// Numbas version: exam_results_page_options {"name": "CA2 Block 1 Q5 2019", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "CA2 Block 1 Q5 2019", "tags": [], "metadata": {"description": "

Simple Indefinite Integrals

\n

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Solve the following indefinite integrals, using $C$ to represent an unknown constant.

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Have you differentiated this rather than integrating?\",0],\n [\"x^3/3+x^2/4\", \"Check the second term. It looks like you have differentiated the second term rather than integrating.\",0],\n [\"x^3/3+x^(-2)+C\", \"Check the second and third terms. Second term: It looks like you have differentiated the second term rather than integrating. Third term: $\\\\frac{x}{2} \\\\neq x^{-2}$. Instead think of $\\\\frac{x}{2}$ as $\\\\frac{1}{2}x$. What do you get when you integrate a number $\\\\times x$?\",0],\n [\"x^3/3+x^(-2)\", \"Check the second and third terms. Second term: It looks like you have differentiated the second term rather than integrating. Third term: $\\\\frac{x}{2} \\\\neq x^{-2}$. Instead think of $\\\\frac{x}{2}$ as $\\\\frac{1}{2}x$. What do you get when you integrate a number $\\\\times x$?\",0],\n [\"x^3/3-x^(-1)+C\", \"Check the second and third terms. Second term: It looks like you have differentiated the second term rather than integrating. Third term: $\\\\frac{x}{2} \\\\neq x^{-2}$. Instead think of $\\\\frac{x}{2}$ as $\\\\frac{1}{2}x$. What do you get when you integrate a number $\\\\times x$?\",0],\n [\"x^3/3-x^(-1)\", \"Check the second and third terms. Second term: It looks like you have differentiated the second term rather than integrating. Third term: $\\\\frac{x}{2} \\\\neq x^{-2}$. Instead think of $\\\\frac{x}{2}$ as $\\\\frac{1}{2}x$. What do you get when you integrate a number $\\\\times x$?\",0],\n [\"x^3/3+ln(2)x+C\", \"Check the second and third terms. Second term: It looks like you have differentiated the second term rather than integrating. Third term: $\\\\int \\\\frac{x}{2} dx \\\\neq x \\\\times \\\\ln 2$. The $\\\\ln$ rule only works if you have $x$ (to the power of 1) underneath the line. Instead think of $\\\\frac{x}{2}$ as $\\\\frac{1}{2}x$. What do you get when you integrate a number $\\\\times x$?\",0],\n [\"x^3/3+ln(2)x\", \"Check the second and third terms. Second term: It looks like you have differentiated the second term rather than integrating. Third term: $\\\\int \\\\frac{x}{2} dx \\\\neq x \\\\times \\\\ln 2$. The $\\\\ln$ rule only works if you have $x$ (to the power of 1) underneath the line. Instead think of $\\\\frac{x}{2}$ as $\\\\frac{1}{2}x$. What do you get when you integrate a number $\\\\times x$?\",0],\n [\"x^3/3-6x+x^(-2)+C\", \"Check the third term. $\\\\frac{x}{2} \\\\neq x^{-2}$. Instead think of $\\\\frac{x}{2}$ as $\\\\frac{1}{2}x$. What do you get when you integrate a number $\\\\times x$?\",0],\n [\"x^3/3-6x+x^(-2)\", \"Check the third term. $\\\\frac{x}{2} \\\\neq x^{-2}$. Instead think of $\\\\frac{x}{2}$ as $\\\\frac{1}{2}x$. What do you get when you integrate a number $\\\\times x$?\",0],\n [\"x^3/3-6x-x^(-1)\", \"Check the third term. $\\\\frac{x}{2} \\\\neq x^{-2}$. Instead think of $\\\\frac{x}{2}$ as $\\\\frac{1}{2}x$. What do you get when you integrate a number $\\\\times x$?\",0],\n [\"x^3/3-6x-x^(-1)+C\", \"Check the third term. $\\\\frac{x}{2} \\\\neq x^{-2}$. Instead think of $\\\\frac{x}{2}$ as $\\\\frac{1}{2}x$. What do you get when you integrate a number $\\\\times x$?\",0],\n [\"x^3/3-6x+ln(2)x+C\", \"Check the third term. $\\\\int \\\\frac{x}{2} dx \\\\neq x \\\\times \\\\ln 2$. The $\\\\ln$ rule only works if you have $x$ (to the power of 1) underneath the line. Instead think of $\\\\frac{x}{2}$ as $\\\\frac{1}{2}x$. What do you get when you integrate a number $\\\\times x$?\",0],\n [\"x^3/3-6x+ln(2)x\", \"Check the third term. $\\\\int \\\\frac{x}{2} dx \\\\neq x \\\\times \\\\ln 2$. The $\\\\ln$ rule only works if you have $x$ (to the power of 1) underneath the line. Instead think of $\\\\frac{x}{2}$ as $\\\\frac{1}{2}x$. What do you get when you integrate a number $\\\\times x$?\",0]\n ]\n\n\nparsed_malrules: \n map(\n [\"expr\":parse(x[0]),\"feedback\":x[1],\"credit\":x[2]],\n x,\n malrules\n )\n\nagree_malrules (Do the student's answer and the expected answer agree on each of the sets of variable values?):\n map(\n len(filter(not x ,x,map(\n try(\n resultsEqual(unset(question_definitions,eval(studentexpr,vars)),unset(question_definitions,eval(malrule[\"expr\"],vars)),settings[\"checkingType\"],settings[\"checkingAccuracy\"]),\n message,\n false\n ),\n vars,\n vset\n )))$\\int(x^2-6+\\frac{x}{2})\\mathrm{dx}$