When Dorothy visited Munchkin Land, it's a good thing for her that she arrived when she did. For on the day she arrived, the InterOz Highway Act was still recent history. Had Dorothy arrived only a few months earlier, the Yellow Brick Road would still have been under construction, and Auntie Em would have had to await its completion before her niece could journey to The Emerald City and find her way back to Kansas.
The Emerald City Public Works Commission had contracted the firm of Follup and Gollup Munchkin Construction to lay the section of the Yellow Brick Road that passed through Munchkin Land. The contract was very explicit. Not only did it require all the pavement to be of yellow brick, but the commission had, in its bureaucratic wisdom, also required that each course be precisely 25 bricks laid lengthwise across and that there be exactly 100 bricks per square meter of Yellow Brick Road.
At first it seemed easy for Follup and Gollup to comply. They would just make the road 5 meters wide and use bricks that were 20 centimeters by 5 centimeters. So each meter of road would have 20 courses of 25 bricks each. Every 5 square meters of road had 500 bricks  it met all the requirements. And without another thought they went and ordered bricks to those specifications.
But then Follup and Gollup read deeper into the contract and found a clause written in at the behest of the Greater Oz Environmental Preservation Committee. It required that no tree of greater than 35 centimeter diameter be cut down to make way for the new road. The problem was that the plan for the road took it right through the old growth forest of Munchkin Land, where quite often there was only 4 or even 3 meters between thick, gnarled trees.
"Well," said Follup, "we'll just have to make the road narrower where we have to squeeze between trees. That means fewer bricks in each course in the bottlenecks."
"No no no!" Gollup complained. "The spec is very clear. It says 25 bricks across. What we'll do instead is cut the bricks shorter so the road can be narrower and still get our 25 bricks across."
"That won't work either," said Follup. "If we do it that way we'll have more than the 100 bricks per square meter that the spec demands."
Alas, they had already spent the money to buy the bricks, and since they could think of no way to meet the specification through the old growth forest, they resigned themselves to losing money on the project. The only productive course of action they both could agree on now was to drown their sorrows. So to the tavern the two of them shuffled.
But who should they see at the bar but Grundleck, the drunken old sorcerer who had once been married to the Wicked Witch of the East till she had thrown him out of her castle a century and a half earlier. Follup and Gollup bought the sodden geezer a pitcher of ale and then proceeded to explain their problem to him.
"We would gladly pay you a handsome sum in gold if you could enchant some of our bricks to make them shorter and fatter," Follup offered.
"Aw poo!" Grundleck grunted. "I'll do it for free. Just give me one pallet of them bricks so I can make a footpath in front of the witch's castle."
"Really," said Gollup. "I thought you hated her."
"You're dern right. I'll put a spell on them suckers so that when the old bag walks on 'em they'll turn into bats and flutter around under her dress. That'll make my day."
Follup and Gollup weren't completely sure it would make their day, especially if the witch ever found out they had anything to do with it. But they didn't see that they had much choice. So they agreed, hoping that old Grundleck would stay too drunk ever to lay a footpath anywhere.
Grundleck still knew enough magic to make the bricks destined for the forest take on the consistency of sillyputty. Follup and Gollup could squeeze them shorter and fatter to narrow the road through the bottlenecks. Yet no matter how much they squeezed a brick, its width times its breadth remained at exactly 100 square centimeters. It worked out perfectly. And to top it all off, the bricks grew hard again once the first rain fell on them.
And the Wicked Witch? Well one fine sunny day Grundleck did finally have his prank on her. Then while the witch was shaking the last of the bats out of her dress, somebody dropped a house on her.
The figure to the right shows an incredible coincidence of laying yellow bricks onto different shaped areas. The two areas shown, though different in shape, are identical in area. In each case the area is exactly 81/16, just or 1/16 more than exactly 5 boxes in the graph. So imagine that each of the yellow bricks occupies 1/16 of the area of one box on the graph. The Munchkins would need 81 such bricks to pave each of the areas shown in the figure. 

But how are the bricks dimensioned? You have plenty of choices. You could have each brick be 1/4 by 1/4. Or you could have each one be 1/2 by 1/8. And you can certainly see there are an infinitude of other combinations of width and breadth of a brick whose area will multiply out to be exactly 1/16 of a box in area.
Suppose that to pave the area under the blue function
(that is
Now I want to pave the area under the green function
(that is Just like the Munchkins, it looks like I'm going to have to squeeze some bricks. Figure 11.33 shows the arrangement of squeezed bricks, except that I had to expand your view of it vertically by a factor of 4 because the bricks on the right are squeezed too thin to show using the original scaling. 

Now suppose you didn't know how to find the indefinite integral of
The trick, though, is to know exactly by how much you have to squeeze the bricks in each column.
Since we do know how to find antiderivatives for both f(x) and g(u), lets do so. From the last section you can use equations 11.24f and 11.23 to work these antiderivatives. Work both of them out for yourself, then scroll down a bit to see if you got the correct results.
You ought to have gotten
F(x) = 
1 ^{ } 1 
G(u) = 
1 ^{ } 1 
Here is figure 11.31 again.
Look at how the x axis that the blue function is plotted on
is related to the u axis that the green function is plotted on.
The relationship is precisely
We know that g(u) must be the derivative of G(u) with respect to u (why?).
But what happens if you assume that u is a function of x
(i.e., that we have u(x)) and you want to know the derivative of G(u(x))
with respect to x? You use


dG(u(x)) dG du 1Notice that the==x (2x) eq. 11.32 dx du dx 8
f(x) = g(u(x)) u'(x) eq. 11.33And I found it by observing that
duThat is one factor of f(x) (the 2x part) is the derivative of a part of the other factor (the x^{2} part). Now multiply both sides by dx and you have= 2x eq. 11.34a dx
du = 2x dx eq. 11.34bNow let's look again at the problem of taking the indefinite integral.
1 
1 
1 
1 ^{ } 1 ^{ } 1 
_______ x Ö x^{2} + 1 dx eq. 11.36a 
2 _______ 
1 
_______ 2x Ö x^{2} + 1 dx eq. 11.36c 
u(x) = x^{2} + 1 eq. 11.37awhich makes
duand then multiplying both sides by dx we get= 2x eq. 11.37b dx
du = 2x dx eq. 11.37cThese substitutions tell us just how we need to squeeze our bricks to make the integral in equation 11.36c a whole lot simpler. We substitute u for
1 
_ Öu du eq. 11.38a 
1 
_ 1 Öu du = 
u^{3/2} + C eq. 11.38b 
1 
_ 1 Öu du = 
u^{3/2} + C = 
1 
(x^{2} + 1)^{3/2} + C eq. 11.38c 
______ x Öx^{2} + 1
Let's try one that's a little trickier this time.
cot(q) dq eq. 11.39a 
cos(q) cot(q) =So we rewrite the original integral aseq. 11.39b sin(q)
cos(q) dq 
duand when you multiply both sides by dq, you find= cos(q) eq. 11.310a dq
du = cos(q) dq eq. 11.310bSo now substitute sin(q) with u and substitute
cos(q) dq 
du 
cos(q) dq 
du 
cot(q) dq = lnsin(q) + C eq. 11.311c 
Here are the steps you go through find the indefinite integral of a function using simple substitution of variables.
Step 1) Look for one part of the integrand that is the derivative of another. This is the only difficult part of this procedure. Sometimes you will have to algebraically munge the integrand to see it. For example, the integral we worked earlier might have been presented as
_______ Öx^{4} + x^{2} dx 
Remember that when you do simple substitution, you are doing the
dx 
2x dx 
du 
u^{1/2} du 
Step 2) This is where you make the substitution. You found one piece of the integrand that is the derivative of another piece. So you let u be equal to that other piece. When you find the derivative of u with respect to x and you multiply through by dx, you will see what expression you have to substitute with du. And whatever expression of x your u is, you substitute that expression with u throughout the rest of the integrand. It must substitute smoothly all the way through. That is, when you are done substituting both the u and the du, you must not have any x's left in the integrand. Here is an example of one that doesn't work because you can't substitute all the way through:
______ 3x^{2} Öx^{3} + x dx 
______ 3x^{2} Öx^{3} + 9 dx 
_ Öu du 
Step 4) Now that you've made the substitution to the new variable (we've been using u for that), you should have a new integrand that is easier to integrate than the original. So go ahead and integrate it in the new variable. But remember that this is not your final answer. To get your final answer, you must ...
Step 5) Substitute back to the original variable. We've been using x for that. In step 2 you came up with u (your new variable) as a function of x. So in the expression you got by integrating, substitute every occurrence of u with the expression in x that you used to make the original subsitution. That will give you your final answer.
Suppose you were asked to do the following definite integral:
p/4 q=0 
sin(q)cos(q) dq eq. 11.312a 
p/4 q=0 
sin(q)cos(q) dq = 
_ Ö2/2 u=0 
u du eq. 11.312b 
p/4 q=0 
sin(q)cos(q) dq = 
_ Ö2/2 u=0 
u du = 
u^{2} 
_ Ö2/2 u=0 
eq. 11.312c 
p/4 q=0 
sin(q)cos(q) dq = 
_ Ö2/2 u=0 
u du = 
u^{2} 
_ Ö2/2 u=0 
1 = 
The point is that when you do a definite integral using simple substitution, you can skip the step of substituting back at the end, but only provided that you remembered to substitute the limits of integration according to the same substitution equation you used on the rest of it. This is a useful shortcut, especially in a whole raft of basic physics problems.
Solutions will be online soon
1) Use simple substitution of variables to find the indefinite integral of
x e^{x2/2} dx 
2) Use simple substitution of variables to find the indefinite integral of
__________ sin(q) Ö1  cos(q) dq 
3) Use simple substitution of variables to find the indefinite integral of
cos^{3}(q) dq 
4) Use simple substitution of variables to find the indefinite integral of
2q cos(q^{2}) sin(q^{2}) dq 
5) Use simple substitution of variables to find the indefinite integral of
(3x^{2}  6x + 2) dx 
Move on to (not available yet)
email me at hahn@netsrq.com