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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/as-level/maths/edexcel/as-pure-mathematics-13-integration/flashcards","label":"Integration","subjectCode":"as_level_maths","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"5a0c8088-60a0-4f50-a8b8-e714cbdc7267","title":"Integration Flashcards","topicId":"cd143d21-1eb5-48d5-8705-960f6d41cada","topicSlug":"as-pure-mathematics-13-integration","cards":[{"id":"0fd02baa-eefe-46c2-8198-cb3bda39cceb","cardType":"basic","points":1,"front":"What is the relationship between integration and differentiation?","back":"Integration is reverse differentiation (finding the antiderivative).","mcqOptions":[{"text":"Integration is reverse differentiation (finding the antiderivative).","correct":true,"explanation":"Differentiation finds a gradient function from an original function, while integration reverses this process by finding an antiderivative whose derivative matches the given expression."},{"text":"Integration is the same as differentiation (finding the gradient function).","correct":false,"explanation":"These processes are inverse operations, not the same operation. Integration finds an antiderivative, whereas differentiation finds the gradient function of an original function."},{"text":"Integration is reverse multiplication (finding the factor).","correct":false,"explanation":"Multiplication is not reversed by integration. Integration reverses differentiation by finding an antiderivative, which differentiates back to the original expression."},{"text":"Integration is reverse division (finding the quotient).","correct":false,"explanation":"Division is not the inverse process involved here. The correct relationship is that integration reverses differentiation and recovers an antiderivative from its gradient function."}]},{"id":"24431881-20bd-4c5c-8acf-1e90739adb45","cardType":"cloze","points":null,"front":"To integrate a power of $x$ using the reverse power rule: [ ], then divide by the new power.","back":"To integrate a power of $x$ using the reverse power rule: increase the power by 1, then divide by the new power.","mcqOptions":[{"text":"increase the power by 1","correct":true,"explanation":"The reverse power rule changes $x^n$ to a term involving $x^{n+1}$, so the exponent is increased by one before dividing by that new power."},{"text":"decrease the power by 1","correct":false,"explanation":"Decreasing the exponent gives the pattern used in differentiation, not integration. The reverse power rule increases the power by one before division."},{"text":"leave the power unchanged","correct":false,"explanation":"Keeping the exponent unchanged misses the essential exponent change in integration. The correct method increases the power by one, then divides by the resulting exponent."},{"text":"multiply the power by 1","correct":false,"explanation":"Multiplying an exponent by one leaves it unchanged and does not produce the reverse power rule. Integration requires increasing the power by one before dividing."}]},{"id":"926a09bf-9130-42be-9fd6-ddd6bcd90038","cardType":"cloze","points":null,"front":"To integrate a power of $x$ using the reverse power rule: increase the power by 1, then [ ].","back":"To integrate a power of $x$ using the reverse power rule: increase the power by 1, then divide by the new power.","mcqOptions":[{"text":"divide by the new power","correct":true,"explanation":"After increasing the exponent from $n$ to $n+1$, division by $n+1$ compensates for the factor produced when the integrated expression is differentiated."},{"text":"multiply by the new power","correct":false,"explanation":"Multiplication would introduce an extra factor of the new exponent when differentiating. The reverse power rule divides by the new power instead."},{"text":"divide by the original power","correct":false,"explanation":"The denominator must match the exponent after it has been increased, not the original exponent. Using the original power gives the wrong coefficient."},{"text":"subtract the new power","correct":false,"explanation":"Subtracting an exponent is unrelated to the coefficient adjustment required by the power rule. The integrated term must be divided by its new power."}]},{"id":"f0a4386f-b108-42ff-8517-940c22058d4b","cardType":"basic","points":1,"front":"State the general rule for $\\int ax^n \\, dx$ where $n \\neq -1$.","back":"$$$\\frac{a}{n+1}x^{n+1}+C$$","mcqOptions":[{"text":"$$$\\frac{a}{n+1}x^{n+1}+C$$","correct":true,"explanation":"Increasing the power from $n$ to $n+1$ and dividing by $n+1$ gives an antiderivative whose derivative is $ax^n$; the constant $C$ represents any unknown constant."},{"text":"$$$\\frac{a}{n}x^n+C$$","correct":false,"explanation":"This leaves the power unchanged and divides by the original exponent, so differentiating does not generally recover $ax^n$. The correct rule uses $\\frac{a}{n+1}x^{n+1}+C$."},{"text":"$$$\\frac{a}{n+1}x^n+C$$","correct":false,"explanation":"The coefficient adjustment is correct, but the exponent has not been increased. The correct antiderivative raises the power to $n+1$ as well as dividing by $n+1$."},{"text":"$$$a(n+1)x^{n+1}+C$$","correct":false,"explanation":"Multiplying by $n+1$ reverses the required coefficient adjustment and produces an excessive factor on differentiation. The rule divides by $n+1$, giving $\\frac{a}{n+1}x^{n+1}+C$."}]},{"id":"8ee538f1-df0f-4b48-8e66-c5bc50623500","cardType":"basic","points":2,"front":"Why must a constant of integration ($C$) be added to indefinite integrals?","back":"Differentiating any constant results in zero, so the original constant cannot be known from the gradient function alone.","mcqOptions":[{"text":"Differentiating any constant results in zero, so the original constant cannot be known from the gradient function alone.","correct":true,"explanation":"All constant terms disappear when differentiated, so different functions can have the same gradient. The arbitrary constant $C$ records this missing information in an indefinite integral."},{"text":"Differentiating any constant results in one, so the original constant can be found from the gradient function alone.","correct":false,"explanation":"The derivative of a constant is zero, not one, so constant terms disappear from the gradient. Therefore an arbitrary $C$ must be included because the original constant is unknown."},{"text":"Differentiating any constant results in the constant itself, so the original constant can be found from the gradient function alone.","correct":false,"explanation":"Constants do not remain visible after differentiation; their derivatives are zero. Since the gradient cannot reveal the original constant, indefinite integrals require an added $C$."},{"text":"Differentiating any constant results in an undefined value, so the original constant cannot be used in the gradient function.","correct":false,"explanation":"The derivative of a constant is defined and equals zero. This loss of the constant term, rather than an undefined result, is why an arbitrary constant $C$ is added."}]},{"id":"911743d4-7393-4814-9e7c-96981e67d5f6","cardType":"basic","points":1,"front":"How do you verify if an integration result is correct?","back":"Differentiate the answer; it should return the original expression.","mcqOptions":[{"text":"Differentiate the answer; it should return the original expression.","correct":true,"explanation":"Differentiation reverses integration, so differentiating the proposed antiderivative should reproduce the original integrand. The constant disappears because its derivative is zero."},{"text":"Integrate the answer again; it should return the original expression.","correct":false,"explanation":"Integrating again produces a further antiderivative rather than checking the result. Differentiating the answer is the verification method because it should recover the original integrand."},{"text":"Substitute zero into the answer; it should return the original expression.","correct":false,"explanation":"Substitution at one value only tests a single output and cannot verify the whole function. Differentiating the answer checks whether its gradient equals the original expression everywhere."},{"text":"Set the answer equal to zero; it should return the original expression.","correct":false,"explanation":"Setting an antiderivative equal to zero tests a particular equation, not whether it differentiates correctly. The answer should instead be differentiated to recover the original integrand."}]},{"id":"d6b3c8b6-0e04-483f-929b-70b9bcc3e4a0","cardType":"cloze","points":null,"front":"When integrating an expression like $(x+5)(x-2)$, you must [ ] before applying the power rule.","back":"When integrating an expression like $(x+5)(x-2)$, you must expand the brackets before applying the power rule.","mcqOptions":[{"text":"expand the brackets","correct":true,"explanation":"The power rule applies directly to individual powers of $x$, so the product must first be expanded into a polynomial before each term is integrated."},{"text":"differentiate the brackets","correct":false,"explanation":"Differentiating the product would change the task and does not prepare the expression for direct integration. The brackets should be expanded into separate powers of $x$."},{"text":"factorise the brackets","correct":false,"explanation":"Factorising does not produce the separate powers needed for straightforward use of the power rule. The correct preliminary step is to expand the brackets."},{"text":"set the brackets equal to zero","correct":false,"explanation":"Solving for the roots does not integrate the function or create terms suitable for the power rule. Expand the brackets first, then integrate the resulting polynomial terms."}]},{"id":"b7ab0c28-68c9-449e-b9aa-abe8c9101bec","cardType":"basic","points":1,"front":"How is a constant term $k$ integrated with respect to $x$?","back":"It becomes a multiple of $x$: $kx + C$.","mcqOptions":[{"text":"It becomes a multiple of $x$: $kx + C$.","correct":true,"explanation":"A constant $k$ can be viewed as $kx^0$. Increasing the power to one and dividing by one gives $kx$, with the arbitrary constant $C$ added."},{"text":"It remains a constant: $k + C$.","correct":false,"explanation":"Differentiating $k+C$ gives zero, so it cannot be an antiderivative of the constant $k$. Integrating $k$ with respect to $x$ gives $kx+C$."},{"text":"It becomes a quadratic term: $kx^2 + C$.","correct":false,"explanation":"A quadratic differentiates to a term involving $x$, not a constant. A constant integrates to a linear term, namely $kx+C$."},{"text":"It becomes a reciprocal term: $\\frac{k}{x} + C$.","correct":false,"explanation":"A reciprocal differentiates to a term involving $x^{-2}$, not a constant. Since the derivative of $kx$ is $k$, the integral is $kx+C$."}]},{"id":"bce4ce8b-f1ee-41c5-80dd-3ee4a86ae3d0","cardType":"cloze","points":null,"front":"To integrate roots or fractions, rewrite them as powers:

$\\sqrt{x} = $ $\\text{[...]}$
$\\frac{1}{x^n} = $ $x^{-n}$

","back":"To integrate roots or fractions, rewrite them as powers:

$\\sqrt{x} = $ $\\boxed{x^{\\frac{1}{2}}}$
$\\frac{1}{x^n} = $ $x^{-n}$

","mcqOptions":[{"text":"$$x^{\\frac{1}{2}}$","correct":true,"explanation":"A square root is equivalent to raising a number to the power of one half, so rewriting it as $x^{\\frac{1}{2}}$ allows the reverse power rule to be applied."},{"text":"$$x^2$","correct":false,"explanation":"Squaring gives $x^2$, not the square root. The correct fractional-power form is $x^{\\frac{1}{2}}$, because the root index becomes the denominator of the exponent."},{"text":"$$x^{-\\frac{1}{2}}$","correct":false,"explanation":"The negative exponent represents the reciprocal of the square root, $\\frac{1}{\\sqrt{x}}$. The square root itself is $x^{\\frac{1}{2}}$, with a positive exponent."},{"text":"$$x^{\\frac{1}{3}}$","correct":false,"explanation":"An exponent of $\\frac{1}{3}$ represents a cube root, not a square root. The square root corresponds to the exponent $\\frac{1}{2}$."}]},{"id":"1fc972be-f5f6-43ce-97b1-a9fd548cf7e1","cardType":"cloze","points":null,"front":"To integrate roots or fractions, rewrite them as powers:

$\\sqrt{x} = $ $x^{\\frac{1}{2}}$
$\\frac{1}{x^n} = $ $\\text{[...]}$

","back":"To integrate roots or fractions, rewrite them as powers:

$\\sqrt{x} = $ $x^{\\frac{1}{2}}$
$\\frac{1}{x^n} = $ $\\boxed{x^{-n}}$

","mcqOptions":[{"text":"$$x^{-n}$","correct":true,"explanation":"A denominator containing $x^n$ can be moved to the numerator by applying a negative exponent, giving $\\frac{1}{x^n}=x^{-n}$."},{"text":"$$x^n$","correct":false,"explanation":"Removing the denominator without changing the exponent loses the reciprocal relationship. The correct rewrite is $x^{-n}$, because a negative exponent represents division by $x^n$."},{"text":"$$x^{\\frac{1}{n}}$","correct":false,"explanation":"The exponent $\\frac{1}{n}$ represents an nth root, not the reciprocal of $x^n$. The correct form for $\\frac{1}{x^n}$ is $x^{-n}$."},{"text":"$$x^{-\\frac{1}{n}}$","correct":false,"explanation":"This combines a reciprocal exponent with a root, giving the reciprocal of an nth root. Only the sign changes when rewriting $\\frac{1}{x^n}$, so the answer is $x^{-n}$."}]},{"id":"706486ba-6ae7-447b-9556-c4c6515b4968","cardType":"basic","points":1,"front":"What specific power of $x$ cannot be integrated using the reverse power rule?","back":"$$x^{-1}$ (or $\\frac{1}{x}$)","mcqOptions":[{"text":"$$x^{-1}$ (or $\\frac{1}{x}$)","correct":true,"explanation":"The reverse power rule would require division by the new power, which is impossible when the original power is $-1$. This case integrates to $\\ln|x|+C$ instead."},{"text":"$$x^0$","correct":false,"explanation":"The power rule works for $x^0=1$, whose integral is $x+C$. The exceptional power is $x^{-1}$, which integrates to $\\ln|x|+C$."},{"text":"$$x^1$","correct":false,"explanation":"The reverse power rule integrates $x$ normally to $\\frac{x^2}{2}+C$. The excluded power is $x^{-1}$, because increasing it by one produces zero."},{"text":"$$x^{-2}$","correct":false,"explanation":"The reverse power rule works for $x^{-2}$, producing $-x^{-1}+C$. It fails specifically for $x^{-1}$, whose integral is logarithmic."}]},{"id":"eca902cc-b794-43e8-a31e-131f4da92bad","cardType":"basic","points":3,"front":"Outline the steps to find the specific constant of integration $C$.","back":"

Integrate the derivative to find the general solution.
Substitute the coordinates of a known point $(x, y)$.
Solve the resulting equation for $C$.

","mcqOptions":[{"text":"

Integrate the derivative to find the general solution.
Substitute the coordinates of a known point $(x, y)$.
Solve the resulting equation for $C$.

","correct":true,"explanation":"Integration first introduces the arbitrary constant, then a known point supplies an equation involving $C$, which can be rearranged to determine its specific value."},{"text":"

Differentiate the derivative to find the general solution.
Substitute the coordinates of a known point $(x, y)$.
Solve the resulting equation for $C$.

","correct":false,"explanation":"Differentiating moves away from the required antiderivative and does not introduce the integration constant. The first step should be to integrate the derivative, then use the known point."},{"text":"

Integrate the derivative to find the general solution.
Substitute the coordinates of the origin $(0, 0)$.
Solve the resulting equation for $C$.

","correct":false,"explanation":"The given point may not be the origin, so substituting $(0,0)$ can produce the wrong constant. Use the coordinates of the known point provided in the question."},{"text":"

Integrate the derivative to find the general solution.
Substitute the coordinates of a known point $(x, y)$.
Differentiate the resulting equation for $C$.

","correct":false,"explanation":"Differentiation cannot determine a numerical constant; it removes constants instead. After substituting the known point, solve the resulting equation algebraically for $C$."}]},{"id":"ced1e014-7613-4e61-9b61-e818e037d839","cardType":"basic","points":2,"front":"What is the primary difference between the output of an indefinite integral and a definite integral?","back":"

Indefinite: A function (family of curves).
Definite: A number (often representing area).

","mcqOptions":[{"text":"

Indefinite: A function (family of curves).
Definite: A number (often representing area).

","correct":true,"explanation":"An indefinite integral has no limits and therefore gives a family of antiderivative functions. A definite integral has limits and evaluates to one numerical value."},{"text":"

Indefinite: A number (often representing area).
Definite: A function (family of curves).

","correct":false,"explanation":"These outputs have been reversed. An indefinite integral gives a family of functions, whereas evaluating a definite integral between limits gives a number."},{"text":"

Indefinite: A function (single curve).
Definite: A number (often representing gradient).

","correct":false,"explanation":"An indefinite integral includes $C$, so it represents a family rather than one curve. A definite integral commonly represents signed area, not gradient."},{"text":"

Indefinite: A number (constant of integration).
Definite: A function (family of curves).

","correct":false,"explanation":"The constant $C$ is part of an indefinite integral but does not make the result a number alone. Definite integration removes $C$ and produces a numerical value."}]},{"id":"74d4d01a-1cad-4fc3-87e6-98a03df09362","cardType":"cloze","points":null,"front":"The notation $[F(x)]_a^b$ indicates the calculation $\\text{[...]}$.","back":"The notation $[F(x)]_a^b$ indicates the calculation $\\boxed{F(b) - F(a)}$.","mcqOptions":[{"text":"$$F(b) - F(a)$","correct":true,"explanation":"The notation means evaluate the antiderivative at the upper limit $b$, evaluate it at the lower limit $a$, and subtract the lower result from the upper result."},{"text":"$$F(a) - F(b)$","correct":false,"explanation":"This reverses the order of the limits and changes the sign of the result. The correct calculation is upper value minus lower value: $F(b)-F(a)$."},{"text":"$$F(b) + F(a)$","correct":false,"explanation":"Definite integration requires subtracting the antiderivative values, not adding them. The correct calculation is $F(b)-F(a)$, corresponding to upper limit minus lower limit."},{"text":"$$F(a) + F(b)$","correct":false,"explanation":"Adding the endpoint values ignores the subtraction built into evaluating limits. The notation requires the upper endpoint value minus the lower endpoint value: $F(b)-F(a)$."}]},{"id":"410c2614-b8b8-42f1-a324-9bf8265c27c2","cardType":"basic","points":1,"front":"Geometrically, what does the definite integral $\\int_a^b f(x) \\, dx$ represent if the curve is above the $x$-axis?","back":"The area under the curve between $x=a$ and $x=b$.","mcqOptions":[{"text":"The area under the curve between $x=a$ and $x=b$.","correct":true,"explanation":"When the curve lies above the x-axis, the definite integral accumulates positive values of $f(x)$ and therefore equals the area between the curve and the x-axis."},{"text":"The gradient of the curve between $x=a$ and $x=b$.","correct":false,"explanation":"Gradient describes the rate of change at a point, usually found using differentiation. A definite integral measures accumulated area, which here is the area under the curve."},{"text":"The length of the curve between $x=a$ and $x=b$.","correct":false,"explanation":"Arc length requires a different formula involving the derivative. A definite integral of $f(x)$ represents the area under the curve when it remains above the x-axis."},{"text":"The volume beneath the curve between $x=a$ and $x=b$.","correct":false,"explanation":"Volume requires rotating a region or integrating an appropriate cross-sectional area. A single integral of $f(x)$ here gives the area under the curve."}]},{"id":"13721334-eeb0-4327-a67a-8ca6808b35d5","cardType":"basic","points":2,"front":"How do you determine the limits of integration for the area enclosed between two curves?","back":"Find the $x$-coordinates of the intersection points by setting the two equations equal to each other.","mcqOptions":[{"text":"Find the $x$-coordinates of the intersection points by setting the two equations equal to each other.","correct":true,"explanation":"At an intersection, both curves have the same $y$-value for the same $x$. Equating their equations finds the boundary x-values, which become the integration limits."},{"text":"Find the $y$-coordinates of the intersection points by setting the two equations equal to each other.","correct":false,"explanation":"Setting the equations equal first gives the x-values where the curves meet, not directly their y-values. Those x-coordinates provide the limits of integration."},{"text":"Find the $x$-coordinates of the turning points by setting the derivative equal to zero.","correct":false,"explanation":"Turning points are found using derivatives, but they do not generally mark where two curves meet. Integration limits come from the x-coordinates of the intersection points."},{"text":"Find the $x$-coordinates of the intercepts by setting each equation equal to zero.","correct":false,"explanation":"Intercepts show where a curve crosses an axis, not necessarily where two curves meet. Set the two curve equations equal to find the intersection x-coordinates."}]},{"id":"5dd1906b-9cdc-4872-aac4-afb04fa64cf1","cardType":"basic","points":1,"front":"State the formula for the area between two curves where $f(x)$ is above $g(x)$ between $a$ and $b$.","back":"$$$\\text{Area} = \\int_a^b (f(x) - g(x)) \\, dx$$","mcqOptions":[{"text":"$$$\\text{Area} = \\int_a^b (f(x) - g(x)) \\, dx$$","correct":true,"explanation":"Because $f(x)$ is above $g(x)$ throughout the interval, the vertical height is $f(x)-g(x)$. Integrating that height from $a$ to $b$ gives the area."},{"text":"$$$\\text{Area} = \\int_a^b (g(x) - f(x)) \\, dx$$","correct":false,"explanation":"This reverses the vertical height, producing a negative value when $f(x)$ is above $g(x)$. The correct setup subtracts the lower curve from the upper curve: $\\int_a^b(f(x)-g(x))\\,dx$."},{"text":"$$$\\text{Area} = \\int_a^b (f(x) + g(x)) \\, dx$$","correct":false,"explanation":"Adding the functions does not represent the vertical distance between the curves. Since $f(x)$ is higher, the required height is their difference, giving $\\int_a^b(f(x)-g(x))\\,dx$."},{"text":"$$$\\text{Area} = \\int_a^b \\frac{f(x)}{g(x)} \\, dx$$","correct":false,"explanation":"A quotient measures neither the vertical separation nor the enclosed area between the curves. The area uses the upper-minus-lower difference, so the correct formula is $\\int_a^b(f(x)-g(x))\\,dx$."}]},{"id":"5ccfcbbd-9d01-4443-ad10-588f7e0f61ec","cardType":"cloze","points":null,"front":"If a curve crosses the $x$-axis, you must [ ] at the roots and treat the area below the axis as positive.","back":"If a curve crosses the $x$-axis, you must split the integral at the roots and treat the area below the axis as positive.","mcqOptions":[{"text":"split the integral","correct":true,"explanation":"The roots divide the interval into parts where the curve may lie above or below the axis. Splitting there allows each region’s geometric area to be counted positively."},{"text":"combine the roots","correct":false,"explanation":"Combining the roots ignores the points where the curve changes sides of the axis, so signed areas may cancel. The correct method is to split the integral at each root."},{"text":"differentiate the roots","correct":false,"explanation":"Differentiating does not separate regions with different signs, and roots are values of $x$, not functions to differentiate. The integral must be split at the roots instead."},{"text":"average the roots","correct":false,"explanation":"Averaging the roots gives only a midpoint and does not identify every sign change. The correct approach is to split the integral at the roots, then handle each region separately."}]},{"id":"eff9eed3-a341-4a30-b77a-a463bf80ea31","cardType":"cloze","points":null,"front":"If a curve crosses the $x$-axis, you must split the integral at the roots and treat the area [ ] as positive.","back":"If a curve crosses the $x$-axis, you must split the integral at the roots and treat the area below the axis as positive.","mcqOptions":[{"text":"below the axis","correct":true,"explanation":"A definite integral below the $x$-axis is negative because its height is negative. For geometric area, that contribution must be treated as positive by reversing its sign."},{"text":"above the axis","correct":false,"explanation":"Regions above the axis already contribute positively to a definite integral, so they do not need this correction. It is the area below the axis that must be made positive."},{"text":"along the axis","correct":false,"explanation":"A region along the axis has zero vertical height and contributes no area. The sign issue arises for regions below the axis, whose integral is negative but whose geometric area is positive."},{"text":"between the curves","correct":false,"explanation":"The phrase describes a different area setup and does not identify the region with a negative integral. The area below the axis must be treated as positive instead."}]},{"id":"4d0d030b-5595-4561-aecd-2336937a9059","cardType":"basic","points":1,"front":"What does the $dx$ in the symbol $\\int f(x) \\, dx$ signify?","back":"The variable you are integrating with respect to.","mcqOptions":[{"text":"The variable you are integrating with respect to.","correct":true,"explanation":"The $dx$ identifies $x$ as the variable of integration. It indicates that the function is being integrated using infinitesimal changes in $x$."},{"text":"The function you are integrating with respect to.","correct":false,"explanation":"The function is $f(x)$, while $dx$ identifies the input variable whose small changes are accumulated. Therefore, the correct description is the variable of integration."},{"text":"The limit you are integrating with respect to.","correct":false,"explanation":"Limits determine the interval only in a definite integral; they are not what $dx$ represents. The notation specifies the variable of integration, namely $x$."},{"text":"The gradient you are integrating with respect to.","correct":false,"explanation":"A gradient describes a rate of change and is not identified by $dx$. This notation indicates that integration is performed with respect to the variable $x$."}]},{"id":"6d52fb82-0af1-4f50-b22f-d0814d1e8383","cardType":"cloze","points":null,"front":"Before integrating $\\frac{4}{x^2}$, you should rewrite it as $\\text{[...]}$.","back":"Before integrating $\\frac{4}{x^2}$, you should rewrite it as $\\boxed{4x^{-2}}$.","mcqOptions":[{"text":"$$4x^{-2}$","correct":true,"explanation":"Since $x^2$ is in the denominator, the reciprocal rule gives $1/x^2=x^{-2}$. Multiplying by 4 therefore produces $4x^{-2}$, which can be integrated using the power rule."},{"text":"$$4x^2$","correct":false,"explanation":"Moving a factor from the denominator to the numerator requires changing the exponent’s sign, not retaining it. The correct rewrite is $4x^{-2}$, because $x^{-2}=1/x^2$."},{"text":"$$\\frac{1}{4}x^{-2}$","correct":false,"explanation":"The coefficient 4 remains a multiplier when the denominator is rewritten with a negative power. Thus $4/x^2$ becomes $4x^{-2}$, not one quarter of that expression."},{"text":"$$4x^{-\\frac{1}{2}}$","correct":false,"explanation":"The exponent changes sign when $x^2$ moves from the denominator, but its magnitude stays 2. The correct power is therefore $x^{-2}$, giving $4x^{-2}$."}]},{"id":"e60244e2-81d7-48fc-acb6-54a2c62c924d","cardType":"basic","points":1,"front":"In a definite integral calculation, when is the constant of integration $C$ omitted?","back":"Always; it cancels out during the subtraction $F(b) - F(a)$.","mcqOptions":[{"text":"Always; it cancels out during the subtraction $F(b) - F(a)$.","correct":true,"explanation":"For a definite integral, the antiderivative is evaluated at both limits. The same constant appears in $F(b)$ and $F(a)$, so subtracting them makes the constants cancel."},{"text":"Only when the integrand is positive throughout the interval.","correct":false,"explanation":"The sign of the integrand does not affect cancellation of the constant. The same constant appears at both endpoints, so it is always omitted in a definite integral."},{"text":"Only when the limits are integers.","correct":false,"explanation":"Whether the limits are integers, fractions, or other valid values makes no difference. The constant cancels because it is added to both endpoint evaluations."},{"text":"Never; it must be included in the final answer.","correct":false,"explanation":"Including the constant at both endpoint evaluations would still make it cancel during subtraction. Therefore, a definite integral is written without $C$ in the final result."}]},{"id":"8c7fede0-fc64-45a4-ac02-be690e8a589d","cardType":"cloze","points":null,"front":"To decide which curve is \"top\" for an area calculation, [ ] between the limits into both functions and compare the $y$-values.","back":"To decide which curve is \"top\" for an area calculation, substitute an x-value between the limits into both functions and compare the $y$-values.","mcqOptions":[{"text":"substitute an x-value","correct":true,"explanation":"Choosing any $x$-value strictly between the limits and evaluating both functions gives their $y$-values at the same position. The larger value identifies the top curve there."},{"text":"differentiate each function","correct":false,"explanation":"Differentiation compares gradients, not the curves’ vertical positions. To identify the top curve, use the same interior $x$-value in both functions and compare their outputs."},{"text":"set the functions equal","correct":false,"explanation":"Equating the functions finds intersection points, which may be the limits or internal crossings, but does not identify which curve is higher between them. Test an interior $x$-value instead."},{"text":"integrate the limits","correct":false,"explanation":"Limits are input values rather than functions whose integrals reveal vertical order. Substitute an interior $x$-value into both functions and compare the resulting $y$-values."}]}]}],"examCountdown":{"examDate":"2027-05-13","paperLabel":"Paper 1"},"heading":"Integration","title":"Integration Flashcards","topbar":{"backHref":"/as-level/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"cd143d21-1eb5-48d5-8705-960f6d41cada","walled":"$L39"}],"$L3a","$L3b"]