Techniques & Applications of Integration (DP IB Analysis & Approaches (AA))

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  • What is integral sin x space straight d x?

    integral sin x space straight d x equals negative cos x plus C

    This is in the exam formula booklet.

  • What is integral cos x space straight d x?

    integral cos x space straight d x equals sin x plus C

    This is in the exam formula booklet.

  • True or False?

    integral sin open parentheses a x plus b close parentheses equals negative 1 over a cos open parentheses a x close parentheses plus C

    False.

    integral sin open parentheses a x plus b close parentheses equals negative 1 over a cos open parentheses a x plus b close parentheses plus C

    This is not in the exam formula booklet.

  • What is integral straight e to the power of x space straight d x?

    integral straight e to the power of x space straight d x equals straight e to the power of x plus C

    This is in the exam formula booklet.

  • What is integral 1 over x space straight d x?

    integral 1 over x space straight d x equals ln open vertical bar x close vertical bar plus C

    This is in the exam formula booklet.

  • True or False?

    integral straight e to the power of a x plus b end exponent space straight d x equals straight e to the power of a x plus b end exponent plus C.

    False.

    integral straight e to the power of a x plus b end exponent space straight d x equals 1 over a straight e to the power of a x plus b end exponent plus C.

    This is not in the exam formula booklet.

  • What is integral cos open parentheses a x plus b close parentheses space straight d x?

    integral cos open parentheses a x plus b close parentheses space straight d x equals 1 over a sin open parentheses a x plus b close parentheses plus C

    This is not in the exam formula booklet.

  • What is integral fraction numerator 1 over denominator a x plus b end fraction space straight d x?

    integral fraction numerator 1 over denominator a x plus b end fraction space straight d x equals 1 over a ln open vertical bar a x plus b close vertical bar plus C

    This is not in the exam formula booklet.

  • True or False?

    integral open parentheses a x plus b close parentheses to the power of n space straight d x equals fraction numerator 1 over denominator a open parentheses n plus 1 close parentheses end fraction open parentheses a x plus b close parentheses to the power of n plus 1 end exponent plus C, so long as n not equal to negative 1.

    True.

    integral open parentheses a x plus b close parentheses to the power of n space straight d x equals fraction numerator 1 over denominator a open parentheses n plus 1 close parentheses end fraction open parentheses a x plus b close parentheses to the power of n plus 1 end exponent plus C, so long as n not equal to negative 1.

    This is not in the exam formula booklet.

    If n not equal to negative 1, use integral fraction numerator 1 over denominator a x plus b end fraction space straight d x equals 1 over a ln open vertical bar a x plus b close vertical bar plus C.

  • True or False?

    When integrating trigonometric functions, angles should be measured in degrees.

    False.

    When integrating trigonometric functions, angles must be measured in radians.

  • What is integral sec squared x space straight d x?

    integral sec squared x space straight d x equals tan x plus C

    This is not in the exam formula booklet, but the standard derivative f open parentheses x close parentheses equals tan x space rightwards double arrow space f to the power of apostrophe open parentheses x close parentheses equals sec squared x is in the formula booklet. The integral is just the inverse of that.

  • What is integral sec squared open parentheses a x plus b close parentheses space straight d x?

    integral sec squared open parentheses a x plus b close parentheses space straight d x equals 1 over a tan open parentheses a x plus b close parentheses plus C

    This is not in the exam formula booklet.

  • What is reverse chain rule?

    Reverse chain rule, also known as integrating by inspection, is spotting that the chain rule could have been used to differentiate another function and turn it into the function you are trying to integrate.

  • True or false?

    An integral in the form integral fraction numerator f to the power of apostrophe open parentheses x close parentheses over denominator f open parentheses x close parentheses end fraction space straight d x can be integrated with the reverse chain rule.

    True.

    An integral in the form integral fraction numerator f to the power of apostrophe open parentheses x close parentheses over denominator f open parentheses x close parentheses end fraction space straight d x can be integrated with the reverse chain rule, by using integral fraction numerator f to the power of apostrophe open parentheses x close parentheses over denominator f open parentheses x close parentheses end fraction space straight d x equals ln vertical line f open parentheses x close parentheses vertical line plus C.

    E.g. integral fraction numerator 2 x plus 7 over denominator x squared plus 7 x minus 12 end fraction space straight d x equals ln open vertical bar x squared plus 7 x minus 12 close vertical bar plus C

    This is not in the exam formula booklet.

  • How can you integrate an integral of the form integral g to the power of apostrophe open parentheses x close parentheses f apostrophe open parentheses g open parentheses x close parentheses close parentheses space straight d x with the reverse chain rule?

    You can integrate an integral of the form integral g to the power of apostrophe open parentheses x close parentheses f apostrophe open parentheses g open parentheses x close parentheses close parentheses space straight d x with the reverse chain rule by using integral g to the power of apostrophe open parentheses x close parentheses f apostrophe open parentheses g open parentheses x close parentheses close parentheses space straight d x equals f open parentheses g open parentheses x close parentheses close parentheses plus C.

    E.g. integral 2 x cos open parentheses x squared close parentheses space straight d x equals sin open parentheses x squared close parentheses plus C.

    This is not in the exam formula booklet.

  • What is the "adjust and compensate" method for integration?

    The adjust and compensate method is a technique used in reverse chain rule integration to deal with coefficients that don't match exactly.

    E.g. rewriting integral x cos open parentheses x squared close parentheses space straight d x as 1 half integral 2 x cos open parentheses x squared close parentheses space straight d x.

    • the thing inside the integral has been adjusted by multiplying by 2

    • 1 half is placed in front of the integral to compensate

    2 x cos open parentheses x squared close parentheses can be integrated directly using reverse chain rule.

  • What is integration by substitution?

    Integration by substitution is a method used (e.g. when reverse chain rule is difficult to spot or awkward to use) to simplify an integral by rewriting it in terms of an alternative variable.

    E.g. an integral in x might be rewritten as a simpler integral in u.

  • What is the first step in integration by substitution?

    The first step in integration by substitution is to identify the substitution to be used.

    When integrating a composite function, f open parentheses g open parentheses x close parentheses close parentheses, the substitution will frequently be u equal to the secondary (or 'inside') function in the composite function, i.e u equals g open parentheses x close parentheses.

  • True or False?

    In integration by substitution, fraction numerator straight d u over denominator straight d x end fraction can be treated like a fraction.

    True.

    In integration by substitution, fraction numerator straight d u over denominator straight d x end fraction can be treated like a fraction.

    (Note: it is not actually a fraction, it is a gradient. But it works to treat it like a fraction in integrations like this.)

  • What would be a suitable substitution to use when integrating integral open parentheses 2 x plus 5 close parentheses straight e to the power of x squared plus 5 x minus 3 end exponent space straight d x?

    When integrating integral open parentheses 2 x plus 5 close parentheses straight e to the power of x squared plus 5 x minus 3 end exponent space straight d x, a suitable substitution would be u equals x squared plus 5 x minus 3.

    Then fraction numerator straight d u over denominator straight d x end fraction equals 2 x plus 5 space space rightwards double arrow space space straight d u equals open parentheses 2 x plus 5 close parentheses straight d x, so the integral can be rewritten as integral straight e to the power of u space straight d u.

  • What should be done with the integration limits when using substitution for a definite integral?

    When using substitution for a definite integral, the integration limits should be changed from x-values to u-values.

  • What is the final step in integration by substitution for an indefinite integral?

    The final step in integration by substitution for an indefinite integral is to substitute x (or whatever the original variable is) back in.

    The final answer should always be in terms of the original variable.

  • True or False?

    When using substitution for a definite integral, you must always substitute x back in before evaluating.

    False.

    When using substitution for a definite integral, you can evaluate using the u limits without substituting x back in.

  • What is a definite integral?

    A definite integral is an integral with specified upper and lower integration limits.

    A definite integral is represented in the form integral subscript a superscript b f open parentheses x close parentheses space straight d x

    Where:

    • a is the lower integration limit

    • b is the upper integration limit

  • A definite integral can be evaluated using the equation integral subscript a superscript b f open parentheses x close parentheses space straight d x equals F open parentheses b close parentheses minus F open parentheses a close parentheses.

    What is the function F open parentheses x close parentheses, whose values appear on the right-hand side of the equation?

    In the equation integral subscript a superscript b f open parentheses x close parentheses space straight d x equals F open parentheses b close parentheses minus F open parentheses a close parentheses, used to evaluate a definite integral, the function F open parentheses x close parentheses is an antiderivative of f open parentheses x close parentheses.

    This is a result of the Fundamental Theorem of Calculus.

  • True or False?

    The constant of integration is needed in definite integration.

    False.

    The constant of integration is not needed in definite integration.

  • True or false?

    integral subscript a superscript b k f open parentheses x close parentheses space straight d x equals k integral subscript a superscript b f open parentheses x close parentheses space straight d x

    True.

    integral subscript a superscript b k f open parentheses x close parentheses space straight d x equals k integral subscript a superscript b f open parentheses x close parentheses space straight d x

    This definite integrals property means that constant factors can be taken outside a definite integral.

  • What is integral subscript a superscript a f open parentheses x close parentheses space straight d x equal to?

    integral subscript a superscript a f open parentheses x close parentheses space straight d x equals 0

    The value of a definite integral with equal upper and lower limits is zero

  • True or False?

    Swapping the limits of a definite integral doesn't change the result.

    False.

    Swapping the limits of a definite integral changes the sign of the result.

    I.e. integral subscript b superscript a f open parentheses x close parentheses space straight d x equals negative integral subscript a superscript b f open parentheses x close parentheses space straight d x

  • Assuming a less or equal than b less or equal than c, what is integral subscript a superscript b f open parentheses x close parentheses space straight d x plus integral subscript b superscript c f open parentheses x close parentheses space straight d x equal to?

    Assuming a less or equal than b less or equal than c, then integral subscript a superscript b f open parentheses x close parentheses space straight d x plus integral subscript b superscript c f open parentheses x close parentheses space straight d x equals integral subscript a superscript c f open parentheses x close parentheses space straight d x.

    This definite integral property allows definite integrals to be 'split' into a sum of integrals over smaller intervals.

  • True or False?

    integral subscript a superscript b f open parentheses x close parentheses space straight d x equals integral subscript a plus k end subscript superscript b plus k end superscript f open parentheses x plus k close parentheses space straight d x

    False.

    integral subscript a superscript b f open parentheses x close parentheses space straight d x equals integral subscript a minus k end subscript superscript b minus k end superscript f open parentheses x plus k close parentheses space straight d x.

    This definite integral property has to do with the horizontal translations of functions in definite integrals.

  • True or False?

    integral subscript a superscript b open square brackets f open parentheses x close parentheses plus g open parentheses x close parentheses close square brackets space straight d x equals integral subscript a superscript b f open parentheses x close parentheses space straight d x plus integral subscript a superscript b g open parentheses x close parentheses space straight d x

    True.

    integral subscript a superscript b open square brackets f open parentheses x close parentheses plus g open parentheses x close parentheses close square brackets space straight d x equals integral subscript a superscript b f open parentheses x close parentheses space straight d x plus integral subscript a superscript b g open parentheses x close parentheses space straight d x

    This definite integral property means you can rewrite the integral of a sum as a sum of integrals.

  • True or False?

    integral subscript a superscript b f open parentheses x close parentheses g open parentheses x close parentheses space straight d x equals open parentheses integral subscript a superscript b f open parentheses x close parentheses space straight d x close parentheses open parentheses integral subscript a superscript b g open parentheses x close parentheses space straight d x close parentheses

    False.

    integral subscript a superscript b f open parentheses x close parentheses g open parentheses x close parentheses space straight d x is not equal to open parentheses integral subscript a superscript b f open parentheses x close parentheses space straight d x close parentheses open parentheses integral subscript a superscript b g open parentheses x close parentheses space straight d x close parentheses.

    You cannot rewrite the integral of a product as a product of integrals.

  • What does negative integral refer to?

    A negative integral is a definite integral that results in a negative value.

    This occurs when the area being calculated by the integral is below the x-axis.

  • True or False?

    The area under a curve is always positive.

    True.

    The area under a curve is always positive, even if the definite integral is negative.

  • What is the formula for finding the area under a curve using the modulus function?

    The formula for finding the area under a curve using the modulus function is A equals integral subscript a superscript b open vertical bar y close vertical bar space straight d x

    Where:

    • A is the area being calculated

    • a and b are the integration limits

    • y is the equation of the curve in terms of x

    This formula is in the exam formula booklet.

  • What should be done when finding the area under a curve that is partially below the x-axis?

    When finding the area under a curve that is partially below the x-axis,

    • split the area into parts above and below the x-axis,

    • calculate each integral separately,

    • and sum the absolute values.

    If you are using a GDC to calculate the integral, then you can use the modulus version of the area under a curve integral. There is no need to split the area in this case.

  • True or False?

    When finding the area between a curve and a line, you always subtract the area under the line from the area under the curve.

    False.

    When finding the area between a curve and a line, you may need to add or subtract areas depending on their relative positions.

  • True or False?

    When finding the area between a curve and a line, you always need to use integration to find any relevant areas.

    False.

    When finding the area between a curve and a line, you don't always need to use integration to find relevant areas.

    For example, you may use basic area formulae (trapezoid, right triangle) to find the area under the line.

  • What is the general form of the integral used to find the area between two curves?

    The general form of the integral used to find the area between two curves is integral subscript a superscript b open parentheses y subscript 1 minus y subscript 2 close parentheses space straight d x

    Where

    • y subscript 1 is the upper function (the function 'on the top')

    • y subscript 2 is the lower function (the function 'on the bottom')

    This formula is not given in the exam formula booklet.

  • True or False?

    When finding the area between two curves, which curve is the 'upper' curve and which curve is the 'lower' curve never changes.

    False.

    When finding the area between two curves, which curve is the 'upper' curve and which is the 'lower' curve can change in different regions of the graph.

    An example of calculating area between two curves.  Which is the 'upper curve' and which is the 'lower curve' is different for different areas. The integrals used to find the areas are also shown.
  • What should you do if no diagram is provided when solving area problems using integrals?

    If no diagram is provided when solving area problems using integrals, you should sketch a diagram, even if the curves are not completely accurate.