I

Example 3 Evaluate (4( —---h 2x\dx, (x + - 1). The indefinite integral is

jC(t

= (In 5 + 16) - (In 1 + 0) = In 5 + 16 [since In 1 = 0]

It is important to realize that the limits of integration a and b both refer to values of the variable x. Were we to use the substitution-of-variables technique (Rules VI and VII) during integration and introduce a variable, u, care should be taken not to consider a and b as the limits of u. The next example will illustrate this point.

Example 4 Evaluate f2(2x3 - 1)2(6x2) dx. Let u = 2x3 - 1; then du/dx =

6x2, or du = 6x2 dx. Now notice that, when x = 1, u will be 1 but that, when x = 2, u will be 15; in other words, the limits of integration in terms of the variable u should be 1 (lower) and 15 (upper). Rewriting the given integral in u will therefore give us not j u2 du but

0 0