 # WARM UP EXERCISE For an average person, the rate of change of weight W with respect to height h is given approximately by dW/dh = 0.0015 h2 Find W (h),

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WARM UP EXERCISE For an average person, the rate of change of weight W with respect to height h is given approximately by dW/dh = h2 Find W (h), if W (60) = 108 pounds. Also, find the weight of a person who is 70 inches tall.

§12.2 Integration by Substitution.
The student will learn about: integration by substitution, some substitution techniques, and applications.

Reversing the Chain Rule.
Recall the chain rule: This implies that: This means that in order to integrate g (x) its derivative, g ‘ (x), must be present.

General Integral Formulas
1. ∫ [f (x) ] n f ‘(x) dx = 2. ∫ e f (x) f ‘(x) dx = e f (x) + C 3.

Example ∫ [f (x) ] n f ‘(x) dx =
Note that the derivative of x 5 – 2 , (i.e. 5x 4 ), is present and the integral appears to be in the chain rule form ∫ [f (x) ] n f ‘(x) dx, with f (x) = x 5 – 2 . ∫ [f (x) ] n f ‘(x) dx = It follows that:

Differential If y = f (x) is a differentiable function, then
1. The differential dx of the independent variable x is any arbitrary real number. The differential dy of the dependent variable y is defined as the product of f ‘ (x) and dx – that is, as dy = f ‘ (x) dx

Examples If y = f (x) = x 5 – 2 , then dy = f ‘ (x) dx =
dy = f ‘ (x) dx = 5x 4 dx 2. If y = f (x) = e 5x , then dy = f ‘ (x) dx = 5e 5x dx 2. If y = f (x) = e 5x , then dy = f ‘ (x) dx = 3. If y = f (x) = ln (3x -5), then dy = f ‘ (x) dx =

Integration by Substitution
By example (Substitution): Find ∫ (x2 + 1)5 (2x) dx For our substitution let u = x2 + 1, then du/dx = 2x, and du = 2x dx and the integral becomes ∫ u5 du which is = u6/6 + C and reverse substitution yields (x2 + 1)6/6 + C.

General Indefinite Integral Formulas.
∫ un du = Very Important ∫ e u du = e u + C

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.)

Example 2 ∫ (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 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

Example 3 ∫ (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 Step 3 – Integrate. ∫ u 1/2 du = u 3/2/(3/2) + c = 2/3 u 3/2 + c Step 4 – Express the answer in terms of x. 2/3 u 3/2 + c = 2/3 (x2 + 5) 3/2 + c

Substitution Technique 1 – Example 1
1. ∫ (x3 - 5)4 (x2) dx Let u = x3 – 5 then du = 3x2 dx ∫ (x3 - 5)4 (x2) dx = 1/3 ∫ (x3 - 5)4 (3x2) dx = 1/3 ∫ u4 du = 1/3 u5/5 = 1/15 u5 = Note – we need a 3. 1/15 (x3 – 5)5 + c 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. 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.

Substitution Technique 1 – Example 2
2. Let u = 4x3 then du = 12x2 dx Note – we need a 12. 1/12 ∫ eu du = 1/12 eu + c = 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.

Substitution Technique 1 – Example 3
3. Let u = 5 – 2x2 then du = - 4x dx Note – we need a - 4. -1/4 · u - 4 /-4 = 1/16 (5 – 2x2) c = In this problem we had to insert a multiple - 4 in order to get things to work out.

Substitution Technique 2 – Example 1
4. Let u = x + 6 then du = dx ∫ x · (x + 6)8 dx = ∫ x u8 du = and this is ok, but we need to get rid of the x. However, since u = x + 6, x = u – 6, so the integral becomes ∫ (u – 6) · u8 du = ∫ u9 – 6u8 du = u10/10 – 6 u9/9 + c = 1/10 · (x + 6)10 – 2/3 · (x + 6)9 + c In this problem we had to be somewhat creative because of the extra x.

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 Continued on next slide.

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 and du = 3 dx Continued on next slide.

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, becomes Continued on next slide.

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 That 75 bottles sell for \$1.60 per bottle. and c = 2 Continued on next slide.

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. So

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

Practice Problems §6.2; 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 43, 47, 51, 55, 59, 63, 67, 69, 71, 77, 79.

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