4. To compute a definite integral using u-substitution, you need to make sure that the numbers you plug in for the limits of integration corlrespond to the variable you are working with. For instance, if we were computing f 8z(z* + 7)* dz with the substi- 0 1 2 z=1 tution u = z% + 7, we would get f du du = u\"' K There are two ways to fix the =0 %= mismatch: e We can convert the indefinite integral back into a function of z before plugging in the limits of integration: =] u'*r =+ 7)4| = 8! 74 = 1695. = z=0 e We can convert the limits of integration into values of u. When z =0, u = 7, and when z = 1, u = 8, so we have u==8 =8 7" = 1695. u=T ud In fact, we can even do the shift before finishing the integral: =1 u=8 / 8z(z? + 7 dz = [ 44 du. & u=T =0 w2 (a) Use both methods to compute / sin(z) cos(z) dz. You should get the same 0 answer both ways. (b) The integral [ 2ze #+7 4z looks hard, and the indefinite integral / 22V H dg = really is hard, but one of the two methods makes the definite integral very easy. Evaluate this definite integral. In(3) e* dr (c) Evaluate / e using whichever method you prefer. 0 eI