CONSTANTS OF INTEGRATION
Under this section we require to address a couple of sections about the constant of integration. During most calculus class we play pretty quick and loose with this and since of that many students do not really know it or how this can be significant.
Firstly, let's address how we play fast and loose with this. Recall that technically while we integrate a sum or difference we are really doing multiple integrals. For illustration,
∫15 x4 - 9 x-2 dx = ∫15 x4 dx - ∫9 x-2 dx
Upon calculating each of these integrals we must find a constant of integration for each integral as we actually are doing two integrals.
∫15 x4 - 9 x-2 dx = ∫15 x4 dx - ∫9 x-2 dx
= 3x5 + c + 9x-1 + k
= 3x5 + 9x-1 + c + k
Since there is no reason to think that the constants of integration will be the same from each integral we use different constants for each integral.
Here, both c and k are unknown constants and therefore the sum of two unknown constants is only an unknown constant and we acknowledge this through simply writing the sum like a c.
Therefore, the integral is then,
∫15 x4 - 9 x-2 dx =3x5 + 9x-1 + c
We also are liable to play fast and loose along with constants of integration in several substitution rule problems. Notice the following problem,
∫cos(1 + 2x) + sin(1 + 2x) dx = ½ ∫cosu + sin u du u = 1 + 2x
Technically while we integrate we must find,
∫cos(1 + 2x) + sin(1 + 2x) dx = ½ (sin u - cos u + c)
Because the complete integral is multiplied with ½, the entire answer, containing the constant of integration, must be multiplied with 1/2. Upon multiplying the ½ with the answer we determine as,
∫cos(1 + 2x) + sin(1 + 2x) dx = ½ sin u - ½ cos u + (c/2)
Though, as the constant of integration is an unknown constant dividing this with 2 isn't going to change this fact therefore we tend to just write the fraction like a c.
∫cos(1 + 2x) + sin(1 + 2x) dx = ½ sin u - ½ cos u + c
Generally, we don't really require worrying about how we have played fast and loose along with the constant of integration in either of the two illustrations above.
The real problem though is that as we play fast and loose along with these constants of integration most students don't actually have a good grasp of them and don't know that there are times where the constants of integration are significant and which we require to be careful with them.
To notice how a lack of understanding about the constant of integration can reason problems see the following integral.
∫1/(2x) dx
It is a really simple integral. Though, there are two simple ways to integrate this and which is where the problem arises.
The first integration way is to just break-up the fraction and perform the integral.
∫1/(2x) dx = ∫ ½ (1/x) dx = ½ In|x| + c
The second way is to utilize the subsequent substitution.
u = 2x du = 2dx => dx = ½ du
∫1/(2x) dx = ½ ∫(1/u) du = ½ In|u| + c = ½ In|2x| + c
Can you notice the problem? We integrated similar function and found very different answers. It doesn't make any meaning. Integrating similar function must provide us the same answer. We only used various methods to do the integral and both are perfectly legitimate integration methods. Therefore, how can using various methods produce various answers?
The first thing which we must notice is that since we used various method for each there is no cause to think that the constant of integration will actually be similar number and therefore we really must use various letters for each.
More suitable answers would be as,
∫1/(2x) dx =½ In|x| + c
∫1/(2x) dx =½ In|2x| + k
Here, let's take the other look at the second answer. By using a property of logarithms we can write the solution to the second integral as given here,
∫1/(2x) dx =½ In|2x| + k
= ½ (In2 + In|x|) + k
= ½ In|x| + ½ In 2 = k
Upon doing it we can notice that the answers actually aren't that different finally. Actually they only are different by a constant and we can even get a relationship among c and k. This looks like,
c = ½ In 2 = k
Therefore, without a specific understanding of the constant of integration, particularly using different integration techniques in similar integral will likely make a different constant of integration; we may never understand why we determined "different" solution for the integral.
Consider as well that getting answers that are different by a constant doesn't violate any principles of calculus. Actually, we've seen a fact which suggested as it might occur. We saw a fact into the Mean Value Theorem section which said as if f'(x) = g'(x) then f(x) = g(x) +c. conversely, if two functions have similar derivative then they can be different by no more than a constant.
It is exactly what we have got now. The two functions,
f(x) = ½ In|x| g(x) = ½ In|2x|
have exactly similar derivative,
1/(2x).
and as we have demonstrated they really only be different by a constant.
There is the other integral which also exhibits this behavior. See,
∫sin(x) cos(x) dx