1. More on Substitution Technique (1/27/06) Remember that you may try it but it may not work. Very likely it won’t! Here’s what to look for: – Is there a “chunk” (“inside function”) in the integrand? If so, what is its derivative? – Is the rest of the integrand (besides the chunk and its outer function) that derivative except perhaps for a constant multiplier? – If so, substitution should work!! Let u = “chunk”, compute du, and rebuild the integral in terms of u.

2. An example Try to find  x 2 sin(2x 3 + 7) dx – Is there a “chunk” (an inner function)? Yes. 2x 3 +7 – Besides the chunk and the outer function (sin), is the rest of the integrand within a constant multiplier of derivative of the chunk? YES! Because the derivative of 2x 3 +7 is 6x 2. We’re in! – Now let u = 2x 3 +7, so du = 6x 2 dx, and replacing equals with equals, the integrand is rebuilt as (1/6)  sin(u) du. We have successfully replaced a complicated integrand with a simple one.

3. Example Continued Okay, we now want (1/6)  sin(u) du. But this is easy. Since the anti-derivative of the sin(u) with respect to u is – cos(u), the answer to our problem is just -(1/6)cos(u) + C = -(1/6)cos(2x 3 + 7) + C Got it ??? Note that when you check by taking the derivative, you use the Chain Rule!!

4. Concerning Definite Integrals If you use substitution and the Fundamental Theorem to evaluate a definite integral, there are two possible approaches: – Go back to the original variable and evaluate at the endpoints as usual, or – Never return to the original variable! Instead, change the endpoints to correspond to your new variable, and then stay with that variable.

5. An example If the previous example were a definite integral, say then the second option is to use u = 2x 3 + 7 to get that if x = 0 then u = 7 and if x = 1 then u = 9. so now our problem becomes

6. Assignment For Monday, we meet in the MCS Lab (Harder 209) for Lab #1 on Integration. For Wednesday, catch up on the exercises on substitution technique and also do 59-65 odd, 71 and 73 on page 421.