1. 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.

2. §12.2 Integration by Substitution.
The student will learn about:
integration by substitution,
some substitution techniques, and
applications.

3. 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.

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

5. 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:

6. 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

7. 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 =

8. 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.

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

10. 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.)

11. 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

12. 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

13. Substitution Technique 1 – Example 11.
∫ (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.

14. Substitution Technique 1 – Example 22.
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.

15. Substitution Technique 1 – Example 33.
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.

16. Substitution Technique 2 – Example 14.
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.

17. 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.

18. 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.

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.
With u = 3x + 25,
becomes
Continued on next slide.

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.
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.

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. So

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

23. 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.