Further Integration (DP IB Analysis & Approaches (AA))

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Cards in this collection (38)

  • 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. However the inverse of this is in the formula booklet, i.e. 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.

  • True or False?

    integral cosec squared x space straight d x equals cot x plus C

    False.

    integral cosec squared x space straight d x equals negative cot x plus C

    This is not in the exam formula booklet. However the inverse of this is in the formula booklet, i.e. the standard derivative f open parentheses x close parentheses equals cot x space rightwards double arrow space f to the power of apostrophe open parentheses x close parentheses equals negative cosec squared x.

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

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

    This is not in the exam formula booklet. However the inverse of this is in the formula booklet, i.e. the standard derivative f open parentheses x close parentheses equals sec x space rightwards double arrow space f to the power of apostrophe open parentheses x close parentheses equals sec x tan x.

  • What is integral cosec x cot x space straight d x?

    integral cosec x cot x space straight d x equals negative cosec x plus C

    This is not in the exam formula booklet. However the inverse of this is in the formula booklet, i.e. the standard derivative f open parentheses x close parentheses equals cosec x space rightwards double arrow space f to the power of apostrophe open parentheses x close parentheses equals negative cosec x cot x.

  • What is integral fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction space straight d x?

    integral fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction space straight d x equals arcsin x plus straight C

    This is not in the exam formula booklet. However the inverse of this is in the formula booklet, i.e. the standard derivative f open parentheses x close parentheses equals arcsin x space rightwards double arrow space f to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction.

  • True of False?

    integral fraction numerator 1 over denominator 1 plus x squared end fraction space straight d x can be found using reverse chain rule.

    False.

    integral fraction numerator 1 over denominator 1 plus x squared end fraction space straight d x can not be found using reverse chain rule.

    integral fraction numerator 1 over denominator 1 plus x squared end fraction space straight d x equals arc tan x plus straight C

    This is not in the exam formula booklet. However the inverse of this is in the formula booklet, i.e. the standard derivative f open parentheses x close parentheses equals arctan x space rightwards double arrow space f to the power of apostrophe open parentheses x close parentheses equals fraction numerator 1 over denominator 1 plus x squared end fraction.

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

    integral a to the power of x space straight d x equals fraction numerator 1 over denominator ln a end fraction a to the power of x plus C

    This is in the exam formula booklet.

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

    integral fraction numerator 1 over denominator a squared plus x squared end fraction space straight d x equals 1 over a arctan open parentheses x over a close parentheses plus C

    This is in the exam formula booklet.

  • What is integral fraction numerator 1 over denominator square root of a squared minus x squared end root end fraction space straight d x?

    integral fraction numerator 1 over denominator square root of a squared minus x squared end root end fraction space straight d x equals arc sin open parentheses x over a close parentheses plus C comma space space open vertical bar x close vertical bar less than a

    This is in the exam formula booklet.

  • True or False?

    Completing the square may be necessary when integrating with inverse trigonometric functions.

    True.

    Completing the square may be necessary when integrating with inverse trigonometric functions.

    E.g. rewriting integral fraction numerator 1 over denominator square root of 5 minus x squared plus 4 x end root end fraction space d x as integral fraction numerator 1 over denominator square root of 9 minus open parentheses x minus 2 close parentheses squared end root end fraction space d x. Then the integral is in a form which can be integrated using the standard integral resultintegral fraction numerator 1 over denominator square root of a squared minus x squared end root end fraction space straight d x equals arc sin open parentheses x over a close parentheses plus C from the exam formula booklet.

  • True or False?

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

    True.

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

    E.g. u equals x squared space rightwards double arrow space fraction numerator straight d u over denominator straight d x end fraction equals 2 x space rightwards double arrow space straight d u equals 2 x straight d x

    fraction numerator straight d u over denominator straight d x end fraction is not actually a fraction (it's a derivative, i.e. gradient function), but treating it like a fraction can give correct answers when integrating by substitution.

  • integral x open parentheses 1 plus 3 x close parentheses to the power of 8 space end exponent d x can be integrated by using the substitution u equals 1 plus 3 x.

    It follows from this that x equals fraction numerator u minus 1 over denominator 3 end fraction, and also that fraction numerator straight d u over denominator straight d x end fraction equals 3 space rightwards double arrow space straight d x equals 1 third straight d u.

    What is the next step in solving the integral?

    integral x open parentheses 1 plus 3 x close parentheses to the power of 8 space end exponent d x can be integrated by using the substitution u equals 1 plus 3 x.

    It follows from this that x equals fraction numerator u minus 1 over denominator 3 end fraction, and also that fraction numerator straight d u over denominator straight d x end fraction equals 3 space rightwards double arrow space straight d x equals 1 third straight d u.

    The next step in solving the integral is to substitute terms in u for all the terms in x in the integral:

    integral x open parentheses 1 plus 3 x close parentheses to the power of 8 space end exponent d x equals integral open parentheses fraction numerator u minus 1 over denominator 3 end fraction close parentheses u to the power of 8 open parentheses 1 third straight d u close parentheses equals 1 over 9 integral open parentheses u to the power of 9 minus u to the power of 8 close parentheses straight d u

  • State the formula for integration by parts.

    The formula for integration by parts is integral u fraction numerator straight d v over denominator straight d x end fraction space straight d x equals u v minus integral v fraction numerator straight d u over denominator straight d x end fraction straight d x, which can also be written as integral u space straight d v equals u v minus integral v space straight d u.

    Both versions of the formula are in the exam formula booklet.

  • True or False?

    In integration by parts, the function that becomes simpler when differentiated should be assigned to fraction numerator straight d v over denominator straight d x end fraction.

    False.

    In integration by parts, the function that becomes simpler when differentiated should be assigned to u, so that it can then be differentiated.

  • True or False?

    Integration by parts may need to be applied more than once to solve a problem.

    True.

    Integration by parts may need to be applied more than once to solve a problem. This is known as repeated integration by parts.

  • True or False?

    Integrating expressions like x squared cos x or straight e to the power of x sin x requires repeated integration by parts.

    True.

    Integrating expressions like x squared cos x or straight e to the power of x sin x requires repeated integration by parts.

  • When using integration by parts to integrate ln x, what should you choose for u and fraction numerator straight d v over denominator straight d x end fraction?

    When using integration by parts to integrate ln x, you should choose u equals ln x and fraction numerator straight d v over denominator straight d x end fraction equals 1.

    Then integral ln x space straight d x equals x ln x minus integral x open parentheses 1 over x close parentheses straight d x equals x ln x minus integral straight d x equals x ln x minus x plus C.

  • True or False?

    In integration by parts, only one overall constant of integration is required.

    True.

    In integration by parts, only one overall constant of integration is required.

    Even though more than one integral may appear in your working, only one constant of integration is required in the final answer.

  • True or False?

    Partial fractions can be used for integration when the denominator of a quotient is of quadratic form.

    E.g. to integrate integral fraction numerator x minus 13 over denominator x squared minus 5 x minus 6 end fraction space straight d x.

    True.

    Partial fractions can be used for integration when the denominator of a quotient is of quadratic form.

    E.g. to integrate integral fraction numerator x minus 13 over denominator x squared minus 5 x minus 6 end fraction space straight d x, first use partial fractions to rewrite the integral as integral open parentheses fraction numerator 2 over denominator x plus 1 end fraction minus fraction numerator 1 over denominator x minus 6 end fraction close parentheses space straight d x.

  • Why should you not use partial fractions to integrate something like integral fraction numerator 2 x minus 5 over denominator x squared minus 5 x minus 6 end fraction space straight d x?

    If the numerator of what you are trying to integrate is the derivative (or a multiple of the derivative) of the denominator, then partial fractions should not be used because it is much quicker to use the reverse chain rule result integral fraction numerator f apostrophe open parentheses x close parentheses over denominator f open parentheses x close parentheses end fraction space straight d x equals ln open vertical bar f open parentheses x close parentheses close vertical bar plus C.

    E.g. integral fraction numerator 2 x minus 5 over denominator x squared minus 5 x minus 6 end fraction space straight d x equals ln open vertical bar x squared minus 5 x minus 6 close vertical bar plus C

  • True or False?

    If a quadratic denominator does not factorise, then you should use inverse trigonometric functions to integrate.

    True.

    If a quadratic denominator does not factorise, then you should use inverse trigonometric functions to integrate.

  • True or False?

    The result of integrating with partial fractions always involves natural logarithms.

    True.

    The result of integrating with partial fractions always involves natural logarithms.

  • How can you further simplify an indefinite integral answer like ln open vertical bar fraction numerator x plus 1 over denominator x plus 3 end fraction close vertical bar plus c, where c is the constant of integration?

    To further simplify an indefinite integral answer like ln open vertical bar fraction numerator x plus 1 over denominator x plus 3 end fraction close vertical bar plus c, where c is the constant of integration:

    • Write c as a logarithm: c equals ln k

    • Substitute into the answer: ln open vertical bar fraction numerator x plus 1 over denominator x plus 3 end fraction close vertical bar plus ln k

    • Use laws of logarithms to combine the two logs: ln open vertical bar k open parentheses fraction numerator x plus 1 over denominator x plus 3 end fraction close parentheses close vertical bar

  • What is the area between a curve and the y-axis?

    The area between a curve and the y-axis is the region bounded by the graph of y=f(x), the y-axis, and two horizontal lines y=a and y=b.

    Graph showing example of an 'area between a curve and the y-axis'.
  • What is the formula for finding the area between a curve and the y-axis?

    The formula for finding the area between a curve and the y-axis is A equals integral subscript a superscript b open vertical bar x close vertical bar straight d y

    Where:

    • A is the area to be found

    • y equals a and y equals b are the equations of the horizontal lines bounding the area

    • x equals g open parentheses y close parentheses is the equation of the curve expressed as a function of y

    This is in the exam formula booklet.

  • True or False?

    The function y=f(x) describing the curve must be rearranged into the form x=g(y) when finding the area between a curve and the y-axis.

    True.

    The function y=f(x) describing the curve must be rearranged into the form x=g(y) when finding the area between a curve and the y-axis.

  • What is a volume of revolution?

    A volume of revolution is the volume of a solid formed when an area bounded by a function y=f(x) is rotated 2π radians (or 360°) around an axis.

    An illustration of a volume of revolution formed by rotating a function y=f(x) about the x-axis.
  • State the formula for a volume of revolution around the x-axis.

    The formula for the volume of revolution around the x-axis is V equals integral subscript a superscript b pi y squared straight d x

    Where:

    • V is the volume to be found

    • x equals a and x equals b are the equations of the vertical lines bounding the area to be rotated

    • y squared is the square of the equation of the curve y equals f open parentheses x close parentheses

    This is in the exam formula booklet.

  • True or False?

    The limits a and b to be used in the volume of revolution formula are always given directly in the question.

    False.

    The limits a and b to be used in the volume of revolution formula are not always given directly in the question.

    You may need to work the limits out. They could involve the y-axis (x=0) or a root of y=f(x).

  • State the formula for a volume of revolution around the y-axis.

    The formula for the volume of revolution around the y-axis is V equals integral subscript a superscript b pi x squared straight d y

    Where:

    • V is the volume to be found

    • y equals a and y equals b are the equations of the horizontal lines bounding the area to be rotated

    • x squared is the square of the equation of the curve x equals g open parentheses y close parentheses

    This is in the exam formula booklet.

    (Note that an equation of the curve in y equals f open parentheses x close parentheses form will need to be rearranged into x equals g open parentheses y close parentheses form to use this formula.)

  • True or False?

    When calculating the volume of a revolution about the y-axis, the equation of the curve must always be rearranged from y=f(x) form to x=g(y) form.

    True.

    When calculating the volume of a revolution about the y-axis, the equation of the curve must always be rearranged from y=f(x) form to x=g(y) form.

  • True or False?

    The constant π in the volume of revolution formulas can be pulled out in front of the integral as a multiplier.

    True.

    The constant π in the volume of revolution formulas (like any constant inside an integral) can be pulled out in front of the integral as a multiplier.

    This can be useful when the integral needs to be evaluated by hand. (Just don't forget to put π back into the final answer at the end!)

  • What is a key modelling assumption when using volumes of revolution?

    A key modelling assumption when using volumes of revolution is that the thickness of the 'walls' of the solid is negligible relative to the size of the object.

  • True or False?

    The volume of revolution always includes all parts of an object, such as the handle of a bucket.

    False.

    The volume of revolution does not always include all parts of an object, such as the handle of a bucket.

    The volume of revolution usually only includes the main shape of the body of the object, not additional parts like the handle of a bucket.

  • What is the unit conversion between litres and cm3 that might be needed when calculating the capacity of an object modelled as a volume of revolution?

    The unit conversion between litres and cm3 that might be needed when calculating the capacity of an object modelled as a volume of revolution is 1 litre = 1000 cm3.

  • True or False?

    Adding or subtracting volumes of revolution is sometimes necessary when modelling complex shapes.

    True.

    Adding or subtracting volumes of revolution is sometimes necessary when modelling complex shapes.

  • What is a criticism of the volume of revolution model for real-world objects?

    A criticism of the volume of revolution model for real-world objects is that it doesn't account for the thickness of the material the object is made from.

  • True or False?

    The volume found by a volume of revolution model always accurately represents the capacity of the real object.

    False.

    The volume found by a volume of revolution model may not always accurately represents the capacity of the real object, due to various factors like material thickness or internal components.