Definite Integration (Cambridge International A-Level Mathematics)

Definite integration

What is definite integration?

• Definite Integration occurs in an alternative version of the Fundamental Theorem of Calculus
• This version of the Theorem is the one referred to by most AS/A level textbooks/websites

• a and b are called limits

  • a is the lower limit
  • b is the upper limit

• f’(x) is the derivative of f(x)

Why is there no constant of integration in definite integration?

• “+c” would appear in both f(a) and f(b)

  • Since we then calculate f(b) – f(a) they cancel each other out
  • There would be a “+c” from f(b) and a –“+c” from f(a)

• So “+c” is not included with definite integration

How do I find the value of a definite integral?

• STEP 1: If not given a name, call the integral

  • This saves you having to rewrite the whole integral every time!

• STEP 2: If necessary rewrite the integral into a more easily integrable form

  • Not all functions can be integrated directly

• STEP 3: Integrate without applying the limits

  • Notation: use square brackets [ ] with limits placed after the end bracket

• STEP 4: Substitute the limits into the function and calculate the answer

How do I find a definite integral on my calculator?

• Advanced scientific calculators can work out the values of definite integrals
• The button will look similar to:

• (Note how the calculator did not return the exact value $( \frac{1256}{3} )$ of the integral)

Exam Hint

• Look out for questions that ask you to find an indefinite integral in one part (so “+c” needed), then in a later part use the same integral as a definite integral (where “+c” is not needed).

Worked Example Find the value of

$\int_{2}^{4} 3 x ( x^{2} - 2 ) \mathrm{d} x$

Start by expanding the brackets inside the integral

$\int_{2}^{4} ( 3 x^{3} - 6 x ) \mathrm{d} x$

Integrate as usual (here it's a 'powers of $x$' integration)

Write the answer in square brackets with the integration limits outside

$\begin{gathered} \int_{2}^{4} ( 3 x^{3} - 6 x ) \mathrm{d} x & = [ 3 ( \frac{x^{3 + 1} }{3 + 1} ) - 6 ( \frac{x^{1 + 1} }{1 + 1} ) ]_{2}^{4} \\ = [ \frac{3}{4} x^{4} - 3 x^{2} ]_{2}^{4} \end{gathered}$

Now substitute 4 into that function And subtract from it the function with 2 substituted in

$\begin{array}{rcl} [ \frac{3}{4} x^{4} - 3 x^{2} ]_{2}^{4} & = ( \frac{3}{4} ( 4 )^{4} - 3 ( 4 )^{2} ) - ( \frac{3}{4} ( 2 )^{4} - 3 ( 2^{2} ) ) \\ = ( 192 - 48 ) - ( 12 - 12 ) \\ = 144 - 0 \\ = 144 \end{array}$

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