3x2+5x+7, 3x2+5x+19, 3x2+5x?3.19, etc. plus more functions of the same sort. One of the difficulties, also, is that the same letter "C" is customarily used for all such arbitrary constants; but one computation may involve several different arbitrary constants. It would be more accurate to put subscripts on the C’s, to differentiate one of them from another – i.e., write
òu(x)v’(x)dx = u(x)v(x) - òu’(x)v(x)dx.
ò(1/x)(1)dx = (1/x)(x) - ò(-1/x2)(x)dx
òab u(x)v¢(x)dx = u(b)v(b) ? u(a)v(a) ? òab u¢(x)v(x)dx.
Loss or misuse of constants of integration. The indefinite integral of a function involves an "arbitrary constant", and this causes confusion for many students, because the notation doesn’t convey the concept very well. An expression such as
"3x2+5x+C" really is supposed to represent an infinite collection of functions – it represents all of the functions
Here is an example. The formula for Integration By Parts, in its briefest form, is
òudv = uv - òvdu; that can be understood more easily as
Now, that formula is correct, but it can easily be mishandled and can lead to errors. Here is one particularly amusing error: Plug
u(x)=1/x and v(x)=x into the formula above. We get
which simplifies to
Now, regardless of what you think is the value of
ò (1/x)dx, you just have to subtract that amount from both sides of the preceding equation, to obtain 0=1. Wait, how can that be???? Well, if we’re very careful, we realize that the two ò(1/x)dx’s on the two sides of the last equation are not actually the same. What that last equation really says is
That is a true equation, if we choose the constants
C1 and C2 appropriately – i.e., if we choose them so that C1?C2=1. Thus, the two constants are not independent of each other – they are not completely "arbitrary". Perhaps a more accurate explanation is this: The two expressions [ln|x| + C1] and 1+[ln|x| + C2] do not actually represent individual functions; rather, each of those expressions represents a set of functions.
Some students manage to make this kind of error even with definite integrals. They start from the formula
ò (1/x)dx = 1 + ò (1/x)dx, which is correct; but then when they "switch to definite integrals", they get the formula òab (1/x)dx = 1 + òab (1/x)dx, which is not correct. If you really want to "switch to definite integrals", you need to think of that constant 1 as a special sort of function. When you switch to definite integrals, any function p(x) gets replaced by p(b)?p(a). In particular, the constant function 1 is the function given by p(x)=1 for all x. So p(b)?p(a) becomes 1?1, or 0.
Some students may understand this better if we do the whole thing with definite integrals, right from the start. Let’s use the formula
Note that this formula has one more term than my previous boxed formula – when we convert
u(x)v(x) to the definite integral version, we replace it with u(b)v(b)?u(a)v(a). Now plug in u(x)=1/x and v(x)=x. We get
which (assuming 0 is not in the interval
[a,b]) simplifies to
which is true – i.e., there is no contradiction here.
Some students may be puzzled by the differences between the two versions of the Integration by Parts formula (in boxes, in the last few paragraphs). I will describe in a little more detail how you get from the definite integral formula (in the last box) to the indefinite integral formula (in the first box in this section). Think of a as a constant and b as a variable, and you’ll get something like this:
Note that the
u(b)v(b) term gets replaced by u(x)v(x), and the u(a)v(a) term "disappears" because it is constant. Finally, we can "absorb" the arbitrary constants into the indefinite integrals – i.e., we don’t need to write C1, C2, C3, because any indefinite integral is only determined up to adding or subtracting a constant anyway. Thus, we arrive at the briefer formula ò u(x)v¢(x)dx = u(x)v(x) ? ò u¢(x)v(x)dx.
Handling constants of integration gets even more complicated in the first course on differential equations, and there are even more kinds of errors possible. I won’t try to list all of them here, but here is the simplest and most common error that I’ve seen: In calculus, some students get the idea that you can just omit the "+C" in your intermediate computations, and then tack it on at the end of your answer, if you know which kinds of problems require an arbitrary constant. That will usually work in calculus, but it doesn’t work in differential equations, because in differential equations the "C" can show up anywhere – not necessarily as a "+C" at the end of the answer.
Here’s a simple example: Let’s solve the differential equation xy¢+7=y (where y¢ means dy/dx). One way to solve it is by the following steps: