Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm doing basic problems on antiderivatives and there seems to be an inconsistency in my book.

The instructions for these problems are:

Find the most general antiderivative of the function.

Number 11 is:

11. $f(x) = \dfrac{10}{x^9}$

So, I naturally wrote down $F(x) = -\dfrac{5}{4x^8} + C$.

The solution manual says this is wrong. The function has domain $(-\infty, 0) \cup (0, \infty)$, so $F(x) = \begin{cases} -\dfrac{5}{4x^8}+C_1 & \text{if } x < 0 \\ -\dfrac{5}{4x^8}+C_2 & \text{if } x > 0 \end{cases}$

Okay, I thought. That makes sense.

I then come to number 13, which is:

13. $f(x) = \dfrac{u^4 + 3\sqrt{u}}{u^2}$.

I figured the domain is $(-\infty, 0) \cup (0, \infty)$, so I wrote down $F(x) = \begin{cases} \frac{1}{3}u^3-6u^{-1/2}+C_1 & \text{if } u > 0 \\ \frac{1}{3}u^3-6u^{-1/2}+C_2 & \text{if } u < 0 \end{cases}$.

Well, the solution book doesn't mention the domain and just says $F(x) = \frac{1}{3}u^3-6u^{-1/2}+C$.

I later realized that the $\sqrt{u}$ in the numerator and the $u^2$ in the denominator must limit the domain to the positive numbers, so the antiderivative doesn't need to be defined for anything but positive numbers. So, okay, I think I get it.

Next number 19 is:

19. $f(x) = \dfrac{x^5-x^3+2x}{x^4} = x - \dfrac{1}{x} + \dfrac{2}{x^3}$

So, again, since the domain appears to be $(-\infty, 0) \cup (0, \infty)$, I wrote down $F(x) = \begin{cases} \frac{1}{2}x^2 - \ln|x| - \dfrac{1}{x^2} + C_1 & \text{if } x > 0 \\ \frac{1}{2}x^2 - \ln|x| - \dfrac{1}{x^2} + C_2 & \text{if } x < 0 \end{cases}$.

But the book again doesn't mention the domain and just says that $F(x) = \frac{1}{2}x^2 - \ln|x| - \dfrac{1}{x^2} + C$.

I am confused. This doesn't seem consistent, especially between #11 and #19. Why is my answer not right for #19?

share|improve this question
5  
Unless I'm missing something, I think you're right. You can chose a different constant for each piece (connected component) of the domain. –  Joel Cohen Sep 5 '11 at 20:25
4  
The person who wrote out the solutions presumably thought that the absolute value sign takes care of the problem. It doesn't. Congratulations on reading with care. –  André Nicolas Sep 5 '11 at 20:32
    
Okay, thanks guys! I was mostly wondering if I had a misunderstanding about functions and their domains. –  Matt Gregory Sep 5 '11 at 20:39
1  
Some solution writers wouldn't have been so careful on #11. Might the solutions to #11 and #19 have been written by different people? –  Jonas Meyer Sep 5 '11 at 20:39
1  
I assume $u$ and $x$ are supposed to be the same variable? You shouldn't use both in the same equation. –  Robert Israel Sep 5 '11 at 21:17
show 2 more comments

1 Answer

up vote 3 down vote accepted

The person who wrote out the solution may have thought that putting absolute value signs around $x$ takes care of the problem. It doesn't. Congratulations on reading with care.

The answer for Question $19$ ought to be something like $$F(x) = \begin{cases} \frac{1}{2}x^2 - \ln x - \dfrac{1}{x^2} + C_1 & \text{if } x > 0 \\ \frac{1}{2}x^2 - \ln(-x) - \dfrac{1}{x^2} + C_2 & \text{if } x < 0 \end{cases}.$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.