- #1
Cosmophile
- 111
- 2
We are going over antiderivatives in my calculus course and reached a question regarding ##f(x) = \frac {1}{x}##.
My instructor went on to say that ##\int \frac {1}{x}dx = \ln |x| + C##. This makes sense to me, but only to a certain point. For ##f(x) = \frac {1}{x}##, ##f## is defined ##\forall x \neq 0##. So we should have two interavls which we are looking at: ##x < 0## and ##x > 0##. Because of this, we would then have:
[tex] \int \frac {1}{x}dx = \ln |x| + C_1 \qquad x > 0 [/tex]
[tex] \text {and} [/tex]
[tex] \int \frac {1}{x}dx = \ln (-x) + C_2 = \ln |x| + C_2 \qquad x < 0 [/tex]
The second integral comes into being because ##-x > 0## when ##x<0##. I brought this up to my teacher and he said that it made no sense and served no purpose to look at it this way. The argument I brought up was that the constants of integration could be different for the two intervals. I was hoping some of you may be able to help me explain why this is the case, or, if I am wrong, explain to me why I am.
Thanks!
My instructor went on to say that ##\int \frac {1}{x}dx = \ln |x| + C##. This makes sense to me, but only to a certain point. For ##f(x) = \frac {1}{x}##, ##f## is defined ##\forall x \neq 0##. So we should have two interavls which we are looking at: ##x < 0## and ##x > 0##. Because of this, we would then have:
[tex] \int \frac {1}{x}dx = \ln |x| + C_1 \qquad x > 0 [/tex]
[tex] \text {and} [/tex]
[tex] \int \frac {1}{x}dx = \ln (-x) + C_2 = \ln |x| + C_2 \qquad x < 0 [/tex]
The second integral comes into being because ##-x > 0## when ##x<0##. I brought this up to my teacher and he said that it made no sense and served no purpose to look at it this way. The argument I brought up was that the constants of integration could be different for the two intervals. I was hoping some of you may be able to help me explain why this is the case, or, if I am wrong, explain to me why I am.
Thanks!