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Section 2 Integration by Substitution

Chapter 5 Integration Section 2 Integration by Substitution

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Learning Objectives for Section 5.2 Integration by Substitution

The student will be able to integrate by reversing the chain rule and by using integration by substitution. The student will be able to use additional substitution techniques. The student will be able to solve applications.

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Reversing the Chain Rule

Recall the chain rule: Reading it backwards, this implies that

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Special Cases

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Example Note that the derivative of x5 – 2 , (i.e. 5×4 ), is present and the integral appears to be in the chain rule form with f (x) = x5 – 2 and n = 3. It follows that

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Differentials If y = f (x) is a differentiable function, then

The differential dx of the independent variable x is any arbitrary real number. The differential dy of the dependent variable y is defined as dy = f ´(x) dx

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Examples If y = f (x) = x5 – 2 , then dy = f ´(x) dx = 5×4 dx

If y = f (x) = e5x , then dy = f ´(x) dx = 5e5x dx If y = f (x) = ln (3x – 5), then dy = f ´(x) dx =

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Integration by Substitution Example

Example: Find ∫ (x2 + 1)5 (2x) dx

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Example (continued) Example: Find (x2 + 1)5 (2x) dx Solution:

For our substitution, let u = x2 + 1, then du/dx = 2x, and du = 2x dx. The integral becomes ∫ u5 du = u6/6 + C, and reverse substitution yields (x2 + 1)6/6 + C.

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General Indefinite Integral Formulas

Very Important!

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Integration by 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. 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.)

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Example ∫(x3 – 5)4 (3×2) dx Step 1 – Select u. Let u = x3 – 5, then du = 3×2 dx Step 2 – Express integral in terms of u. ∫(x3 – 5)4 (3×2) dx = ∫ u4 du Step 3 – Integrate. ∫ u4 du = u5/5 + C Step 4 – Express the answer in terms of x. u5/5 + C = (x3 – 5)5/5 + C

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Example ∫(x2 + 5)1/2 (2x) dx Step 1 – Select u. Let u = x2 + 5, then du = 2x dx Step 2 – Express integral in terms of u. ∫ (x2 + 5)1/2 (2x) dx = ∫ u1/2 du Step 3 – Integrate. ∫ u1/2 du = (2/3)u3/2 + C Step 4 – Express the answer in terms of x. (2/3)u3/2 + C = (2/3)(x2 + 5)3/2 + C

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Example ∫(x3 – 5)4 x2 dx Let u = x3 – 5, then du = 3×2 dx

We need a factor of 3 to make this work. ∫(x3 – 5)4 x2 dx = (1/3) ∫(x3 – 5)4 (3×2) dx = (1/3) ∫ u4 du = (1/3) u5/5 + C = (x3 – 5)5/15 + C In this problem we had to insert a factor of 3 in order to get things to work out. Caution – a constant can be adjusted, but a variable cannot.

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Example Let u = 4×3, then du = 12×2 dx

We need a factor of 12 to make this work.

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Example Let u = 5 – 2×2, then du = –4x dx We need a factor of (–4)

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Example Let u = x + 6, then du = dx, and

We need to get rid of the x, and express it in terms of u: x = u – 6, so Section 6-2, # 68

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Application 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.

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Application (continued)

Solution: We need to find Let u = 3x + 25, so du = 3 dx.

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Applications (continued)

Now we need to find C using the fact that 75 bottles sell for $1.60 per bottle. We get C = 2, so

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