Let’s look at two integrals that look kind of alike, but one is straight forward, another requires different techniques: In the first one, the substitution u = x 2 + 1 works because it has the factor x needed for du. In the second one, the substitution u = x 2 + 1 doesn’t work because it doesn’t have the factor x needed for du. Fortunately we can just expand and treat as a polynomial.
In other words, we can do a u-substitution that doesn’t require back substitution at the end if we also change our upper and lower limits to match what we are changing with the u-substitution.
Ex 8, p. 301 Change of variables If x = 0, then u = 0 2 + 1= 1. If x = 1, then u = 1 2 + 1 = 2 Now to achieve du inside the integral, I want to multiply inside by 2 and multiply outside by ½. With a rewrite, we get Were you paying attention? I changed the limits of integration and DID NOT have to back substitute to get things in terms of x!
Ex 9 p. 302 Change of variables, a little more difficult! Will it work to let u = 2x - 1? Then du = 2dx. What will we do with that x on top? Let’s try again. Won’t du have a radical? Rewrite! When x = 1, u = 1. When x = 5, u = 3 Geometrically, two different regions have the same area!
When evaluating definite integrals by substitution, it is possible for the upper limit of integration to end up smaller than the lower limit of integration. Just keep plugging through without rearranging the limits. Check that out with If x = 0, u = 1. If x = 1, u = 0 Rewrite!
Even with changes of variables, integration can be difficult. So whatever helps you can get, you will want to use.
Ex 10, p. 303 Integration of a function I think this is an odd function: put (-y) in for y, and (-x) in for x It is an odd function, so apply Thm 4.15 part 2 Area below x-axis is same size as area above x-axis, so integration produces a cancelling effect.