{b})*e^({c}x) -({a}/{c^2})*e^({c}x) + C}\\\\ &=& \\simplify[std]{(({a}x+{b})/{c}-{a}/{c^2})*e^({c}*x) + C}\\\\ &=& \\simplify[std]{(({a}/{c})x+{b*c-a}/{c^2})*e^({c}*x) + C} \\end{eqnarray} \\]
Hence $\\displaystyle \\simplify[std]{g(x)=({a}/{c})*x+{c*b-a}/{c^2}}$
\n\tb)
\n\tFor this part we choose $u = \\simplify[std]{({a}x+{b})^2}$ and $\\displaystyle \\frac{dv}{dx} = \\simplify[std]{e^({c}x)}$.
\n\tSo $\\displaystyle \\frac{du}{dx}$ = $\\simplify[std]{{2*a}*({a}*(x)+{b})}$ and $\\displaystyle v = \\simplify[std]{(1/{c})*e^({c}*x)}$.
\n\tHence,
\\[ \\begin{eqnarray*}I= \\int \\simplify[std]{({a}*x+{b})^2*e^({c}*x)} dx &=& uv - \\int v \\frac{du}{dx} dx \\\\ &=& \\simplify[std]{({1}/{c})*({a}x+{b})^2*e^({c}x) - (1/{c})*Int({2*a}*({a}x+{b})*e^({c}x),x)} \\\\ &=& \\simplify[std]{(1/{c})*({a}x+{b})^2*e^({c}x) -({2*a}/{c})*Int(({a}x+{b})*e^({c}x),x)}\\dots (*) \\end{eqnarray*}\\]
But in Part a we have aready worked out $\\displaystyle \\simplify[std]{Int(({a}x+{b})*e^({c}*x),x)=(({a}/{c})*x+({c*b-a}/{c^2}))*e^({c}*x)+C}$
\n\tSo on substituting this in equation (*) we find:
\\[ \\begin{eqnarray*}I&=& \\simplify[std]{(1/{c})*({a}x+{b})^2*e^({c}x) -({2*a}/{c})*(({a}/{c})*x+({c*b-a}/{c^2}))*e^({c}*x)+C}\\\\ &=& \\simplify[std]{({a^2}/{c}*x^2+{2*a*b*c-2*a^2}/{c^2}*x+{b^2*c^2-2*a*b*c+2*a^2}/{c^3})*e^({c}x) +C} \\end{eqnarray*}\\]
Hence $\\displaystyle \\simplify[std]{h(x)={a^2}/{c}*x^2+{2*a*b*c-2*a^2}/{c^2}*x+{b^2*c^2-2*a*b*c+2*a^2}/{c^3}}$
\n\t", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}