Karl's Calculus Forum: March 2003

Now let  v = 1 - u,  (same as  v + 1 = u)  and  -dv = du.  That gives you a single variable under the square root.

Finally let  w² = v,  and  2w dw = dv.  That will convert it to something that you can integrate by trig substitution.

These methods are all covered in the More Substitutions section.
Karl <Click to Send Email to Karl>
USA - Sunday, March 30, 2003 at 16:44:50 (EST)

      1
-------------  dx
sqrt(1-e^(-x))

Thanks in advance.

   2

 u(1 - u)
          du
  1 + u

On the second one, if  u⁶ = x,  then  6u⁵ du = dx.  The integral becomes

  6

   u5
         du
 u3 + u2

The integral of

1-sqrt(x)
--------- dx      (Hint: Substitute u = sqrt(x).)
1+sqrt(x)

and

The integral of

      1
--------------- dx   (Hint: Substitute x=u^(1/6)
sqrt(x)+(x)^1/3

x dx
_____
4+x^2

Looks like many people are at the "integration by parts" stage of the semester. This is another parts style integration.

Look at how you could solve this integral using integration by parts. You have to split the integrand into a u and a dv part. Whatever you pick for u, you'll have to take the derivative of, and whatever you pick for dv, you'll have to take the integral of. Since the integrand is only a single part (namely, just arccos(x)), you can't split that, so something is going to have to be arccos(x), and something is going to have to be dx.

How do you split it? Take a minute and think about it. You're trying to find arccos(x)'s integral, so I don't think you want to make dv=arccos(x), do you? That means it has to be u, and then dv=dx. Work it out and you're on your way to the answer.

Good luck!
Jason K.

Ugly looking integral. When in doubt, first try to make it simple(r). What is common in the integrand? Well, there are two occurences of [sin(x)+cos(x)] in there. It might make things easier if you get rid of those.

I'd take a look at setting u=sin(x)+cos(x), and then rewrite the integral completely in terms of u. That means you need to figure out du, and you'll also want to rewrite the upper and lower bounds of the integral. (To find out the new bounds, just plug in 0 and p into your formula for u, and you'll get the new bound you want.) Once you do all this, you should find that the integral becomes really, really easy to do!

Good luck!
Jason K.

The solution is trivial. When you take the derivative of an integral of a function f(x), you will get the exact same thing: f(x). (Note: you will NOT get the same f(x) if you take the integral of the derivative of f(x). Can you see why?)

Since your f(u) in the integrand is f(u)=u²e^u, we know that no matter what the integral of f(u) is, if we take the derivative, we'll get f(u) back. Now, remember what I said before about the derivative of a constant/number - it will always be 0. So, the lower bound evaluation of the integral (at 3), will end up being some number, and when we take the derivative of it, it goes to 0. You can ignore that bound.

If you pretend F(u) is the theoretical integral of f(u), and you evaluate it at the upper bound of 'x', then you get a function F(x). Take the derivative of that to get f'(x). If you've followed what I've mentioned up to now, you'll see the quick and simple answer ahead of you. If not, you can work the integral out the long way by parts, and then do the derivative.

Good luck!
Jason K.

This is one of those problems that I swore I had the answer to as soon as I read it, but I'm suspicious about. It seems too easy to me, so I'm not sure I should be trusted! :-)

If you have a definite integral, that is evaluated at two numerical (ie, non-variable) extents, then it would stand to reason that the answer will be numerical as well. No matter what the original function is, when you evaluate the integral, you'll be evaluating it at both bounds, and so the variable of integration will resolve out of the function, and you'll be left with constants.

So, what my thinking is if you define your f(x) in terms of an integral of u, which is integrated over a range from 3 to p, then you have two choices: work out the integral by using integration by parts, and then plug in the bounds like always, and you'll get an answer, which you will then differentiate to get f'(x); your other choice is more intuitive: recognize that any definite integral that you evaluate from a number to a number will result in a number, and any time you take the derivative of a constant (number), you get 0.

The reason I'm suspicious about this easy of an answer is because you often see this sort of a problem when you start studying the Fundamental Theory of Calculus, and often one of the bounds you are integrating over (either upper or lower) has 'x' in it somewhere, and thus you learn the "tricks" of taking a derivative of an integral. Since the problem you gave didn't have an 'x' anywhere, I think this is the right answer.

Good luck!
Jason K.

I'm going to take a crack at helping you, but based on what you typed, I have to make a handful of assumptions. You mention in the information you gave that all dollars are expressed in hundreds of thousands of dollars (ie. 4 ~ 400,000), but when you use the variable 'x', you don't specify if x represents units in singles, hundreds, thousands, etc. Given the scope of the numbers involved, I am going to assume that x is in hundreds of thousands of units as well, though be aware of how that will change the problem/formulas if it's not. (I only picked hundreds of thousands of units since if x refers to single units, then the max profit question is laughably easy to answer...)
Anyway, first you need to get a profit function. You were close, but I think you overlooked one vital piece of the puzzle. Profit is simply expressed as Total Revenue - Total Cost...what you made hopefully will be more than it cost you to make/sell, otherwise you have negative profit, otherwise known as debt. (Like my credit card bills...)
What contributes to the company's costs, first? Well, outright every year they have $3 in costs, whether they make anything or not. (I say $3, but it's supposed to mean $300,000 - I'm just abbreviating.) The average cost per item is given by the problem, and that's 15x. But, keep in mind, this is the average cost per item, not the total cost of making x items in a year. That's partly where you were close, but not right on. So, given this, we should be able to get what we need:

Total Costs = Annual Base Cost + Total Cost Of Producing X Items
            = 3 + (Cost Per Item) * (Total Number Items Made)
            = 3 + (15x) * (x)
            = 3 + 15x2

Total Revenues = (Sales Price Per Item) * (Total Number Items Sold)
               = (x2+72) * (x)
               = x3+72x

Total Profit = (Total Revenues) - (Total Costs)
             = (x3+72x) - (3+15x2)
             = x3-15x2+72x-3

• When you are looking for times when the profit is increasing or decreasing, think of the derivative. (In word problems, "buzz" words like increasing/decreasing/changing/rate should clue you to the potential for needing a derivative somewhere.) In order to see how the profit is changing for a given number of items x, you will want to find f'(x), which would be the derivative of the profit function we just discussed.
• For what intervals the profit function is increasing, you will want to examine the profit derivative, and look to see where it is positive. I would suggest you first look to see where it is zero, in other words, where f'(x)=0. These "spots" of x will be critical spots where, in general, the profit function has either bottomed out (worst profits), or topped off (best profits). You'll need to look at the second derivative, or f''(x), to know which case it will be. Given those two pieces of information, you should have an idea what the profit function is doing around those points. (Think along the lines of 0 units --> rising or falling --> some critical --> rising or falling --> some critical --> rising or falling --> 800000 max units.)
• When asked over what intervals the decreasing/increasing rate itself is increasing, think of a derivative. It's just that you're looking for the derivative of the derivative itself, which as I said before, is the second derivative, or f''(x). Use that to answer those type of questions.
• To answer how many items should be made to get the best profit, you will want to look for any maximums of the profit function in the range you have to work with. This could be where the derivative is 0, like I said above, it could be at x=0 (when no units are made), or it could be at x=800000 (when most units can be made). Where the derivative is 0, it could be either a high or a low, so use f''(x) to find out (not to mention you should evaluate the profit f(x) at that point too!).
• When looking for what intervals will there be no profit, remember that in order to have a profit, you have to make money. So look for wherever the function is "making money", or positive. Negative profit is a bad, bad thing. By the time you get to this question, you should already have found out exactly what you need to know to answer it.
• One note! I'm still unsure of whether x in your problem is single units, or hundreds of thousands, so watch yourself! Since your last question asks for "approximate" intervals, I think I was okay by saying x=1 would mean 100,000. Given that guess, let's say you find out x=3.345 is an interesting point (I'm not saying it is!) I would then say on your test you would say approximately 334500 units, if you catch my drift.

a) The problem becomes markedly more difficult than before, but not uncontrollably so. Again, you have to work with the volume formula, namely p[f(x)]² for the normal x-axis rotation. This complicates the integral you'll have to deal with to the following:

p                 p                   p
 p [f(x)]2 dx = p [x(sin x)]2 dx = p x2sin2x dx
0                 0                   0

b)This problem is far trickier than the earlier problem since we can't take advantage of any symmetry in the function itself. Since sin(x) is was symmetrical around x=p/2 on either side, we could rotate it about the y-axis rather than the line x=p, and get the exact same shape. (Can you see why that was?)
However, the function x*sin(x) is clearly not symmetrical about it's middle, so the same trick will not work here, which means we have to rethink the setup. I would suggest you use a graphing calculator, or hand-drawn graph, to look at the function you'll be rotating. There are two solutions here: express f(x) in terms of y, and then rotate about the y-axis using the standard rotation formula of p[g(y)]², or use the "shells" method. Since it might be too fun to try and turn y=x*sin(x) into a function of y, I'll go the "shells" route.
Deriving the formular for volume of rotation via the "shells" method is too in depth to go into here, so I'll just repeat the formula. (I encourage you to look it up on the Internet, and understand how it is derived, though.) For a volume rotated about an axis perpendicular to the x-axis (such as our x=p), the volume can be found as follows:

b
 2p(radius)*f(x) dx
a

p                        p
 2p(p-x)(x*sin(x)) dx =  (2p2-2px)(x*sin(x)) dx =
0                        0

p                 p
 2p2xsin(x) dx -  2px2sin(x) dx
0                 0

Good luck!
Jason K.

a) First, you need to know how to set up the problem. I would suggest graphing the function and it's rotation about the x-axis. This can help you visualize what you are working with. Since you asked about the integration by parts method, I will assume that you are aware of how to at least set up the integral. In case you do not, recall that you can derive the formula for a volume of a rotated area by "carving" the area under the curve into circular discs, and then summing them over the range, while making the disc slices smaller and smaller. In any case, the formula for the volume of a rotation of a function, f(x), about the x-axis from x₁=a to x₂=b is:

b
 p[f(x)]2 dx
a

 p                 p                  p
  p[sin(x)]2 dx =  p sin2(x) dx = p  sin2(x) dx
 0                 0                  0

u  = sin(x)       dv = sin(x)dx
du = cos(x)dx     v  = -cos(x)

sin2(x) = 1/2 - [cos(2x)]/2

b
 2px[f(x)] dx
a

 p                    p
  2px[sin(x)] dx = 2p x sin(x) dx
 0                    0

Good luck!
Jason K.
Jason K.