shcards","topicId":"63afc081-a04e-4456-9c2e-f1c3f6473884","topicSlug":"integration-6","cards":[{"id":"32b53d82-54e3-4225-864b-5c46c387f565","cardType":"basic","front":"Why must an indefinite integral include $+C$?","back":"Differentiation makes every constant disappear.","points":2,"mcqOptions":null},{"id":"cc8c0eaf-ee3f-4fc8-b0e5-61f733813a9c","cardType":"basic","front":"How can you check an indefinite integral?","back":"Differentiate the answer and recover the integrand.","points":1,"mcqOptions":null},{"id":"f7073029-38a0-4b50-a9a8-2bbc4f6b7254","cardType":"cloze","front":"For $k\\neq0$, $\\int e^{kx}\\,dx=$ $\\text{[...]}$.","back":"For $k\\neq0$, $\\int e^{kx}\\,dx=$ $\\boxed{\\frac{1}{k}e^{kx}+C}$.","points":null,"mcqOptions":null},{"id":"7082aa9b-f052-4af7-acc9-04a13496929c","cardType":"cloze","front":"For $k\\neq0$, $\\int \\cos kx\\,dx=$ $\\text{[...]}$, while $\\int \\sin kx\\,dx=$ $-\\frac{1}{k}\\cos kx+C$.","back":"For $k\\neq0$, $\\int \\cos kx\\,dx=$ $\\boxed{\\frac{1}{k}\\sin kx+C}$, while $\\int \\sin kx\\,dx=$ $-\\frac{1}{k}\\cos kx+C$.","points":null,"mcqOptions":null},{"id":"f94cae0a-818d-41e7-8514-9ac8cefdf527","cardType":"cloze","front":"For $k\\neq0$, $\\int \\cos kx\\,dx=$ $\\frac{1}{k}\\sin kx+C$, while $\\int \\sin kx\\,dx=$ $\\text{[...]}$.","back":"For $k\\neq0$, $\\int \\cos kx\\,dx=$ $\\frac{1}{k}\\sin kx+C$, while $\\int \\sin kx\\,dx=$ $\\boxed{-\\frac{1}{k}\\cos kx+C}$.","points":null,"mcqOptions":null},{"id":"8b1c7852-98d7-4825-942d-a748b92504bc","cardType":"basic","front":"Why can the power rule not integrate $x^{-1}$?","back":"It would require division by zero.","points":2,"mcqOptions":null},{"id":"0b610f82-da1d-4ce3-b816-198517b40907","cardType":"cloze","front":"On an interval not crossing zero, $\\int\\frac{1}{x}\\,dx=$ $\\text{[...]}$.","back":"On an interval not crossing zero, $\\int\\frac{1}{x}\\,dx=$ $\\boxed{\\ln|x|+C}$.","points":null,"mcqOptions":null},{"id":"a4324da8-175b-4af3-b157-5300491bd40b","cardType":"basic","front":"Why are modulus signs used in $\\ln|x|$?","back":"They allow positive and negative values of $x$.","points":2,"mcqOptions":null},{"id":"71d3db67-f9c7-4b30-9b14-8cce92bd7435","cardType":"basic","front":"Why can $\\int\\frac{1}{x}\\,dx=\\ln|x|+C$ not be applied across $x=0$?","back":"$$\\frac{1}{x}$ is undefined at $x=0$.","points":2,"mcqOptions":null},{"id":"d6f990b7-1355-4c32-b08f-fb9e4410eb40","cardType":"basic","front":"Find $\\int \\frac{7}{x}\\,dx$.","back":"$7\\ln|x|+C$","points":1,"mcqOptions":null},{"id":"0a2396bf-c881-4d82-8070-48dec47f2ad7","cardType":"basic","front":"Find $\\int e^{-2x}\\,dx$.","back":"$-\\frac{1}{2}e^{-2x}+C$","points":1,"mcqOptions":null},{"id":"cabf9b77-b537-4431-ad36-8bf09ddcdabf","cardType":"basic","front":"Find $\\int 8\\sin(2x-1)\\,dx$.","back":"$-4\\cos(2x-1)+C$","points":1,"mcqOptions":null},{"id":"2e86c945-bb7a-4c4d-a520-8e5846341d17","cardType":"cloze","front":"If $F'=f$ and $a\\neq0$, then $\\int f(ax+b)\\,dx=$ $\\text{[...]}$. The shift $b$ contributes no additional factor because its derivative is zero.","back":"If $F'=f$ and $a\\neq0$, then $\\int f(ax+b)\\,dx=$ $\\boxed{\\frac{1}{a}F(ax+b)+C}$. The shift $b$ contributes no additional factor because its derivative is zero.","points":null,"mcqOptions":null},{"id":"4b6c39e9-64cb-4955-a1cf-de5198de16a4","cardType":"cloze","front":"If $F'=f$ and $a\\neq0$, then $\\int f(ax+b)\\,dx=$ $\\frac{1}{a}F(ax+b)+C$. The shift $b$ contributes [ ] because its derivative is zero.","back":"If $F'=f$ and $a\\neq0$, then $\\int f(ax+b)\\,dx=$ $\\frac{1}{a}F(ax+b)+C$. The shift $b$ contributes no additional factor because its derivative is zero.","points":null,"mcqOptions":null},{"id":"708d79e8-3266-445f-a7d0-1470ce0ea40c","cardType":"basic","front":"Find $\\int(3e^{2x}-4e^{-x})\\,dx$.","back":"$\\frac{3}{2}e^{2x}+4e^{-x}+C$","po

3.7.1 Integration (A-level only) Flashcards

3.7.1 Integration (A-level only) Flashcards