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abc

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Calculating the definite integral $\\int_{n_1}^{n_2}ax^b dx$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate \\[ \\simplify{defint({a}x^{b},x,{n_1},{n_2})} . \\] 

", "advice": "

Integrating a function of the form  \\[ f(x)=x^n \\] has the integral \\[ \\int_a^b x^n dx  =  \\left[\\frac{x^{n+1}}{n+1}\\right]_a^b,\\]

\n

and \\[\\int_a^b kf(x) dx = k \\int_a^b f(x) dx.\\]

\n

Therefore, 

\n

{advice}

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\\\\[ \\\\begin{split}\\\\simplify{defint({a}x^{b},x,{n_1},{n_2})} = \\\\simplify{{a}defint(x^{b},x,{n_1},{n_2})}&\\\\,= \\\\left[\\\\simplify[all,fractionNumbers]{{a}x^{b+1}/{b+1}}\\\\right]_\\\\var{n_1}^\\\\var{n_2},\\\\\\\\ \\\\\\\\&\\\\,= \\\\left[\\\\frac{\\\\var{a}\\\\times\\\\var{n_2}^\\\\var{b+1}}{\\\\var{b+1}} \\\\right] - \\\\left[\\\\frac{\\\\var{a}\\\\times\\\\var{n_1}^\\\\var{b+1}}{\\\\var{b+1}} \\\\right]\\\\\\\\ &\\\\,= \\\\simplify[!noLeadingMinus,fractionNumbers]{{a*n_2^(b+1)/(b+1)}-{a*n_1^(b+1)/(b+1)}}, \\\\\\\\ &\\\\,=\\\\simplify[all]{{a}({n_2}^{b+1}-{n_1}^{b+1})/{b+1}} \\\\\\\\&\\\\,=\\\\var{sol1}  \\\\end{split}  \\\\]

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\\\\[ \\\\begin{split}\\\\simplify{defint({a}x^{b},x,{n_1},{n_2})} = \\\\simplify{{a}defint(x^{b},x,{n_1},{n_2})}&\\\\,= \\\\left[\\\\simplify[all,!collectNumbers,simplifyFractions]{{a}x^{b+1}/{b+1}}\\\\right]_\\\\var{n_1}^\\\\var{n_2},\\\\\\\\ \\\\\\\\&\\\\,= \\\\left[\\\\frac{\\\\var{a}\\\\times\\\\var{n_2}^\\\\var{b+1}}{\\\\var{b+1}} \\\\right] - \\\\left[\\\\frac{\\\\var{a}\\\\times\\\\var{n_1}^\\\\var{b+1}}{\\\\var{b+1}} \\\\right]\\\\\\\\ &\\\\,= \\\\simplify[!noLeadingMinus,fractionNumbers]{{a*n_2^(b+1)/(b+1)}-{a*n_1^(b+1)/(b+1)}}, \\\\\\\\ &\\\\,=\\\\simplify[all]{{a}({n_2}^{b+1}-{n_1}^{b+1})/{b+1}} \\\\end{split}  \\\\]

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[[0]]

\n

(Give your answers to two decimal places where necessary)

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Calculating the definite integral $\\int_{n_1}^{n_2}a_1x^{b_1}+a_2x^{b_2}+a_3x^{b_3} dx$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate \\[ \\int_{\\var{n_1}}^{\\var{n_2}}\\simplify[unitFactor, unitPower, fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}} \\,dx.\\]

\n

", "advice": "

Integrating a function of the form  \\[ f(x)=x^n \\] has the integral \\[ \\int_a^b x^n dx  =  \\left[\\frac{x^{n+1}}{n+1}\\right]_a^b,\\]

\n

and \\[\\int_a^b kf(x) dx = k \\int_a^b f(x) dx.\\]

\n

Additionally, the integral of the sum or difference of two or more functions is equal to the sum or difference of the integrals of each function: \\[ \\int(f(x)\\pm g(x))\\, dx = \\int f(x)\\, dx  \\pm \\int g(x) \\, dx.\\]

\n

\n

Therefore,

\n

\\[ \\begin{split}\\simplify[unitFactor,unitPower]{defint({a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3},x,{n_1},{n_2})} &\\,= \\simplify{{a_1}defint(x^{b_1},x,{n_1},{n_2})+{a_2}defint(x^{b_2},x,{n_1},{n_2})+{a_3}defint(x^{b_3},x,{n_1},{n_2})} \\\\ &\\,= \\left[\\simplify[all,fractionNumbers]{{a_1}x^{b_1+1}/{b_1+1}+{a_2}x^{b_2+1}/{b_2+1}+{a_3}x^{b_3+1}/{b_3+1}}\\right]_\\var{n_1}^\\var{n_2} \\\\ &\\,= \\left[\\simplify[all,fractionNumbers,!collectNumbers]{{a_1*n_2^(b_1+1)}/{b_1+1}+{a_2*n_2^(b_2+1)}/{b_2+1}+{a_3*n_2^(b_3+1)}/{b_3+1}}\\right] -\\left[\\simplify[all,fractionNumbers,!collectNumbers]{{a_1*n_1^(b_1+1)}/{b_1+1}+{a_2*n_1^(b_2+1)}/{b_2+1}+{a_3*n_1^(b_3+1)}/{b_3+1}}\\right] \\\\ &\\,= \\simplify[!collectNumbers]{{eval2a}-{eval1a}} \\\\ &\\,=\\var{sol1} \\end{split} \\]

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[[0]] (Give answers to 2 decimal places where necessary)

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Calculating the definite integral $\\int_{n_1}^{n_2}a \\sin(bx) dx$, where $n_1$ and $n_2$ are multiples of $\\frac{\\pi}{12}$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate \\[ \\int_\\simplify{{n_1}pi/12}^\\simplify{{n_2}pi/12}\\simplify{{a}sin({b}x)}dx\\]

", "advice": "

Integrating a function of the form  \\[ f(x)=\\sin(kx) \\] has the integral \\[ \\int_a^b \\sin(kx) dx  =  \\left[-\\frac{1}{k}\\cos(kx)\\right]_a^b,\\]

\n

and \\[\\int_a^b kf(x) dx = k \\int_a^b f(x) dx.\\]

\n

\n

Therefore,

\n

\\[ \\begin{split} \\simplify{defint({a}sin({b}x),x,{n_1}pi/12,{n_2}pi/12)} &\\,= \\simplify{{a}defint(sin({b}x),x,{n_1}pi/12,{n_2}pi/12)}, \\\\ &\\,= \\left[ \\simplify{-{a}/{b}cos({b}x)}\\right]_\\simplify{{n_1}pi/12}^\\simplify{{n_2}pi/12}, \\\\ &\\,= \\left[\\simplify[all,!trig,fractionNumbers]{-{a/b}cos({simp2}pi)}\\right]-\\left[\\simplify[all,!trig,fractionNumbers]{-{a/b} cos({simp1}pi)}\\right] \\\\ &\\,=\\left[\\simplify[all,fractionNumbers]{-{a*frac2/b}}\\right]-\\left[\\simplify[all,fractionNumbers]{-{a*frac1/b}}\\right]\\\\ &\\,=\\simplify[all,fractionNumbers]{{(-a*frac2+a*frac1)/b}} \\end{split} \\]

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[[0]] (Give your answer as a fraction or to 2 decimal places, where necessary)

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Calculating the definite integral $\\int_{n_1}^{n_2}a \\sin(bx) dx$, where $n_1$ and $n_2$ are multiples of $\\frac{\\pi}{12}$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate \\[ \\int_\\simplify{{n_1}pi/12}^\\simplify{{n_2}pi/12}\\simplify{{a}cos({b}x)}dx\\]

", "advice": "

Integrating a function of the form  \\[ f(x)=\\cos(kx) \\] has the integral \\[ \\int_a^b \\cos(kx) dx  =  \\left[\\frac{1}{k}\\sin(kx)\\right]_a^b,\\]

\n

and \\[\\int_a^b kf(x) dx = k \\int_a^b f(x) dx.\\]

\n

\n

Therefore,

\n

\\[ \\begin{split} \\simplify{defint({a}cos({b}x),x,{n_1}pi/12,{n_2}pi/12)} &\\,= \\simplify{{a}defint(cos({b}x),x,{n_1}pi/12,{n_2}pi/12)}, \\\\ &\\,= \\left[ \\simplify{{a}/{b}sin({b}x)}\\right]_\\simplify{{n_1}pi/12}^\\simplify{{n_2}pi/12}, \\\\ &\\,= \\left[\\simplify[all,!trig,fractionNumbers]{{a/b}sin({simp2}pi)}\\right]-\\left[\\simplify[all,!trig,fractionNumbers]{{a/b} sin({simp1}pi)}\\right] \\\\&\\,=\\simplify{{sol}} \\end{split} \\]

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Calculating the area enclosed between a linear function and a quadratic function by integration. The limits (points of intersection) are given in the question.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the area enclosed by $\\simplify{y={m}x+{c_1}}$ and $\\simplify{y=x^2+{b}x+{c_2}}$ between $x=\\var{cp1}$ and $x=\\var{cp2}$.

", "advice": "

When finding the area between two functions it can be helpful to sketch the graph of these functions to have a better understanding of the area you are trying to find. In this case, we have $\\simplify{y={m}x+{c_1}}$ and $\\simplify{y=x^2+{b}x+{c_2}}$:

\n

{geogebra_applet('https://www.geogebra.org/m/cxemr4pd',defs)}

\n

We can see that the area we are being asked to find is bounded above by the line $\\simplify{y={m}x+{c_1}}$, and bounded below by the curve $\\simplify{y=x^2+{b}x+{c_2}}$. Additionally, the $x$-values we are told to find the area between are the points of intersection between these two functions.

\n

Hence, by finding the difference between the integrals

\n

\\[\\simplify{defint({m}x+{c_1},x,{cp1},{cp2})}\\quad \\text{and} \\quad \\simplify{defint(x^2+{b}x+{c_2},x,{cp1},{cp2})} ,\\]

\n

this will give us the area we are being asked to find.

\n

So,

\n

\\[ \\begin{split}\\simplify{defint({m}x+{c_1},x,{cp1},{cp2})}-\\simplify{defint(x^2+{b}x+{c_2},x,{cp1},{cp2})} &\\,= \\simplify{defint(({m}x+{c_1})-(x^2+{b}x+{c_2}),x,{cp1},{cp2})},\\\\ &\\,=\\simplify[all,!noLeadingMinus]{defint(-x^2+{m-b}x+{c_1-c_2},x,{cp1},{cp2})}. \\end{split} \\]

\n

Hence,

\n

{advice}

\n

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"b": {"name": "b", "group": "Ungrouped variables", "definition": "random(-6..6)", "description": "", "templateType": "anything", "can_override": false}, "c_1": {"name": "c_1", "group": "Ungrouped variables", "definition": "random(-6..6 except 0)", "description": "", "templateType": "anything", "can_override": false}, "cp1": {"name": "cp1", "group": "Ungrouped variables", "definition": "(-(b-m)-sqrt(dis))/2", "description": "", "templateType": "anything", "can_override": false}, "cp2": {"name": "cp2", "group": "Ungrouped variables", "definition": "(-(b-m)+sqrt(dis))/2", "description": "", "templateType": "anything", "can_override": false}, "m": {"name": "m", "group": "Ungrouped variables", "definition": "random(1..5)", "description": "", "templateType": "anything", "can_override": false}, "c_2": {"name": "c_2", "group": "Ungrouped variables", "definition": "random(-8..8 except [0,c_1])", "description": "", "templateType": "anything", "can_override": false}, "dis": {"name": "dis", "group": "Ungrouped variables", "definition": "(b-m)^2-4*(c_2-c_1)", "description": "", "templateType": "anything", "can_override": false}, "sol": {"name": "sol", "group": "Ungrouped variables", "definition": "-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2+(cp1)^3/3-(m-b)cp1^2/2-(c_1-c_2)*cp1", "description": "", "templateType": "anything", "can_override": false}, "sol2dp": {"name": "sol2dp", "group": "Ungrouped variables", "definition": "precround(sol,2)", "description": "", "templateType": "anything", "can_override": false}, "advice": {"name": "advice", "group": "Ungrouped variables", "definition": "if(sol=round(sol),'{simplified}',if(sol=sol2dp,'{simplifyy}','{simplify}'))", "description": "", "templateType": "anything", "can_override": false}, "simplified": {"name": "simplified", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\begin{split}\\\\simplify[all, !noLeadingMinus]{defint(-x^2+{m-b}x+{c_1-c_2},x,{cp1},{cp2})} &\\\\,= \\\\left[\\\\simplify[all, !noLeadingMinus]{-x^3/3+{m-b}x^2/2+{c_1-c_2}x} \\\\right]_\\\\var{cp1}^\\\\var{cp2},\\\\\\\\ &\\\\,=\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp2^3/3}+{(m-b)*cp2^2/2}+{(c_1-c_2)*cp2}}\\\\right]-\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp1^3/3}+{(m-b)*cp1^2/2}+{(c_1-c_2)*cp1}}\\\\right]\\\\\\\\ &\\\\,= \\\\left(\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2}}\\\\right)-\\\\left(\\\\simplify[all, fractionNumbers]{{-(cp1)^3/3+(m-b)cp1^2/2+(c_1-c_2)*cp1}}\\\\right)\\\\\\\\ &\\\\,=\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2+(cp1)^3/3-(m-b)cp1^2/2-(c_1-c_2)*cp1}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "simplify": {"name": "simplify", "group": "Ungrouped variables", "definition": "safe(\"

\\\\[ \\\\begin{split}\\\\simplify[all, !noLeadingMinus]{defint(-x^2+{m-b}x+{c_1-c_2},x,{cp1},{cp2})} &\\\\,= \\\\left[\\\\simplify[all, !noLeadingMinus]{-x^3/3+{m-b}x^2/2+{c_1-c_2}x} \\\\right]_\\\\var{cp1}^\\\\var{cp2},\\\\\\\\ &\\\\,=\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp2^3/3}+{(m-b)*cp2^2/2}+{(c_1-c_2)*cp2}}\\\\right]-\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp1^3/3}+{(m-b)*cp1^2/2}+{(c_1-c_2)*cp1}}\\\\right]\\\\\\\\ &\\\\,= \\\\left(\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2}}\\\\right)-\\\\left(\\\\simplify[all, fractionNumbers]{{-(cp1)^3/3+(m-b)cp1^2/2+(c_1-c_2)*cp1}}\\\\right)\\\\\\\\ &\\\\,=\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2+(cp1)^3/3-(m-b)cp1^2/2-(c_1-c_2)*cp1}}\\\\\\\\ &\\\\,=\\\\var{sol2dp} \\\\,\\\\text{(2 d.p.)}. \\\\end{split} \\\\]

\")", "description": "", "templateType": "long string", "can_override": false}, "simplifyy": {"name": "simplifyy", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\begin{split}\\\\simplify[all,!noLeadingMinus]{defint(-x^2+{m-b}x+{c_1-c_2},x,{cp1},{cp2})} &\\\\,= \\\\left[\\\\simplify[all,!noLeadingMinus]{-x^3/3+{m-b}x^2/2+{c_1-c_2}x} \\\\right]_\\\\var{cp1}^\\\\var{cp2},\\\\\\\\ &\\\\,=\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp2^3/3}+{(m-b)*cp2^2/2}+{(c_1-c_2)*cp2}}\\\\right]-\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp1^3/3}+{(m-b)*cp1^2/2}+{(c_1-c_2)*cp1}}\\\\right]\\\\\\\\ &\\\\,= \\\\left(\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2}}\\\\right)-\\\\left(\\\\simplify[all, fractionNumbers]{{-(cp1)^3/3+(m-b)cp1^2/2+(c_1-c_2)*cp1}}\\\\right)\\\\\\\\ &\\\\,=\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2+(cp1)^3/3-(m-b)cp1^2/2-(c_1-c_2)*cp1}}\\\\\\\\ &\\\\,=\\\\var{sol2dp} \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "defs": {"name": "defs", "group": "Ungrouped variables", "definition": "[\n ['M',m],\n ['D',c_1],\n ['B',b],\n ['C',c_2]\n]", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "cp1=round(cp1) and cp2=round(cp2) and dis>0", "maxRuns": "100"}, "ungrouped_variables": ["m", "c_1", "b", "c_2", "cp1", "cp2", "dis", "sol", "sol2dp", "advice", "simplified", "simplify", "simplifyy", "defs"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

[[0]] (Give your answers to 2 decimal places where necessary)

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Gap 0", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{sol2dp}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Integration: Definite Integrals 5b - Area between 2 graphs (Advanced)", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}], "tags": [], "metadata": {"description": "

Calculating the area enclosed between a linear function and a quadratic function by integration. The limits (points of intersection) are not given in the question and must be calculated.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the area enclosed by $\\simplify{y={m}x+{c_1}}$ and $\\simplify{y=x^2+{b}x+{c_2}}$.

", "advice": "

When finding the area between two functions it can be helpful to sketch the graph of these functions to have a better understanding of the area you are trying to find. In this case, we have $\\simplify{y={m}x+{c_1}}$ and $\\simplify{y=x^2+{b}x+{c_2}}$:

\n

{geogebra_applet('https://www.geogebra.org/m/fyxr2fqj',defs)}

\n

We can see that the area we are being asked to find is bounded above by the line $\\simplify{y={m}x+{c_1}}$, and bounded below by the curve $\\simplify{y=x^2+{b}x+{c_2}}$. To find the points of intersection, we want to set the functions equal to each other and solve for $x$:

\n

\\[ \\begin{split} \\simplify{x^2+{b}x+{c_2}} &\\,= \\simplify{{m}x+{c_1}}, \\\\ \\simplify{x^2+{b-m}x+{c_2-c_1}} &\\,=0, \\\\ \\simplify{(x-{cp1})(x-{cp2})}&\\,=0. \\end{split} \\]

\n

Therefore, the points of intersection are when $\\simplify{x={cp1}}$ and $\\simplify{x={cp2}}$.

\n

Hence, by finding the difference between the integrals

\n

\\[\\simplify{defint({m}x+{c_1},x,{cp1},{cp2})}\\quad \\text{and} \\quad \\simplify{defint(x^2+{b}x+{c_2},x,{cp1},{cp2})} ,\\]

\n

this will give us the area we are being asked to find.

\n

So,

\n

\\[ \\begin{split}\\simplify{defint({m}x+{c_1},x,{cp1},{cp2})}-\\simplify{defint(x^2+{b}x+{c_2},x,{cp1},{cp2})} &\\,= \\simplify{defint(({m}x+{c_1})-(x^2+{b}x+{c_2}),x,{cp1},{cp2})},\\\\ &\\,=\\simplify[all,!noLeadingMinus]{defint(-x^2+{m-b}x+{c_1-c_2},x,{cp1},{cp2})}. \\end{split} \\]

\n

Hence,

\n

{advice}

\n

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\\\\[ \\\\begin{split}\\\\simplify[all, !noLeadingMinus]{defint(-x^2+{m-b}x+{c_1-c_2},x,{cp1},{cp2})} &\\\\,= \\\\left[\\\\simplify[all, !noLeadingMinus]{-x^3/3+{m-b}x^2/2+{c_1-c_2}x} \\\\right]_\\\\var{cp1}^\\\\var{cp2},\\\\\\\\ &\\\\,=\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp2^3/3}+{(m-b)*cp2^2/2}+{(c_1-c_2)*cp2}}\\\\right]-\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp1^3/3}+{(m-b)*cp1^2/2}+{(c_1-c_2)*cp1}}\\\\right]\\\\\\\\ &\\\\,= \\\\left(\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2}}\\\\right)-\\\\left(\\\\simplify[all, fractionNumbers]{{-(cp1)^3/3+(m-b)cp1^2/2+(c_1-c_2)*cp1}}\\\\right)\\\\\\\\ &\\\\,=\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2+(cp1)^3/3-(m-b)cp1^2/2-(c_1-c_2)*cp1}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "simplify": {"name": "simplify", "group": "Ungrouped variables", "definition": "safe(\"

\\\\[ \\\\begin{split}\\\\simplify[all, !noLeadingMinus]{defint(-x^2+{m-b}x+{c_1-c_2},x,{cp1},{cp2})} &\\\\,= \\\\left[\\\\simplify[all, !noLeadingMinus]{-x^3/3+{m-b}x^2/2+{c_1-c_2}x} \\\\right]_\\\\var{cp1}^\\\\var{cp2},\\\\\\\\ &\\\\,=\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp2^3/3}+{(m-b)*cp2^2/2}+{(c_1-c_2)*cp2}}\\\\right]-\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp1^3/3}+{(m-b)*cp1^2/2}+{(c_1-c_2)*cp1}}\\\\right]\\\\\\\\ &\\\\,= \\\\left(\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2}}\\\\right)-\\\\left(\\\\simplify[all, fractionNumbers]{{-(cp1)^3/3+(m-b)cp1^2/2+(c_1-c_2)*cp1}}\\\\right)\\\\\\\\ &\\\\,=\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2+(cp1)^3/3-(m-b)cp1^2/2-(c_1-c_2)*cp1}}\\\\\\\\ &\\\\,=\\\\var{sol2dp} \\\\,\\\\text{(2dp)}. \\\\end{split} \\\\]

\")", "description": "", "templateType": "long string", "can_override": false}, "advice2": {"name": "advice2", "group": "Ungrouped variables", "definition": "if(sol=sol2dp,'{simplifyy}','{simplify}')", "description": "", "templateType": "anything", "can_override": false}, "simplifyy": {"name": "simplifyy", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\begin{split}\\\\simplify[all,!noLeadingMinus]{defint(-x^2+{m-b}x+{c_1-c_2},x,{cp1},{cp2})} &\\\\,= \\\\left[\\\\simplify[all,!noLeadingMinus]{-x^3/3+{m-b}x^2/2+{c_1-c_2}x} \\\\right]_\\\\var{cp1}^\\\\var{cp2},\\\\\\\\ &\\\\,=\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp2^3/3}+{(m-b)*cp2^2/2}+{(c_1-c_2)*cp2}}\\\\right]-\\\\left[\\\\simplify[all, !collectNumbers,fractionNumbers,!noLeadingMinus]{-{cp1^3/3}+{(m-b)*cp1^2/2}+{(c_1-c_2)*cp1}}\\\\right]\\\\\\\\ &\\\\,= \\\\left(\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2}}\\\\right)-\\\\left(\\\\simplify[all, fractionNumbers]{{-(cp1)^3/3+(m-b)cp1^2/2+(c_1-c_2)*cp1}}\\\\right)\\\\\\\\ &\\\\,=\\\\simplify[all, fractionNumbers]{{-(cp2^3/3)+(m-b)cp2^2/2+(c_1-c_2)*cp2+(cp1)^3/3-(m-b)cp1^2/2-(c_1-c_2)*cp1}}\\\\\\\\ &\\\\,=\\\\var{sol2dp} \\\\end{split} \\\\]

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[[0]] (Give your answers to 2 decimal places where necessary)

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Calculating the area enclosed between a cosine function and a quadratic function by integration. The limits (points of intersection) are given in the question.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the area enclosed by $\\simplify{y=cos(pi*x/{a})}$ and $\\simplify{y=x^2-{a^2+1}}$, between $x=\\var{-a}$ and $x=\\var{a}$.

", "advice": "

When finding the area between two functions it can be helpful to sketch the graph of these functions to have a better understanding of the area you are trying to find. In this case, we have $\\simplify{y=cos(pi x/{a})}$ and $\\simplify{y=x^2-{a^2+1}}$:

\n

{geogebra_applet('https://www.geogebra.org/m/wwvmyd78',defs)}

\n

We can see that the lower bound of the area is the curve $\\simplify{y=x^2-{a^2+1}}$ and the upper bound is the curve $\\simplify{y=cos(pi x/{a})}$, and the $x$-values we are evaluating between are where the curves intersect. 

\n

Therefore, by finding the difference between the integrals

\n

\\[\\simplify{defint(cos(pi x/{a}),x,{-a},{a})}\\quad \\text{and} \\quad \\simplify{defint(x^2-{a^2+1},x,{-a},{a})} ,\\]

\n

this will give us the area we are being asked to find.

\n

So,

\n

\\[ \\begin{split}\\simplify{defint(cos(pi x/{a}),x,{-a},{a})}-\\simplify{defint(x^2-{a^2+1},x,{-a},{a})} &\\,= \\simplify{defint((cos(pi x/{a}))-(x^2-{a^2+1}),x,{-a},{a})},\\\\ &\\,=\\simplify{defint(cos(pi x/{a})-x^2+{a^2+1},x,{-a},{a})}. \\end{split} \\]

\n

Hence,

\n

\n

{advice}

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a": {"name": "a", "group": "Ungrouped variables", "definition": "random(2..5)", "description": "", "templateType": "anything", "can_override": false}, "simplified": {"name": "simplified", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\begin{split} \\\\simplify{defint(cos(pi x/{a})-x^2+{a^2+1},x,{-a},{a})} &\\\\,= \\\\left[\\\\simplify{{a}/pi sin(pi x/{a})-x^3/3+{a^2+1}x}\\\\right]_\\\\var{-a}^\\\\var{a}\\\\\\\\ &\\\\,= \\\\left[\\\\simplify[all,fractionNumbers,!zeroTerm,!collectNumbers]{0-{a^3/3}+{(a^2+1)(a)}}\\\\right]-\\\\left[\\\\simplify[all,fractionNumbers,!zeroTerm,!collectNumbers]{0-{(-a)^3/3}+{(a^2+1)(-a)}}\\\\right] \\\\\\\\ &\\\\,=\\\\simplify[all,fractionNumbers]{{2((a^2+1)(a)-(a)^3/3)}}\\\\end{split} \\\\]

\\n

\"", "description": "", "templateType": "long string", "can_override": false}, "simplify": {"name": "simplify", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\begin{split} \\\\simplify{defint(cos(pi x/{a})-x^2+{a^2+1},x,{-a},{a})} &\\\\,= \\\\left[\\\\simplify{{a}/pi sin(pi x/{a})-x^3/3+{a^2+1}x}\\\\right]_\\\\var{-a}^\\\\var{a}\\\\\\\\ &\\\\,= \\\\left[\\\\simplify[all,fractionNumbers,!zeroTerm,!collectNumbers]{0-{a^3/3}+{(a^2+1)(a)}}\\\\right]-\\\\left[\\\\simplify[all,fractionNumbers,!zeroTerm,!collectNumbers]{0-{(-a)^3/3}+{(a^2+1)(-a)}}\\\\right] \\\\\\\\ &\\\\,=\\\\simplify[all,fractionNumbers]{{2((a^2+1)(a)-(a)^3/3)}}\\\\\\\\&\\\\,=\\\\var{soldp}\\\\,\\\\text{(2dp)}\\\\end{split} \\\\]

\\n

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[[0]] (Give your answers to 2 decimal places where necessary)

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