Definite Integrals (Cambridge International IGCSE Further Maths)

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 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)

• The value can be positive, zero or negative

Why do I not need to include a constant of integration for 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
  • So “+c” is not included with definite integration

How do I find a definite integral?

• STEP 1

  • Give the integral a name (if it does not already have one)

    • 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

    • Substitute the top limit first
    • Then substitute the bottom limit
    • Subtract the second value from the first

What are the special properties of definite integrals?

• Some of these have been encountered already and some may seem obvious …

  • taking constant factors outside the integral

    • $\int_{a}^{b} k \mathrm{f} ( x ) \mathrm{d} x = k \int_{a}^{b} \mathrm{f} ( x ) \mathrm{d} x$ where$k$ is a constant
    • useful when fractional and/or negative values involved

  • integrating term by term

    • $integral_{a}^{b} [ \mathrm{f} ( x ) + \mathrm{g} ( x ) ] \mathrm{d} x = integral_{a}^{b} \mathrm{f} ( x ) \mathrm{d} x + integral_{a}^{b} \mathrm{g} ( x ) \mathrm{d} x$
    • the above works for subtraction of terms/functions too

  • equal upper and lower limits

    • $\int_{a}^{a} \mathrm{f} ( x ) d x = 0$
    • on evaluating, this would be a value subtracted from itself!

  • swapping limits gives the same, but negative, result

    • $\int_{a}^{b} \mathrm{f} ( x ) \mathrm{d} x = - \int_{b}^{a} \mathrm{f} ( x ) \mathrm{d} x$
    • compare 8 subtract 5 say, with 5 subtract 8 …

  • splitting the interval

    • $\int_{a}^{b} \mathrm{f} ( x ) \mathrm{d} x = \int_{a}^{c} \mathrm{f} ( x ) \mathrm{d} x + \int_{c}^{b} \mathrm{f} ( x ) \mathrm{d} x$ where$a\leq c\leq b$
    • this is particularly useful for areas under multiple curves or areas under the$x$-axis

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

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

Answer:

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

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

Error converting from MathML to accessible text.