Definite Integrals - SS3 Mathematics Lesson Note

So far, all the integrals we have been evaluating are indefinite integrals because of the presence of arbitrary constants.

The definite integral of a function between a lower and upper limit of \(x\) is the difference between the integrals of the function at those points. Thus, any integral with limits is a definite integral. In general,

\(\int_{a}^{b}{f(x)}\ dx\) is a definite integral of \(f(x)\) under the lower limit \(a\) and upper limit \(b\) by:

1. Evaluating \(\int_{}^{}{f(x)} = g(x)\), omitting the constant

Thus, \(\int_{a}^{b}{f(x)}\ dx = \ {\lbrack g(x)\rbrack}_{a}^{b} = g(b) - g(a)\)

Example: Evaluate \(\int_{- 1}^{2}{(x^{2} - 3x)}\ dx\)

Solution:

\[\int_{- 1}^{2}{(x^{2} - 3x)}\ dx = \ \left\lbrack \frac{x^{3}}{3} - \frac{3x^{2}}{2} \right\rbrack_{- 1}^{2} = \left\lbrack \left\lbrack \frac{{(2)}^{3}}{3} - \frac{3{(2)}^{2}}{2} \right\rbrack - \left\lbrack \frac{{( - 1)}^{3}}{3} - \frac{3{( - 1)}^{2}}{2} \right\rbrack \right\rbrack\]

\[= \left\lbrack \frac{8}{3} - 6 \right\rbrack - \left\lbrack - \frac{1}{3} - \frac{3}{2} \right\rbrack = \frac{9}{6} = \frac{3}{2}\]