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11 The student will learn about: §4.3 Integration by Substitution. integration by substitution. differentials, and

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22 Introduction The use of the “Chain Rule” greatly expanded the range of functions that we could differentiate. We need a similar property in integration to allow us to integrate more functions. The method we are going to use is called integration by substitution. But first lets practice some basic integrals.

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33 Preliminaries. For a differentiable function f (x), the differential df is df = f ′(x) dx This is the derivative. This comes from previous notation: Multiply both sides by dx to get the differential

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44 Examples. Function f (x) = x 3 Differential df df = 3 x 2 dx f (x) = ln x df = 1/x dx f (x) = e 3x df = dx f (x) = x 2 – 3x + 5df = (2x – 3) dx We are taking a derivative now !!!

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55 Reversing the Chain Rule. Recall the chain rule: This means that in order to integrate 5 (x 3 + 3) 4 its derivative, 3x 2, must be present. The “chain” must be present in order to integrate. A differential

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66 General Indefinite Integral Formulas. ∫ u n du = ∫ e u du = e u + C Note the chain “du” is present!

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77 Integration by Substitution Note that the derivative of x 5 – 2, (i.e. 5x 4, the chain), is present and the integral is in the power rule form ∫ u n du. There is a nice method of integration called “integration by substitution” that will handle this problem. This is the power rule form ∫ u n du.

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88 Example by Substitution ∫ u 3 du which is = By example (Substitution): Find ∫ (x 5 – 2) 3 (5x 4 ) dx For our substitution let u = x 5 – 2, 5x 4, and du =5x 4 dxthen du/dx = and the integral becomes For our substitution let u = x 5 – 2, Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? ∫ (x 5 – 2) 3 (5x 4 ) dx = “Power Rule” ∫ u n du = Continued

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99 Example by Substitution ∫ u 3 du which is = By example (Substitution): Find ∫ (x 5 – 2) 3 (5x 4 ) dx u = x 5 – 2, and reverse substitution yields Remember you may differentiate to check your work!

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10 Integration by Substitution. Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? Step 2. Express the integrand entirely in terms of u and du, completely eliminating the original variable. Step 3. Evaluate the new integral, if possible. Step 4. Express the antiderivative found in step 3 in terms of the original variable. (Reverse the substitution.) Step 1. Select a substitution that appears to simplify the integrand. In particular, try to select u so that du is a factor of the integrand. Remember you may differentiate to check your work!

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11 Example ∫ (x 3 – 5) 4 (3x 2 ) dx Step 1 – Select u. Let u = x 3 – 5and then du = Step 2 – Express integral in terms of u. ∫ (x 3 – 5) 4 (3x 2 ) dx =∫ u 4 du Step 3 – Integrate. ∫ u 4 du = Step 4 – Express the answer in terms of x. 3x 2 dx Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? Did I mention you may differentiate to check your work? “Power Rule” ∫ u n du =

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12 Example ∫ (x 2 + 5) 1/2 (2x ) dx Step 1 – Select u. Let u = x 2 + 5and then du = Step 2 – Express integral in terms of u. ∫ (x 2 + 5) 1/2 (2x) dx =∫ u 1/2 du Step 3 – Integrate. ∫ u 1/2 du = Step 4 – Express the answer in terms of x. 2x dx Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? “Power Rule”

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13 Multiplying Inside and Outside by Constants

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14 Multiplying Inside and Outside by Constants If the integral does not exactly match the form ∫u n du, we may sometimes still solve the integral by multiplying by constants.

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15 Substitution Technique – Example 1 1. ∫ (x 3 – 5) 4 (x 2 ) dx = ∫ (x 3 – 5) 4 (x 2 ) dxLet u = ∫ (x 3 – 5) 4 ( x 2 ) dx = In this problem we had to insert a multiple 3 in order to get things to work out. We did this by also dividing by 3 elsewhere. x 3 – 5 then du =3x 2 dx Note – we need a 3. In this problem we had to insert a multiple 3 in order to get things to work out. We did this by also dividing by 3 elsewhere. Caution – a constant can be adjusted but a variable cannot. Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? Remember you may differentiate to check your work! 3 “Power Rule”

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16 2.Let u =4x 3 then du =12x 2 dx Note – we need a 12. In this problem we had to insert a multiple 12 in order to get things to work out. We did this by also dividing by 12 elsewhere. Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? Remember you may differentiate to check your work! Substitution Technique – Example 2 “Exponential Rule” ∫ e u du = e u + c

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17 3.Let u =5 – 2x 2 then du =- 4x dx Note – we need a – 4. In this problem we had to insert a multiple – 4 in order to get things to work out. Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? Substitution Technique – Example 3 “Power Rule”

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18 Applications §6-2, #68. The marginal price of a supply level of x bottles of baby shampoo per week is given by Find the price-supply equation if the distributor of the shampoo is willing to supply 75 bottles a week at a price of $1.60 per bottle. Continued on next slide. To find p (x) we need the∫ p ‘ (x) dx Step 1

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19 Applications – continued The marginal price of a supply level of x bottles of baby shampoo per week is given by Find the price-supply equation if the distributor of the shampoo is willing to supply 75 bottles a week at a price of $1.60 per bottle. Continued on next slide. Let u =3x + 25 3 dx Step 1 and du =

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20 Applications – continued The marginal price of a supply level of x bottles of baby shampoo per week is given by Find the price-supply equation if the distributor of the shampoo is willing to supply 75 bottles a week at a price of $1.60 per bottle. Continued on next slide. With u = 3x + 25, so Remember you may differentiate to check your work! Step 1

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21 Applications – continued The marginal price of a supply level of x bottles of baby shampoo per week is given by Find the price-supply equation if the distributor of the shampoo is willing to supply 75 bottles a week at a price of $1.60 per bottle. Continued on next slide. Now we need to find c using the fact That 75 bottles sell for $1.60 per bottle. and c =2 Step 2

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22 Applications – concluded The marginal price of a supply level of x bottles of baby shampoo per week is given by Find the price-supply equation if the distributor of the shampoo is willing to supply 75 bottles a week at a price of $1.60 per bottle. So Step 2 This problem does not have a step 3!

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23 Summary. 2. 1. ∫ u n du = ∫ e u du = e u + C 3.

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24 ASSIGNMENT §4.3 on my website. 13, 14, 15, 16.

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