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bg-background","children":["$","$L3b",null,{"backHref":"/a-level/maths/ccea/questions","basePath":"/a-level/maths/ccea/integration-6/questions","formulaUrl":"https://ccea.org.uk/downloads/docs/Support/General/2019/Mathematical%20Formulae%20and%20Tables.pdf","initialIndex":22,"initialQuestionWindow":[{"id":"3eacbbf9-f07a-4d1f-8856-766abf8a4f98","revisionId":"d5903e80-90c9-42fc-a7f6-71ec0f546cab","questionNumber":null,"questionType":"LA","specification":"Integrate with respect to $x$:","difficulty":1,"difficultyBand":"easy","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"e442c368-4e39-4f79-bfa4-4eee41ab1a46","label":"a","content":"$$\\sqrt{2x - 7}$","markscheme":"- Rewrites as $(2x-7)^{\\frac{1}{2}}$ and integrates to the form $k(2x-7)^{\\frac{3}{2}}$ M1\n- Simplifies to $\\frac{1}{3}(2x-7)^{\\frac{3}{2}}$ A1\n- Includes constant of integration $+ C$ B1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Attempt to integrate power to obtain $(2x-7)^{\\frac{3}{2}}$.","code":"M"},{"id":"p2","marks":1,"description":"Correct coefficient $\\frac{1}{3}$ or equivalent (from $\\frac{2}{3} \\times \\frac{1}{2}$).","code":"A"},{"id":"p3","marks":1,"description":"Include $+ C$ in the final answer.","code":"B"}],"finalAnswers":["\\frac{1}{3}(2x-7)^{\\frac{3}{2}} + C"],"generatedAt":"2026-05-06T14:04:09.699Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"reverse_chain_rule","label":"Reverse Chain Rule","steps":[{"type":"text","order":0,"title":"Rewrite using index form","content":"To integrate $\\sqrt{2x - 7}$ with respect to $x$, we first rewrite the square root in index (power) form:\n\n$$\\int \\sqrt{2x - 7} \\, dx = \\int (2x - 7)^{\\frac{1}{2}} \\, dx$$","highlights":[{"text":"$$\\sqrt{2x - 7}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$a = \\htmlData{anchor=val-a}{\\color{@sky}{2}}, \\quad n = \\htmlData{anchor=val-n}{\\color{@purple}{\\frac{1}{2}}}$$"},{"latex":"$$$\\int (\\htmlData{anchor=base-form}{ax+b})^n \\, dx = \\frac{(\\htmlData{anchor=base-res}{ax+b})^{n+1}}{\\htmlData{anchor=rule-a}{a}(\\htmlData{anchor=rule-n}{n+1})} + C$$"},{"latex":"$$$\\int (2x-7)^{\\frac{1}{2}} \\, dx = \\frac{(2x-7)^{\\frac{1}{2}+1}}{\\htmlData{anchor=sub-a}{\\color{@sky}{2}}\\left(\\htmlData{anchor=sub-n}{\\color{@purple}{\\frac{1}{2}}}+1\\right)} + C$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"val-a"},{"color":"purple","style":"text-color","anchor":"val-n"},{"color":"sky","style":"text-color","anchor":"sub-a"},{"color":"purple","style":"text-color","anchor":"sub-n"}],"connections":[{"to":"sub-a","from":"val-a","color":"sky","label":null,"style":"solid"},{"to":"sub-n","from":"val-n","color":"purple","label":null,"style":"solid"}]},"order":1,"title":"Apply Reverse Chain Rule","description":"We apply the general integration rule for linear composite functions: \n\n$$\\int (ax+b)^n \\, dx = \\frac{(ax+b)^{n+1}}{a(n+1)} + C$$\n\nHere, the inner function is $2x - 7$ (which has $a = 2$) and the exponent is $n = \\frac{1}{2}$."},{"type":"text","order":2,"title":"Simplify the expression","content":"Now we simplify the power and the denominator:\n\n* **Power:** $\\frac{1}{2} + 1 = \\frac{3}{2}$\n* **Denominator:** $2 \\times \\left(\\frac{1}{2} + 1\\right) = 2 \\times \\frac{3}{2} = 3$\n\nSubstituting these values back into our integration step yields:\n\n$$\\int (2x-7)^{\\frac{1}{2}} \\, dx = \\frac{(2x-7)^{\\frac{3}{2}}}{3} + C$$\n\nWhich can be written as:\n\n$$\\frac{1}{3}(2x-7)^{\\frac{3}{2}} + C$$"},{"type":"text","order":3,"title":"State final mark breakdown","content":"Our final answer is:\n\n$$\\frac{1}{3}(2x-7)^{\\frac{3}{2}} + C$$\n\n**Marks awarded:**\n* **M1**: Rewriting the root as $(2x-7)^{\\frac{1}{2}}$ and integrating to the form $k(2x-7)^{\\frac{3}{2}}$\n* **A1**: Correctly simplifying the coefficient to $\\frac{1}{3}(2x-7)^{\\frac{3}{2}}$\n* **B1**: Including the constant of integration $+ C$"}]},{"id":"substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We want to find the indefinite integral of the given function with respect to $x$:\n\n$$\\int \\sqrt{2x - 7} \\, dx$$\n\nTo make this easier to integrate, we can rewrite the radical using a fractional exponent:\n\n$$\\int (2x - 7)^{\\frac{1}{2}} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"u = \\htmlData{anchor=sub-u}{2x - 7}"},{"text":"Differentiating both sides with respect to $x$ gives:"},{"latex":"\\frac{du}{dx} = 2"},{"text":"Which we can rewrite to find an expression for $dx$:"},{"latex":"dx = \\htmlData{anchor=sub-dx}{\\frac{1}{2} \\, du}"}],"highlights":[{"color":"sky","style":"box","anchor":"sub-u"},{"color":"amber","style":"box","anchor":"sub-dx"}],"connections":[]},"order":1,"title":"Define the substitution","description":"We choose $u = 2x - 7$ to simplify the inner function. Then we find $dx$ in terms of $du$ so we can substitute it into our integral."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\int (\\htmlData{anchor=int-u}{\\color{@sky}{2x - 7}})^{\\frac{1}{2}} \\, \\htmlData{anchor=int-dx}{\\color{@amber}{dx}}"},{"latex":"= \\int (\\htmlData{anchor=res-u}{\\color{@sky}{u}})^{\\frac{1}{2}} \\left( \\htmlData{anchor=res-dx}{\\color{@amber}{\\frac{1}{2} \\, du}} \\right)"},{"latex":"= \\frac{1}{2} \\int u^{\\frac{1}{2}} \\, du"}],"highlights":[{"color":"sky","style":"text-color","anchor":"int-u"},{"color":"sky","style":"text-color","anchor":"res-u"},{"color":"amber","style":"text-color","anchor":"int-dx"},{"color":"amber","style":"text-color","anchor":"res-dx"}],"connections":[{"to":"res-u","from":"int-u","color":"sky","label":"substitute u","style":"solid"},{"to":"res-dx","from":"int-dx","color":"amber","label":"substitute dx","style":"solid"}]},"order":2,"title":"Substitute into the integral","description":"Now we replace $2x - 7$ with $u$, and $dx$ with $\\frac{1}{2} \\, du$ in our integral."},{"type":"text","order":3,"title":"Integrate using power rule","content":"Now we integrate $u^{\\frac{1}{2}}$ with respect to $u$ using the power rule, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1}$ (for $n \\neq -1$):\n\n$$\\begin{aligned} \\frac{1}{2} \\int u^{\\frac{1}{2}} \\, du &= \\frac{1}{2} \\left( \\frac{u^{\\frac{3}{2}}}{\\frac{3}{2}} \\right) + C \\\\ &= \\frac{1}{2} \\cdot \\frac{2}{3} u^{\\frac{3}{2}} + C \\\\ &= \\frac{1}{3} u^{\\frac{3}{2}} + C \\end{aligned}$$\n\n*Note: This earns us the **M1** mark for obtaining the form $k(2x-7)^{\\frac{3}{2}}$ and the **B1** mark for including the constant of integration $+ C$.*"},{"type":"text","order":4,"title":"Substitute back to x","content":"Finally, we substitute our original definition $u = 2x - 7$ back into the expression:\n\n$$\\frac{1}{3}(2x - 7)^{\\frac{3}{2}} + C$$\n\nThis gives us our final answer, earning the final **A1** mark."}]}],"generatedAt":"2026-06-09T04:45:44.374Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"192e5e23-2d30-4e54-b428-4cb49fa5b683","label":"b","content":"$$e^{4x + 2}$","markscheme":"- Integrates exponential to the form $k e^{4x+2}$ M1\n- Correct coefficient $\\frac{1}{4}$ A1\n- Includes constant of integration $+ C$ B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Integrate to obtain $k e^{4x+2}$.","code":"M"},{"id":"p2","marks":1,"description":"Correct coefficient $\\frac{1}{4}$.","code":"A"},{"id":"p3","marks":1,"description":"Include $+ C$ in the final answer.","code":"B"}],"finalAnswers":["\\frac{1}{4}e^{4x+2} + C"],"generatedAt":"2026-05-06T14:04:09.699Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We want to integrate the given function $e^{4x + 2}$ with respect to $x$:\n\n$$\\int e^{4x+2} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int e^{\\htmlData{anchor=orig-exp}{\\color{@sky}{4x+2}}} \\, \\htmlData{anchor=orig-dx}{\\color{@amber}{dx}}$"},{"latex":"$$\\text{Let } \\htmlData{anchor=sub-u}{\\color{@sky}{u}} = 4x + 2 \\implies \\frac{du}{dx} = 4 \\implies \\htmlData{anchor=sub-dx}{\\color{@amber}{dx}} = \\htmlData{anchor=sub-du}{\\color{@purple}{\\frac{1}{4} du}}$"},{"latex":"$$=\\int e^{\\htmlData{anchor=new-u}{\\color{@sky}{u}}} \\, \\htmlData{anchor=new-dx}{\\color{@purple}{\\frac{1}{4} du}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"orig-exp"},{"color":"sky","style":"text-color","anchor":"sub-u"},{"color":"sky","style":"text-color","anchor":"new-u"},{"color":"purple","style":"text-color","anchor":"sub-du"},{"color":"purple","style":"text-color","anchor":"new-dx"}],"connections":[{"to":"sub-u","from":"orig-exp","color":"sky","label":"Substitute exponent","style":"solid"},{"to":"new-u","from":"sub-u","color":"sky","label":"New variable","style":"solid"},{"to":"sub-dx","from":"orig-dx","color":"amber","label":"Differential term","style":"solid"},{"to":"new-dx","from":"sub-du","color":"purple","label":"Substitute dx","style":"solid"}]},"order":1,"title":"Apply substitution","description":"We choose $u = 4x+2$ to simplify the exponent, then find the differential $dx$ in terms of $du$ to perform the complete change of variables."},{"type":"text","order":2,"title":"Integrate with respect to u","content":"Factor out the constant $\\frac{1}{4}$ and integrate the exponential function:\n\n$$\\int e^u \\left(\\frac{1}{4} du\\right) = \\frac{1}{4} \\int e^u \\, du$$\n\nSince the integral of $e^u$ is simply $e^u$:\n\n$$= \\frac{1}{4} e^u + C$$"},{"type":"text","order":3,"title":"Substitute back to x","content":"Finally, replace $u$ with the original expression $4x + 2$ to get our final answer:\n\n$$\\frac{1}{4} e^{4x+2} + C$$"}]},{"id":"inspection","label":"Direct Integration","steps":[{"type":"text","order":0,"title":"Identify the standard rule","content":"To integrate an exponential function where the exponent is a linear function of the form $ax + b$, we can use the direct integration rule:\n\n$$\\int e^{ax+b} \\, dx = \\frac{1}{a} e^{ax+b} + C$$\n\nFor this problem, we want to find:\n\n$$\\int e^{4x + 2} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int e^{\\htmlData{anchor=rule-a}{\\color{@sky}{a}}x+b} \\, dx = \\frac{1}{\\htmlData{anchor=rule-recip}{\\color{@sky}{a}}}e^{ax+b} + C$"},{"latex":"$$\\int e^{\\htmlData{anchor=val-a}{\\color{@sky}{4}}x+2} \\, dx = \\frac{1}{\\htmlData{anchor=val-recip}{\\color{@sky}{4}}}e^{4x+2} + C$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"rule-a"},{"color":"sky","style":"text-color","anchor":"val-a"},{"color":"sky","style":"text-color","anchor":"rule-recip"},{"color":"sky","style":"text-color","anchor":"val-recip"}],"connections":[{"to":"val-a","from":"rule-a","color":"sky","label":"a = 4","style":"solid"},{"to":"val-recip","from":"rule-recip","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Apply standard formula","description":"We identify that the coefficient of $x$ is $a = 4$ and apply the integration rule directly."},{"type":"text","order":2,"title":"Write final answer","content":"Thus, the result of the integration is:\n\n$$\\frac{1}{4}e^{4x + 2} + C$$\n\n### Markscheme Breakdown\n- **M1**: Integrating to the form $k e^{4x+2}$ where $k$ is a constant.\n- **A1**: Finding the correct coefficient $k = \\frac{1}{4}$.\n- **B1**: Including the constant of integration $+ C$."}]}],"generatedAt":"2026-06-09T04:47:36.154Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"e442c368-4e39-4f79-bfa4-4eee41ab1a46:reverse_chain_rule","label":"(a) Reverse Chain Rule","steps":[{"type":"text","order":0,"title":"Rewrite using index form","content":"To integrate $\\sqrt{2x - 7}$ with respect to $x$, we first rewrite the square root in index (power) form:\n\n$$\\int \\sqrt{2x - 7} \\, dx = \\int (2x - 7)^{\\frac{1}{2}} \\, dx$$","highlights":[{"text":"$$\\sqrt{2x - 7}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$a = \\htmlData{anchor=val-a}{\\color{@sky}{2}}, \\quad n = \\htmlData{anchor=val-n}{\\color{@purple}{\\frac{1}{2}}}$$"},{"latex":"$$$\\int (\\htmlData{anchor=base-form}{ax+b})^n \\, dx = \\frac{(\\htmlData{anchor=base-res}{ax+b})^{n+1}}{\\htmlData{anchor=rule-a}{a}(\\htmlData{anchor=rule-n}{n+1})} + C$$"},{"latex":"$$$\\int (2x-7)^{\\frac{1}{2}} \\, dx = \\frac{(2x-7)^{\\frac{1}{2}+1}}{\\htmlData{anchor=sub-a}{\\color{@sky}{2}}\\left(\\htmlData{anchor=sub-n}{\\color{@purple}{\\frac{1}{2}}}+1\\right)} + C$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"val-a"},{"color":"purple","style":"text-color","anchor":"val-n"},{"color":"sky","style":"text-color","anchor":"sub-a"},{"color":"purple","style":"text-color","anchor":"sub-n"}],"connections":[{"to":"sub-a","from":"val-a","color":"sky","label":null,"style":"solid"},{"to":"sub-n","from":"val-n","color":"purple","label":null,"style":"solid"}]},"order":1,"title":"Apply Reverse Chain Rule","description":"We apply the general integration rule for linear composite functions: \n\n$$\\int (ax+b)^n \\, dx = \\frac{(ax+b)^{n+1}}{a(n+1)} + C$$\n\nHere, the inner function is $2x - 7$ (which has $a = 2$) and the exponent is $n = \\frac{1}{2}$."},{"type":"text","order":2,"title":"Simplify the expression","content":"Now we simplify the power and the denominator:\n\n* **Power:** $\\frac{1}{2} + 1 = \\frac{3}{2}$\n* **Denominator:** $2 \\times \\left(\\frac{1}{2} + 1\\right) = 2 \\times \\frac{3}{2} = 3$\n\nSubstituting these values back into our integration step yields:\n\n$$\\int (2x-7)^{\\frac{1}{2}} \\, dx = \\frac{(2x-7)^{\\frac{3}{2}}}{3} + C$$\n\nWhich can be written as:\n\n$$\\frac{1}{3}(2x-7)^{\\frac{3}{2}} + C$$"},{"type":"text","order":3,"title":"State final mark breakdown","content":"Our final answer is:\n\n$$\\frac{1}{3}(2x-7)^{\\frac{3}{2}} + C$$\n\n**Marks awarded:**\n* **M1**: Rewriting the root as $(2x-7)^{\\frac{1}{2}}$ and integrating to the form $k(2x-7)^{\\frac{3}{2}}$\n* **A1**: Correctly simplifying the coefficient to $\\frac{1}{3}(2x-7)^{\\frac{3}{2}}$\n* **B1**: Including the constant of integration $+ C$"}]},{"id":"e442c368-4e39-4f79-bfa4-4eee41ab1a46:substitution","label":"(a) Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We want to find the indefinite integral of the given function with respect to $x$:\n\n$$\\int \\sqrt{2x - 7} \\, dx$$\n\nTo make this easier to integrate, we can rewrite the radical using a fractional exponent:\n\n$$\\int (2x - 7)^{\\frac{1}{2}} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"u = \\htmlData{anchor=sub-u}{2x - 7}"},{"text":"Differentiating both sides with respect to $x$ gives:"},{"latex":"\\frac{du}{dx} = 2"},{"text":"Which we can rewrite to find an expression for $dx$:"},{"latex":"dx = \\htmlData{anchor=sub-dx}{\\frac{1}{2} \\, du}"}],"highlights":[{"color":"sky","style":"box","anchor":"sub-u"},{"color":"amber","style":"box","anchor":"sub-dx"}],"connections":[]},"order":1,"title":"Define the substitution","description":"We choose $u = 2x - 7$ to simplify the inner function. Then we find $dx$ in terms of $du$ so we can substitute it into our integral."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\int (\\htmlData{anchor=int-u}{\\color{@sky}{2x - 7}})^{\\frac{1}{2}} \\, \\htmlData{anchor=int-dx}{\\color{@amber}{dx}}"},{"latex":"= \\int (\\htmlData{anchor=res-u}{\\color{@sky}{u}})^{\\frac{1}{2}} \\left( \\htmlData{anchor=res-dx}{\\color{@amber}{\\frac{1}{2} \\, du}} \\right)"},{"latex":"= \\frac{1}{2} \\int u^{\\frac{1}{2}} \\, du"}],"highlights":[{"color":"sky","style":"text-color","anchor":"int-u"},{"color":"sky","style":"text-color","anchor":"res-u"},{"color":"amber","style":"text-color","anchor":"int-dx"},{"color":"amber","style":"text-color","anchor":"res-dx"}],"connections":[{"to":"res-u","from":"int-u","color":"sky","label":"substitute u","style":"solid"},{"to":"res-dx","from":"int-dx","color":"amber","label":"substitute dx","style":"solid"}]},"order":2,"title":"Substitute into the integral","description":"Now we replace $2x - 7$ with $u$, and $dx$ with $\\frac{1}{2} \\, du$ in our integral."},{"type":"text","order":3,"title":"Integrate using power rule","content":"Now we integrate $u^{\\frac{1}{2}}$ with respect to $u$ using the power rule, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1}$ (for $n \\neq -1$):\n\n$$\\begin{aligned} \\frac{1}{2} \\int u^{\\frac{1}{2}} \\, du &= \\frac{1}{2} \\left( \\frac{u^{\\frac{3}{2}}}{\\frac{3}{2}} \\right) + C \\\\ &= \\frac{1}{2} \\cdot \\frac{2}{3} u^{\\frac{3}{2}} + C \\\\ &= \\frac{1}{3} u^{\\frac{3}{2}} + C \\end{aligned}$$\n\n*Note: This earns us the **M1** mark for obtaining the form $k(2x-7)^{\\frac{3}{2}}$ and the **B1** mark for including the constant of integration $+ C$.*"},{"type":"text","order":4,"title":"Substitute back to x","content":"Finally, we substitute our original definition $u = 2x - 7$ back into the expression:\n\n$$\\frac{1}{3}(2x - 7)^{\\frac{3}{2}} + C$$\n\nThis gives us our final answer, earning the final **A1** mark."}]},{"id":"192e5e23-2d30-4e54-b428-4cb49fa5b683:substitution","label":"(b) Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We want to integrate the given function $e^{4x + 2}$ with respect to $x$:\n\n$$\\int e^{4x+2} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int e^{\\htmlData{anchor=orig-exp}{\\color{@sky}{4x+2}}} \\, \\htmlData{anchor=orig-dx}{\\color{@amber}{dx}}$"},{"latex":"$$\\text{Let } \\htmlData{anchor=sub-u}{\\color{@sky}{u}} = 4x + 2 \\implies \\frac{du}{dx} = 4 \\implies \\htmlData{anchor=sub-dx}{\\color{@amber}{dx}} = \\htmlData{anchor=sub-du}{\\color{@purple}{\\frac{1}{4} du}}$"},{"latex":"$$=\\int e^{\\htmlData{anchor=new-u}{\\color{@sky}{u}}} \\, \\htmlData{anchor=new-dx}{\\color{@purple}{\\frac{1}{4} du}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"orig-exp"},{"color":"sky","style":"text-color","anchor":"sub-u"},{"color":"sky","style":"text-color","anchor":"new-u"},{"color":"purple","style":"text-color","anchor":"sub-du"},{"color":"purple","style":"text-color","anchor":"new-dx"}],"connections":[{"to":"sub-u","from":"orig-exp","color":"sky","label":"Substitute exponent","style":"solid"},{"to":"new-u","from":"sub-u","color":"sky","label":"New variable","style":"solid"},{"to":"sub-dx","from":"orig-dx","color":"amber","label":"Differential term","style":"solid"},{"to":"new-dx","from":"sub-du","color":"purple","label":"Substitute dx","style":"solid"}]},"order":1,"title":"Apply substitution","description":"We choose $u = 4x+2$ to simplify the exponent, then find the differential $dx$ in terms of $du$ to perform the complete change of variables."},{"type":"text","order":2,"title":"Integrate with respect to u","content":"Factor out the constant $\\frac{1}{4}$ and integrate the exponential function:\n\n$$\\int e^u \\left(\\frac{1}{4} du\\right) = \\frac{1}{4} \\int e^u \\, du$$\n\nSince the integral of $e^u$ is simply $e^u$:\n\n$$= \\frac{1}{4} e^u + C$$"},{"type":"text","order":3,"title":"Substitute back to x","content":"Finally, replace $u$ with the original expression $4x + 2$ to get our final answer:\n\n$$\\frac{1}{4} e^{4x+2} + C$$"}]},{"id":"192e5e23-2d30-4e54-b428-4cb49fa5b683:inspection","label":"(b) Direct Integration","steps":[{"type":"text","order":0,"title":"Identify the standard rule","content":"To integrate an exponential function where the exponent is a linear function of the form $ax + b$, we can use the direct integration rule:\n\n$$\\int e^{ax+b} \\, dx = \\frac{1}{a} e^{ax+b} + C$$\n\nFor this problem, we want to find:\n\n$$\\int e^{4x + 2} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int e^{\\htmlData{anchor=rule-a}{\\color{@sky}{a}}x+b} \\, dx = \\frac{1}{\\htmlData{anchor=rule-recip}{\\color{@sky}{a}}}e^{ax+b} + C$"},{"latex":"$$\\int e^{\\htmlData{anchor=val-a}{\\color{@sky}{4}}x+2} \\, dx = \\frac{1}{\\htmlData{anchor=val-recip}{\\color{@sky}{4}}}e^{4x+2} + C$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"rule-a"},{"color":"sky","style":"text-color","anchor":"val-a"},{"color":"sky","style":"text-color","anchor":"rule-recip"},{"color":"sky","style":"text-color","anchor":"val-recip"}],"connections":[{"to":"val-a","from":"rule-a","color":"sky","label":"a = 4","style":"solid"},{"to":"val-recip","from":"rule-recip","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Apply standard formula","description":"We identify that the coefficient of $x$ is $a = 4$ and apply the integration rule directly."},{"type":"text","order":2,"title":"Write final answer","content":"Thus, the result of the integration is:\n\n$$\\frac{1}{4}e^{4x + 2} + C$$\n\n### Markscheme Breakdown\n- **M1**: Integrating to the form $k e^{4x+2}$ where $k$ is a constant.\n- **A1**: Finding the correct coefficient $k = \\frac{1}{4}$.\n- **B1**: Including the constant of integration $+ C$."}]}],"generatedAt":"2026-06-09T04:47:49.337Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"e2a4f4b3-0e2e-4895-ba94-4348ce074c2c","revisionId":"d04e5412-3756-45bf-96e9-9e6a4fbb351b","questionNumber":null,"questionType":"LA","specification":"Find the indefinite integral of each of the following with respect to $x$:","difficulty":1,"difficultyBand":"easy","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"ec4e6030-bcab-412d-b137-3745330a8c4a","label":"a","content":"$$e^{1-4x}$","markscheme":"- Integrates exponential: $-\\frac{1}{4}e^{1-4x} + C$ M1\n- Correct coefficient $-\\frac{1}{4}$ A1\n- Includes constant of integration $+ C$ B1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Evidence of integrating to the form $ke^{1-4x}$.","code":"M"},{"id":"p2","marks":1,"description":"Correct coefficient of $-\\frac{1}{4}$.","code":"A"},{"id":"p3","marks":1,"description":"Includes constant of integration $+ C$.","code":"B"}],"finalAnswers":["-\\frac{1}{4}e^{1-4x} + C"],"generatedAt":"2026-05-06T14:04:15.206Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the substitution","content":"To integrate $\\int e^{1-4x} \\, dx$ using the method of substitution, we start by defining a new variable $u$ to simplify the exponent.\n\nLet:\n$$u = 1 - 4x$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$u = 1 - 4x$$"},{"latex":"$$$\\htmlData{anchor=du-dx}{\\frac{du}{dx}} = \\htmlData{anchor=deriv-val}{-4}$$"},{"latex":"$$$\\htmlData{anchor=dx-eq}{dx} = \\htmlData{anchor=du-sub}{-\\frac{1}{4} \\, du}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"du-dx"},{"color":"amber","style":"text-color","anchor":"deriv-val"},{"color":"purple","style":"text-color","anchor":"dx-eq"},{"color":"purple","style":"text-color","anchor":"du-sub"}],"connections":[{"to":"du-sub","from":"deriv-val","color":"amber","label":"rearrange","style":"solid"}]},"order":1,"title":"Find the differential","description":"Differentiate $u$ with respect to $x$ to find $dx$ in terms of $du$."},{"type":"text","order":2,"title":"Substitute into the integral","content":"Now we substitute $u = 1 - 4x$ and $dx = -\\frac{1}{4} \\, du$ back into our original integral:\n\n$$\\int e^{1-4x} \\, dx = \\int e^u \\left(-\\frac{1}{4} \\, du\\right)$$\n\nWe can factor out the constant $-\\frac{1}{4}$ to make the integration straightforward:\n\n$$= -\\frac{1}{4} \\int e^u \\, du$$"},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Now we perform the integration. Recall that the integral of $e^u$ with respect to $u$ is simply $e^u$:\n\n$$-\\frac{1}{4} \\int e^u \\, du = -\\frac{1}{4} e^u + C$$\n\nwhere $C$ is the constant of integration. \n\n*Note: In exam markschemes, writing down the general integrated exponential term earns you the first method mark (M1), and including the constant $+C$ earns you an independent mark (B1).*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$u = \\htmlData{anchor=orig-val}{\\color{@sky}{1 - 4x}}$$"},{"latex":"$$$-\\frac{1}{4} e^{\\htmlData{anchor=u-var}{\\color{@sky}{u}}} + C$$"},{"latex":"$$$-\\frac{1}{4} e^{\\htmlData{anchor=final-sub}{\\color{@sky}{1 - 4x}}} + C$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"orig-val"},{"color":"sky","style":"text-color","anchor":"u-var"},{"color":"sky","style":"text-color","anchor":"final-sub"}],"connections":[{"to":"final-sub","from":"orig-val","color":"sky","label":"substitute back","style":"solid"}]},"order":4,"title":"Substitute back to get x","description":"Finally, replace $u$ with its original expression in terms of $x$ to get the final answer."},{"type":"text","order":5,"title":"State final answer","content":"The indefinite integral is:\n\n$$-\\frac{1}{4}e^{1-4x} + C$$\n\nGetting the final correct coefficient of $-\\frac{1}{4}$ gives the final accuracy mark (A1)."}]},{"id":"inspection","label":"Reverse Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the integration rule","content":"We need to find the indefinite integral of the exponential function with respect to $x$:\n\n$$\\int e^{1-4x} \\, dx$$\n\nTo integrate an exponential function where the power is a linear function of $x$ (of the form $ax + b$), we can use the **Reverse Chain Rule** rule:\n\n$$\\int e^{ax+b} \\, dx = \\frac{1}{a}e^{ax+b} + C$$\n\nThis rule states that we keep the exponential term the same and divide by the derivative of the linear exponent, $a$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$e^{\\htmlData{anchor=exponent}{\\color{@sky}{1 - 4x}}}$"},{"latex":"$$ax+b = \\htmlData{anchor=linear}{\\color{@sky}{-4x + 1}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"exponent"},{"color":"sky","style":"text-color","anchor":"linear"}],"connections":[{"to":"linear","from":"exponent","color":"sky","label":"linear form","style":"solid"}]},"order":1,"title":"Identify the linear exponent","description":"First, let's identify the linear component in the exponent."},{"type":"text","order":2,"title":"Differentiate the exponent","content":"Now we find the derivative of the exponent, $1 - 4x$, with respect to $x$:\n\n$$\\frac{d}{dx}(1 - 4x) = -4$$\n\nSo, our constant factor $a$ is $-4$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int e^{1-4x} \\, dx = \\frac{1}{\\htmlData{anchor=deriv-val}{\\color{@red}{-4}}} e^{1-4x} + \\htmlData{anchor=const-val}{\\color{@purple}{C}}$"},{"latex":"$$= \\htmlData{anchor=final-coeff}{\\color{@red}{-\\frac{1}{4}}} e^{1-4x} + \\htmlData{anchor=final-const}{\\color{@purple}{C}}$"}],"highlights":[{"color":"red","style":"text-color","anchor":"deriv-val"},{"color":"red","style":"text-color","anchor":"final-coeff"},{"color":"purple","style":"text-color","anchor":"const-val"},{"color":"purple","style":"text-color","anchor":"final-const"}],"connections":[{"to":"final-coeff","from":"deriv-val","color":"red","label":"simplify","style":"solid"}]},"order":3,"title":"Apply the reverse chain rule","description":"According to the Reverse Chain Rule, we divide the original function by the derivative of the exponent (which is $-4$) and add the constant of integration $C$."},{"type":"text","order":4,"title":"State final answer","content":"The final indefinite integral is:\n\n$$-\\frac{1}{4}e^{1-4x} + C$$\n\n### Mark Breakdown:\n* **M1**: Integrating the exponential function to get the form $ke^{1-4x}$.\n* **A1**: Finding the correct coefficient of $-\\frac{1}{4}$.\n* **B1**: Including the constant of integration, $+ C$."}]}],"generatedAt":"2026-06-09T09:19:47.494Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"a6cec0a1-9001-4f65-a141-10f609052bf5","label":"b","content":"$$\\frac{1}{2x + 7}$","markscheme":"- Integrates to natural log: $\\frac{1}{2}\\ln|2x+7| + C$ M1\n- Correct coefficient $\\frac{1}{2}$ A1\n- Includes constant of integration $+ C$ B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Evidence of integrating to the form $k\\ln|2x+7|$.","code":"M"},{"id":"p2","marks":1,"description":"Correct coefficient of $\\frac{1}{2}$.","code":"A"},{"id":"p3","marks":1,"description":"Includes constant of integration $+ C$.","code":"B"}],"finalAnswers":["\\frac{1}{2}\\ln|2x+7| + C"],"generatedAt":"2026-05-06T14:04:15.206Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We want to find the indefinite integral of the given function with respect to $x$:\n\n$$\\int \\frac{1}{2x + 7} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{Let } \\htmlData{anchor=u-def}{\\color{@sky}{u}} = \\htmlData{anchor=u-expr}{\\color{@sky}{2x + 7}}"},{"latex":"\\text{Differentiating both sides gives:}"},{"latex":"\\htmlData{anchor=du-def}{\\color{@amber}{du}} = \\htmlData{anchor=du-expr}{\\color{@amber}{2 \\, dx}} \\implies dx = \\frac{1}{2} \\, du"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"amber","style":"text-color","anchor":"du-def"}],"connections":[{"to":"u-expr","from":"u-def","color":"sky","label":"substitution","style":"solid"}]},"order":1,"title":"Substitute variable u","description":"To make the integration easier, we use substitution. We let $u = 2x + 7$ and find its derivative to replace $dx$."},{"type":"text","order":2,"title":"Rewrite in terms of u","content":"Now we substitute $u$ and $dx = \\frac{1}{2} \\, du$ back into our original integral:\n\n$$\\int \\frac{1}{2x + 7} \\, dx = \\int \\frac{1}{u} \\left(\\frac{1}{2} \\, du\\right)$$\n\nWe can factor out the constant coefficient $\\frac{1}{2}$ to get:\n\n$$\\frac{1}{2} \\int \\frac{1}{u} \\, du$$"},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Recall the standard integration formula from your formula booklet:\n\n$$\\int \\frac{1}{u} \\, du = \\ln|u| + C$$\n\nApplying this standard integral, we obtain:\n\n$$\\frac{1}{2} \\ln|u| + C$$"},{"type":"text","order":4,"title":"Substitute back to x","content":"Finally, we substitute $u = 2x + 7$ back into our result to express our final answer in terms of $x$:\n\n$$\\frac{1}{2} \\ln|2x + 7| + C$$"}]},{"id":"inspection","label":"Reverse Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the standard integral form","content":"We need to find the indefinite integral:\n\n$$\\int \\frac{1}{2x + 7} \\, dx$$\n\nThis expression is a fraction with a linear denominator of the form $ax + b$. We can use the **Reverse Chain Rule** standard integration formula:\n\n$$\\int \\frac{1}{ax+b} \\, dx = \\frac{1}{a}\\ln|ax+b| + C$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{Comparing } \\frac{1}{\\htmlData{anchor=a-term}{\\color{@sky}{2}}x + \\htmlData{anchor=b-term}{\\color{@amber}{7}}} \\text{ to the general form } \\frac{1}{\\htmlData{anchor=a-gen}{\\color{@sky}{a}}x + \\htmlData{anchor=b-gen}{\\color{@amber}{b}}}:"},{"latex":"a = \\htmlData{anchor=a-val}{\\color{@sky}{2}}"},{"latex":"b = \\htmlData{anchor=b-val}{\\color{@amber}{7}}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-term"},{"color":"amber","style":"text-color","anchor":"b-term"}],"connections":[{"to":"a-val","from":"a-term","color":"sky","label":null,"style":"solid"},{"to":"b-val","from":"b-term","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Identify the coefficients","description":"Matching our denominator $2x + 7$ with the general form $ax+b$ allows us to identify the values of the constants $a$ and $b$."},{"type":"text","order":2,"title":"Apply the integration formula","content":"Now, substitute $a = 2$ and $b = 7$ into our standard formula:\n\n$$\\int \\frac{1}{2x+7} \\, dx = \\frac{1}{2}\\ln|2x+7| + C$$\n\nThis simple, direct application of the reverse chain rule avoids the need for a full algebraic substitution."},{"type":"text","order":3,"title":"Understand the marking points","content":"Let's review how the **3 marks** are awarded for this question:\n\n* **M1**: Realising that integrating a linear fraction yields a natural logarithm form, i.e., showing $\\ln|2x+7|$.\n* **A1**: Finding the correct coefficient of $\\frac{1}{2}$ in front of the logarithm.\n* **B1**: Remembering to include the constant of integration, $+ C$.\n\nThus, our final complete answer is:\n\n$$\\frac{1}{2}\\ln|2x+7| + C$$"}]}],"generatedAt":"2026-06-09T09:21:16.200Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"ec4e6030-bcab-412d-b137-3745330a8c4a:substitution","label":"(a) Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the substitution","content":"To integrate $\\int e^{1-4x} \\, dx$ using the method of substitution, we start by defining a new variable $u$ to simplify the exponent.\n\nLet:\n$$u = 1 - 4x$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$u = 1 - 4x$$"},{"latex":"$$$\\htmlData{anchor=du-dx}{\\frac{du}{dx}} = \\htmlData{anchor=deriv-val}{-4}$$"},{"latex":"$$$\\htmlData{anchor=dx-eq}{dx} = \\htmlData{anchor=du-sub}{-\\frac{1}{4} \\, du}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"du-dx"},{"color":"amber","style":"text-color","anchor":"deriv-val"},{"color":"purple","style":"text-color","anchor":"dx-eq"},{"color":"purple","style":"text-color","anchor":"du-sub"}],"connections":[{"to":"du-sub","from":"deriv-val","color":"amber","label":"rearrange","style":"solid"}]},"order":1,"title":"Find the differential","description":"Differentiate $u$ with respect to $x$ to find $dx$ in terms of $du$."},{"type":"text","order":2,"title":"Substitute into the integral","content":"Now we substitute $u = 1 - 4x$ and $dx = -\\frac{1}{4} \\, du$ back into our original integral:\n\n$$\\int e^{1-4x} \\, dx = \\int e^u \\left(-\\frac{1}{4} \\, du\\right)$$\n\nWe can factor out the constant $-\\frac{1}{4}$ to make the integration straightforward:\n\n$$= -\\frac{1}{4} \\int e^u \\, du$$"},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Now we perform the integration. Recall that the integral of $e^u$ with respect to $u$ is simply $e^u$:\n\n$$-\\frac{1}{4} \\int e^u \\, du = -\\frac{1}{4} e^u + C$$\n\nwhere $C$ is the constant of integration. \n\n*Note: In exam markschemes, writing down the general integrated exponential term earns you the first method mark (M1), and including the constant $+C$ earns you an independent mark (B1).*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$u = \\htmlData{anchor=orig-val}{\\color{@sky}{1 - 4x}}$$"},{"latex":"$$$-\\frac{1}{4} e^{\\htmlData{anchor=u-var}{\\color{@sky}{u}}} + C$$"},{"latex":"$$$-\\frac{1}{4} e^{\\htmlData{anchor=final-sub}{\\color{@sky}{1 - 4x}}} + C$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"orig-val"},{"color":"sky","style":"text-color","anchor":"u-var"},{"color":"sky","style":"text-color","anchor":"final-sub"}],"connections":[{"to":"final-sub","from":"orig-val","color":"sky","label":"substitute back","style":"solid"}]},"order":4,"title":"Substitute back to get x","description":"Finally, replace $u$ with its original expression in terms of $x$ to get the final answer."},{"type":"text","order":5,"title":"State final answer","content":"The indefinite integral is:\n\n$$-\\frac{1}{4}e^{1-4x} + C$$\n\nGetting the final correct coefficient of $-\\frac{1}{4}$ gives the final accuracy mark (A1)."}]},{"id":"ec4e6030-bcab-412d-b137-3745330a8c4a:inspection","label":"(a) Reverse Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the integration rule","content":"We need to find the indefinite integral of the exponential function with respect to $x$:\n\n$$\\int e^{1-4x} \\, dx$$\n\nTo integrate an exponential function where the power is a linear function of $x$ (of the form $ax + b$), we can use the **Reverse Chain Rule** rule:\n\n$$\\int e^{ax+b} \\, dx = \\frac{1}{a}e^{ax+b} + C$$\n\nThis rule states that we keep the exponential term the same and divide by the derivative of the linear exponent, $a$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$e^{\\htmlData{anchor=exponent}{\\color{@sky}{1 - 4x}}}$"},{"latex":"$$ax+b = \\htmlData{anchor=linear}{\\color{@sky}{-4x + 1}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"exponent"},{"color":"sky","style":"text-color","anchor":"linear"}],"connections":[{"to":"linear","from":"exponent","color":"sky","label":"linear form","style":"solid"}]},"order":1,"title":"Identify the linear exponent","description":"First, let's identify the linear component in the exponent."},{"type":"text","order":2,"title":"Differentiate the exponent","content":"Now we find the derivative of the exponent, $1 - 4x$, with respect to $x$:\n\n$$\\frac{d}{dx}(1 - 4x) = -4$$\n\nSo, our constant factor $a$ is $-4$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int e^{1-4x} \\, dx = \\frac{1}{\\htmlData{anchor=deriv-val}{\\color{@red}{-4}}} e^{1-4x} + \\htmlData{anchor=const-val}{\\color{@purple}{C}}$"},{"latex":"$$= \\htmlData{anchor=final-coeff}{\\color{@red}{-\\frac{1}{4}}} e^{1-4x} + \\htmlData{anchor=final-const}{\\color{@purple}{C}}$"}],"highlights":[{"color":"red","style":"text-color","anchor":"deriv-val"},{"color":"red","style":"text-color","anchor":"final-coeff"},{"color":"purple","style":"text-color","anchor":"const-val"},{"color":"purple","style":"text-color","anchor":"final-const"}],"connections":[{"to":"final-coeff","from":"deriv-val","color":"red","label":"simplify","style":"solid"}]},"order":3,"title":"Apply the reverse chain rule","description":"According to the Reverse Chain Rule, we divide the original function by the derivative of the exponent (which is $-4$) and add the constant of integration $C$."},{"type":"text","order":4,"title":"State final answer","content":"The final indefinite integral is:\n\n$$-\\frac{1}{4}e^{1-4x} + C$$\n\n### Mark Breakdown:\n* **M1**: Integrating the exponential function to get the form $ke^{1-4x}$.\n* **A1**: Finding the correct coefficient of $-\\frac{1}{4}$.\n* **B1**: Including the constant of integration, $+ C$."}]},{"id":"a6cec0a1-9001-4f65-a141-10f609052bf5:substitution","label":"(b) Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We want to find the indefinite integral of the given function with respect to $x$:\n\n$$\\int \\frac{1}{2x + 7} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{Let } \\htmlData{anchor=u-def}{\\color{@sky}{u}} = \\htmlData{anchor=u-expr}{\\color{@sky}{2x + 7}}"},{"latex":"\\text{Differentiating both sides gives:}"},{"latex":"\\htmlData{anchor=du-def}{\\color{@amber}{du}} = \\htmlData{anchor=du-expr}{\\color{@amber}{2 \\, dx}} \\implies dx = \\frac{1}{2} \\, du"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"amber","style":"text-color","anchor":"du-def"}],"connections":[{"to":"u-expr","from":"u-def","color":"sky","label":"substitution","style":"solid"}]},"order":1,"title":"Substitute variable u","description":"To make the integration easier, we use substitution. We let $u = 2x + 7$ and find its derivative to replace $dx$."},{"type":"text","order":2,"title":"Rewrite in terms of u","content":"Now we substitute $u$ and $dx = \\frac{1}{2} \\, du$ back into our original integral:\n\n$$\\int \\frac{1}{2x + 7} \\, dx = \\int \\frac{1}{u} \\left(\\frac{1}{2} \\, du\\right)$$\n\nWe can factor out the constant coefficient $\\frac{1}{2}$ to get:\n\n$$\\frac{1}{2} \\int \\frac{1}{u} \\, du$$"},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Recall the standard integration formula from your formula booklet:\n\n$$\\int \\frac{1}{u} \\, du = \\ln|u| + C$$\n\nApplying this standard integral, we obtain:\n\n$$\\frac{1}{2} \\ln|u| + C$$"},{"type":"text","order":4,"title":"Substitute back to x","content":"Finally, we substitute $u = 2x + 7$ back into our result to express our final answer in terms of $x$:\n\n$$\\frac{1}{2} \\ln|2x + 7| + C$$"}]},{"id":"a6cec0a1-9001-4f65-a141-10f609052bf5:inspection","label":"(b) Reverse Chain Rule","steps":[{"type":"text","order":0,"title":"Identify the standard integral form","content":"We need to find the indefinite integral:\n\n$$\\int \\frac{1}{2x + 7} \\, dx$$\n\nThis expression is a fraction with a linear denominator of the form $ax + b$. We can use the **Reverse Chain Rule** standard integration formula:\n\n$$\\int \\frac{1}{ax+b} \\, dx = \\frac{1}{a}\\ln|ax+b| + C$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{Comparing } \\frac{1}{\\htmlData{anchor=a-term}{\\color{@sky}{2}}x + \\htmlData{anchor=b-term}{\\color{@amber}{7}}} \\text{ to the general form } \\frac{1}{\\htmlData{anchor=a-gen}{\\color{@sky}{a}}x + \\htmlData{anchor=b-gen}{\\color{@amber}{b}}}:"},{"latex":"a = \\htmlData{anchor=a-val}{\\color{@sky}{2}}"},{"latex":"b = \\htmlData{anchor=b-val}{\\color{@amber}{7}}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-term"},{"color":"amber","style":"text-color","anchor":"b-term"}],"connections":[{"to":"a-val","from":"a-term","color":"sky","label":null,"style":"solid"},{"to":"b-val","from":"b-term","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Identify the coefficients","description":"Matching our denominator $2x + 7$ with the general form $ax+b$ allows us to identify the values of the constants $a$ and $b$."},{"type":"text","order":2,"title":"Apply the integration formula","content":"Now, substitute $a = 2$ and $b = 7$ into our standard formula:\n\n$$\\int \\frac{1}{2x+7} \\, dx = \\frac{1}{2}\\ln|2x+7| + C$$\n\nThis simple, direct application of the reverse chain rule avoids the need for a full algebraic substitution."},{"type":"text","order":3,"title":"Understand the marking points","content":"Let's review how the **3 marks** are awarded for this question:\n\n* **M1**: Realising that integrating a linear fraction yields a natural logarithm form, i.e., showing $\\ln|2x+7|$.\n* **A1**: Finding the correct coefficient of $\\frac{1}{2}$ in front of the logarithm.\n* **B1**: Remembering to include the constant of integration, $+ C$.\n\nThus, our final complete answer is:\n\n$$\\frac{1}{2}\\ln|2x+7| + C$$"}]}],"generatedAt":"2026-06-09T09:21:27.372Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__ccea","summaries":[{"id":"41fb9ee5-982a-4e7b-ba09-a06e0a62b120","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"1ad2598d-4168-4c2d-9266-61dbe77513b6","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"66d1554b-b243-47c0-a69a-9d7efc1f3c97","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"678b8a53-192a-4dc8-8642-f373e1ea4601","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"6a3c5f33-7ee6-4203-9e71-d6f699fe93be","difficultyBand":"easy","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"7242d354-9652-4ce4-b93c-a86a27c4f362","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"94974db3-2c8e-482a-9290-65838478f475","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":false},{"id":"9d98ac71-67b3-4622-a4f2-a035d1537bdf","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"c4e6b947-384c-4f92-8ef3-0ef73e7dafb0","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"e4aae231-ae0a-4dde-89b7-edf0e38302b4","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"e590d6fd-a354-4b0f-b68f-e586836db738","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"e7766021-317c-4299-8e8f-b50a9ec0b559","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"1c062306-1388-4be0-af8a-f4aee2770406","difficultyBand":"easy","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"71396e06-cadf-4d46-b502-86a4b22086a2","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"3b1cdc8b-dd67-45bf-9f55-ab1c7970a359","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"3eacbbf9-f07a-4d1f-8856-766abf8a4f98","difficultyBand":"easy","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"564330e4-3146-4624-92ba-7e0a988422c0","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"82d47be0-9de4-468e-b752-a583b4bc530e","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"946e6230-7a3d-4000-b391-7ce2b0b2fff0","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"bdc8d91d-c276-4450-9b9c-f694f8a23cf2","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"d929e877-ba83-4854-ad2c-efd7438df34b","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"e0b0e605-1cb8-406c-a66b-dd1dcd10a666","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"e2a4f4b3-0e2e-4895-ba94-4348ce074c2c","difficultyBand":"easy","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"e8e07c8c-2869-47ae-b49d-30874035a15e","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":false},{"id":"032b71c5-1f39-4b32-8302-1b7b7935fe15","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"1000ed6d-fe65-4492-9d9e-726a40b78c5a","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"16baf928-c5e9-4af9-85de-ebb2b2243183","difficultyBand":"easy","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"21b2b410-e76f-498c-bc49-84474bad9107","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"2af416db-89f5-4bef-bcc6-a4891fd8c4cf","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"343dd622-7848-4031-bd4e-6d6e5e38de8c","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"3c77626d-a364-4fa7-8697-8aa18fceb88c","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"63ea12ec-34dc-47ab-9af8-68de7fc877e8","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":false},{"id":"681d4c00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Integration (A-level only)","topicId":"53edcf23-243a-46a3-8a7b-03d789a50f35","topicSidebar":"$L3c","topicSlug":"integration-6","topicTitle":"3.6.1 Integration (A-level only)"}]}],"$L3d","$L3e"]}]

3.6.1 Integration (A-level only)

Question 5

3.6.1 Integration (A-level only) Questions

3.6.1 Integration (A-level only)

Question 5

3.6.1 Integration (A-level only) Questions