AP®MathsCollege BoardCalculus ABRevision NotesUnit 6: Integration & Accumulation of ChangeRiemann Sums & Definite IntegralsEvaluating Definite Integrals

Evaluating Definite Integrals (College Board AP® Calculus AB)

Revision Note

Roger B

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Roger B

Expertise

Maths

Evaluating definite integrals

How do I evaluate a definite integral?

  • To evaluate a definite integral like integral subscript a superscript b f open parentheses x close parentheses space d x means to find its numerical value

    • Note this difference between definite and indefinite integrals

      • The answer to a definite integral is a number

      • The answer to an indefinite integral is another function

  • The first fundamental theorem of calculus tells us that if f is a continuous function on the closed interval open square brackets a comma space b close square brackets, and if F is an antiderivative of f, then

    • integral subscript a superscript b f open parentheses x close parentheses space d x equals F open parentheses b close parentheses minus F open parentheses a close parentheses

  • The following notation is often used

    • open square brackets F open parentheses x close parentheses close square brackets subscript a superscript b equals F open parentheses b close parentheses minus F open parentheses a close parentheses

  • This provides a simple way to evaluate a definite integral

    • As long as you can find an antiderivative for the function being integrated!

    • First solve the indefinite integral to find F open parentheses x close parentheses

      • integral f open parentheses x close parentheses space d x equals F open parentheses x close parentheses

    • Then substitute in the integration limits

      • integral subscript a superscript b f open parentheses x close parentheses space d x equals open square brackets F open parentheses x close parentheses close square brackets subscript a superscript b equals F open parentheses b close parentheses minus F open parentheses a close parentheses

    • Note that you don't need to worry about constants of integration when calculating definite integrals

      • They would just cancel out

        • open square brackets F open parentheses x close parentheses plus C close square brackets subscript a superscript b equals open parentheses F open parentheses b close parentheses plus C close parentheses minus open parentheses F open parentheses a close parentheses plus C close parentheses equals F open parentheses b close parentheses minus F open parentheses a close parentheses

  • Remember that

    • If f open parentheses x close parentheses greater than 0 on the interval open square brackets a. b close square brackets

      • then integral subscript a superscript b f open parentheses x close parentheses space d x greater than 0

    • If f open parentheses x close parentheses less than 0 on the interval open square brackets a. b close square brackets

      • then integral subscript a superscript b f open parentheses x close parentheses space d x less than 0

    • If f open parentheses x close parentheses is both positive and negative on the interval open square brackets a comma space b close square brackets

      • the negative parts of the integral will subtract from the positive parts

        • The total value can therefore be positive, negative, or zero

Worked Example

Evaluate the definite integral integral subscript negative pi over 3 end subscript superscript pi over 4 end superscript cos x space d x.

Answer:

integral cos x space d x equals sin x plus C, so F open parentheses x close parentheses equals sin x is the antiderivative we can use to evaluate the definite integral

Use integral subscript a superscript b f open parentheses x close parentheses space d x equals open square brackets F open parentheses x close parentheses close square brackets subscript a superscript b equals F open parentheses b close parentheses minus F open parentheses a close parentheses

table row cell integral subscript negative pi over 3 end subscript superscript pi over 4 end superscript cos x space d x end cell equals cell open square brackets sin x close square brackets subscript negative pi over 3 end subscript superscript pi over 4 end superscript end cell row blank equals cell sin pi over 4 minus sin open parentheses negative pi over 3 close parentheses end cell row blank equals cell fraction numerator square root of 2 over denominator 2 end fraction minus open parentheses negative fraction numerator square root of 3 over denominator 2 end fraction close parentheses end cell end table

integral subscript negative pi over 3 end subscript superscript pi over 4 end superscript cos x space d x equals fraction numerator square root of 2 plus square root of 3 over denominator 2 end fraction

Worked Example

Consider the function f defined by f open parentheses x close parentheses equals integral subscript 0 superscript x t squared open parentheses 2 minus t close parentheses space d t.

(a) Calculate f open parentheses 2 close parentheses and f open parentheses 3 close parentheses and confirm that f open parentheses 3 close parentheses less than f open parentheses 2 close parentheses.

Answer:

Start by expanding the brackets and finding the indefinite integral

table row cell integral t squared open parentheses 2 minus t close parentheses space d t end cell equals cell integral open parentheses 2 t squared minus t cubed close parentheses space d t end cell row blank equals cell 2 over 3 t cubed minus 1 fourth t to the power of 4 plus C end cell end table

That gives the antiderivative that can be used for calculating both f open parentheses 2 close parentheses and f open parentheses 3 close parentheses

table row cell f open parentheses 2 close parentheses end cell equals cell integral subscript 0 superscript 2 t squared open parentheses 2 minus t close parentheses space d t end cell row blank equals cell open square brackets 2 over 3 t cubed minus 1 fourth t to the power of 4 close square brackets subscript 0 superscript 2 end cell row blank equals cell open parentheses 2 over 3 open parentheses 2 close parentheses cubed minus 1 fourth open parentheses 2 close parentheses to the power of 4 close parentheses minus open parentheses 2 over 3 open parentheses 0 close parentheses cubed minus 1 fourth open parentheses 0 close parentheses to the power of 4 close parentheses end cell row blank equals cell 4 over 3 minus 0 end cell row blank equals cell 4 over 3 end cell end table

table row cell f open parentheses 3 close parentheses end cell equals cell integral subscript 0 superscript 3 t squared open parentheses 2 minus t close parentheses space d t end cell row blank equals cell open square brackets 2 over 3 t cubed minus 1 fourth t to the power of 4 close square brackets subscript 0 superscript 3 end cell row blank equals cell open parentheses 2 over 3 open parentheses 3 close parentheses cubed minus 1 fourth open parentheses 3 close parentheses to the power of 4 close parentheses minus open parentheses 2 over 3 open parentheses 0 close parentheses cubed minus 1 fourth open parentheses 0 close parentheses to the power of 4 close parentheses end cell row blank equals cell negative 9 over 4 minus 0 end cell row blank equals cell negative 9 over 4 end cell end table

f open parentheses 2 close parentheses equals 4 over 3 and f open parentheses 3 close parentheses equals negative 9 over 4, so f open parentheses 3 close parentheses less than f open parentheses 2 close parentheses

(b) By considering the properties of the expression being integrated, explain why you would expect f open parentheses 3 close parentheses less than f open parentheses 2 close parentheses to be true.

Answer:

Consider the sign of t squared open parentheses 2 minus t close parentheses between 0 and 3, and recall that negative values of a function being integrated contribute negative quantities to a definite integral

Note that between 0 and 2, t squared and 2 minus t are both positive; while for t>2, t squared is positive and open parentheses 2 minus t close parentheses is negative

Between 0 and 2, t squared open parentheses 2 minus t close parentheses greater than 0, so the value of the definite integral between those values must be positive.

For t>2, t squared open parentheses 2 minus t close parentheses less than 0, so the part of the definite integral from 2 to 3 will be negative, and will subtract from the value found between 0 and 2.

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    Roger B

    Author: Roger B

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.