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§4.3 Integration by Substitution.

§4.3 Integration by Substitution. The student will learn about:. differentials, and. integration by substitution. 1. Introduction.

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§4.3 Integration by Substitution.

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  1. §4.3 Integration by Substitution. The student will learn about: differentials, and integration by substitution. 1

  2. 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. 2

  3. This comes from previous notation: Multiply both sides by dx to get the differential Preliminaries. For a differentiable function f (x), the differentialdf is df = f ′(x)dx This is the derivative. 3

  4. df = dx Examples. Function Differential df df = 3 x 2 dx f (x) = x 3 We are taking a derivative now !!! f (x) = x 2 - 3x + 5 df = (2x – 3) dx f (x) = ln x df = 1/x dx f (x) = e 3x 4

  5. A differential 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. 5

  6. General Indefinite Integral Formulas. ∫ undu = Very Important ∫ e udu = e u + C Note the chain “du” is present! 6

  7. Integration by Substitution This is the power rule form ∫ u n du. 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 ndu. There is a nice method of integration called “integration by substitution” that will handle this problem. 7

  8. ∫ undu = Example by Substitution By example (Substitution): Find ∫ (x5 - 2)3 (5x 4) dx For our substitution let u = x 5 - 2, For our substitution let u = x 5 - 2, then du/dx = 5x 4, and du = 5x4 dx Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? “Power Rule” and the integral becomes ∫ (x5 - 2)3 (5x 4) dx = ∫ u 3 du which is = Continued 8

  9. Example by Substitution By example (Substitution): Find ∫ (x5 - 2)3 (5x 4) dx u = x 5 - 2, ∫ u 3 du which is = and reverse substitution yields Remember you may differentiate to check your work! 9

  10. Integration by Substitution. Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? 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. 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.) 10 Remember you may differentiate to check your work!

  11. ∫ undu = “Power Rule” Example Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? ∫ (x3 - 5)4 (3x2) dx Step 1 – Select u. Let u = x3 - 5 and then du = 3x2 dx Step 2 – Express integral in terms of u. ∫ (x3 - 5)4 (3x2) dx = ∫ u4 du Did I mention you may differentiate to check your work? Step 3 – Integrate. ∫ u4 du = Step 4 – Express the answer in terms of x. 11

  12. Example Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? ∫ (x 2 + 5) 1/2 (2x) dx Step 1 – Select u. Let u = x2 + 5 and then du = 2x dx Step 2 – Express integral in terms of u. ∫ (x2 + 5) 1/2 (2x) dx = ∫ u 1/2 du “Power Rule” Remember you may differentiate to check your work! Step 3 – Integrate. ∫ u 1/2 du = Step 4 – Express the answer in terms of x. 12

  13. Multiplying Inside and Outside by Constants

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

  15. Substitution Technique – Example 1 Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? “Power Rule” 1. ∫ (x3 - 5)4 (x2) dx Let u = x3 – 5 then du = 3x2 dx ∫ (x3 - 5)4 (x2) dx = ∫ (x3 - 5)4 (x2) dx = 3 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. 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. 15 Remember you may differentiate to check your work!

  16. Substitution Technique – Example 2 ∫ e u du = e u + c Is it “Power Rule”, “Exponential Rule” or the “Log Rule”? “Exponential Rule” 2. Let u = 4x3 then du = 12x2 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. 16 Remember you may differentiate to check your work!

  17. Substitution Technique – Example 3 3. Let u = 5 – 2x2 then du = - 4x dx Note – we need a - 4. Remember you may differentiate to check your work! 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”? “Power Rule” 17

  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. To find p (x) we need the ∫ p ‘ (x) dx Step 1 Continued on next slide. 18

  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. Let u = 3x + 25 and du = 3 dx Step 1 Continued on next slide. 19

  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. With u = 3x + 25, Step 1 so Remember you may differentiate to check your work! Continued on next slide. 20

  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. Now we need to find c using the fact Step 2 That 75 bottles sell for $1.60 per bottle. and c = 2 Continued on next slide. 21

  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! 22

  23. Summary. 1. ∫ un du = ∫ e u du = e u + C 2. Very Important 3. 23

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

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