4.5: Integration by Substitution Objectives: Assignment:
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4.5: Integration by Substitution Objectives: Assignment:
4.5: Integration by Substitution Objectives: Assignment: 1. To use a change of variables to evaluate indefinite and definite integrals β’ P. 304-305: 7-33 odd, 3747 odd, 61-67 odd, 70, 71-81 eoo, 83, 87 2. To use symmetry as an aid in integration β’ P. 306: 101-107 odd Warm Up 1 Find the derivative of β π₯ = π₯ 2 + π₯. β π₯ = π₯2 + π₯ ββ² π₯ = ββ² π₯ = 1/2 1 2 π₯ +π₯ 2 2π₯ + 1 β1/2 2π₯ + 1 πβ² π₯ 2 π₯2 + π₯ π π₯ π π₯ = π₯ The Chain Rule If π¦ = π π’ is a differentiable function of π’ and π’ = π π₯ is a differentiable function of π₯, then π¦ = π π π₯ is a differentiable function of π₯ and ππ¦ ππ¦ ππ’ = β ππ₯ ππ’ ππ₯ π π π π₯ ππ₯ = πβ² π π₯ β πβ² π₯ The derivative of the exterior function times the derivative of the interior function Warm Up 2 2π₯+1 β² Find the antiderivative of β π₯ = ββ² π₯ = 2π₯ + 1 2 π₯ 2 +π₯ πβ² π₯ 2 π₯2 + π₯ π π₯ = π₯ π π₯ ββ² π₯ = πβ² π π₯ πβ² π₯ β π₯ =π π π₯ +πΆ = π₯2 + π₯ + πΆ . Objective 1 You will be able to use a change of variables to evaluate indefinite and definite integrals Composite Functions When integrating a composite function π, realize that function π was the product of the Chain Rule. πΉβ² π π₯ πβ² π₯ ππ₯ = πΉ π π₯ +πΆ Antidifferentiation of Composite Functions Let π be a function whose range is an interval πΌ, and let π be a function that is continuous on πΌ. If π is differentiable on its domain and πΉ is an antiderivative of π on πΌ, then π π π₯ πβ² π₯ ππ₯ = πΉ π π₯ +πΆ Substitution Rule If π’ = π(π₯), then ππ’ ππ₯ = πβ² π₯ . ππ’ = πβ² π₯ ππ₯ π π π₯ πβ² π₯ ππ₯ = π π’ ππ’ = πΉ π’ + πΆ This substitution is accomplished with a change of variable, where π’ is a function of π₯. Exercise 1 Find 1 + π₯ 2 2π₯ ππ₯ 2π₯ 1 + π₯ 2 ππ₯ = Let π’ = 1 + π₯ 2 ππ’ = 2π₯ ππ₯ = Always remember that your answer should be in terms of the original variable. ππ’ = 2π₯ ππ₯ π’ ππ’ = π’1/2 ππ’ 2 3/2 2 = π’ + πΆ = 1 + π₯2 3 3 3/2 +πΆ Check your answer by taking its derivative. Protip When changing variables, try letting π’ = Some complicated part of the integrand The inner function of a composite function Some function whose derivative is part of the integrand Use Constant Multiple Rule Except for a constant factor If your first guess doesnβt work, try something different. Exercise 2 Find 2π₯ π₯ 2 + 1 2 ππ₯. Exercise 3 Find π₯ π₯2 + 1 2 ππ₯. Exercise 4 Find 5cos 5π₯ ππ₯. Exercise 5 Find 2π₯ β 1 ππ₯. Exercise 6 Find π₯ 2π₯ β 1 ππ₯. Protip: When changing variables, sometimes you have to use the equation for π’ to solve for π₯ to do another substitution. Exercise 7 Find sin2 3π₯ cos 3π₯ ππ₯. General Power Rule for Integration If π is a differentiable function of π₯, then π+1 π π₯ π π₯ π πβ² π₯ ππ₯ = + πΆ, π+1 If π’ = π π₯ , then π+1 π’ π’π ππ’ = + πΆ, π+1 π β β1 π β β1 Exercise 8 Find the following antiderivatives. 1. 3 3π₯ β 4 4 ππ₯ 2. 2π₯ + 1 π₯ 2 + π₯ 4 ππ₯ Exercise 8 Find the following antiderivatives. 3. 3π₯ 2 π₯3 β 2 ππ₯ 4. β4π₯ ππ₯ 1β2π₯ 2 2 Exercise 9 Find cos 2 π₯ sin π₯ ππ₯. Exercise 10 Evaluate 1 π₯ 0 π₯2 + 1 3 ππ₯. Method 1: π’ = π₯2 + 1 ππ’ = 2π₯ ππ₯ ππ’ = π₯ ππ₯ 2 π π ππ’ 1 3 π’ = 2 2 π π 1 4 3 π’ ππ’ = π’ 8 π π 1 2 = π₯ +1 8 = 15 8 1 4 0 Exercise 10 Evaluate 1 π₯ 0 π₯2 + 1 3 ππ₯. Method 2: π’ = π₯2 + 1 ππ’ = 2π₯ ππ₯ ππ’ = π₯ ππ₯ 2 2 1 ππ’ 1 π’3 = 2 2 New Limits of Integration: π’ = 12 + 1 = 2 π’ = 02 + 1 = 1 2 1 1 π’3 ππ’ = π’4 8 15 = 8 2 1 Definite Integrals If the function π’ = π π₯ has a continuous derivative on the π, π and π is continuous on the range of π, then π π π π π π₯ πβ² π₯ ππ₯ = π π π’ ππ’ π π Exercise 11 Evaluate π΄ = 4 0 2π₯ + 1 ππ₯. Objective 2 You will be able to use symmetry as an aid in integration Exercise 12 Evaluate each definite integral. 1. 2 6 π₯ β2 + 1 ππ₯ 2. 2 6 π₯ 0 + 1 ππ₯ Exercise 13 Evaluate π/3 sin 3π₯ ππ₯. βπ/3 Symmetry Recall that the graphs of even and odd functions have a useful kind of symmetry: Even = π¦-axis Symmetry π βπ₯ = π π₯ Odd = Origin Symmetry π βπ₯ = βπ π₯ Integration of Even/Odd Functions Let π be integrable on βπ, π . If π is an even function, then π π π π₯ ππ₯ = 2 π π₯ ππ₯. βπ 0 If π is an odd function, then π π π₯ ππ₯ = 0. βπ Exercise 14 Evaluate π/2 3 sin π₯ cos π₯ βπ/2 + sin π₯ cos π₯ ππ₯. 4.5: Integration by Substitution Objectives: Assignment: 1. To use a change of variables to evaluate indefinite and definite integrals β’ P. 304-305: 7-33 odd, 3747 odd, 61-67 odd, 70, 71-81 eoo, 83, 87 2. To use symmetry as an aid in integration β’ P. 306: 101-107 odd