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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.

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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.

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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!!

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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.

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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

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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.

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