1. A quick question about integrals

If a problem asks to evaluate
$\int_{3}^{4} (2x-4)^\frac{-1}{2} dx$
and I do this:
$=[(2x-4)^\frac{1}{2}]_3^4$
And I eventually get:
$=\sqrt{4} - \sqrt{2}$
Would I get multiple answers? As:
$=\pm 2$ - $\pm \sqrt{2}$

Because the answers in my textbook say there's only one answer, which is: $2 - \sqrt{2}$
Thanks!

2. Re: A quick question about integrals

Hey winkyinky146.

I'm interested to know why you think there are two answers.

For integrals you get F(b) - F(a) and both F(b) and F(a) have only one value.

It's not the same as solving for roots which is completely different - think in terms of doing each step arithmetically or algebraically - you input a value into the function and get one unique value.

3. Re: A quick question about integrals

Oooh, I see. I have just realised that integrals are determined by subtracting functions, for some reason I was under the misconception that relations could also be used, that's why I thought F(b) could give two values. But does that mean that even though it has only one answer, it can be any one of the two? e.g can the answer also be $-2 + \sqrt{2}$? (which I know is negative).

4. Re: A quick question about integrals

There are two values of x such that $x^2= 3$. They are $x= \sqrt{3}$ and $x= -\sqrt{3}$. The reason we need to write both is that $\sqrt{3}= 3^{1/2}$ itself is the positive root, not both.

5. Re: A quick question about integrals

Originally Posted by HallsofIvy
There are two values of x such that $x^2= 3$. They are $x= \sqrt{3}$ and $x= -\sqrt{3}$. The reason we need to write both is that $\sqrt{3}= 3^{1/2}$ itself is the positive root, not both.
I'm sorry, I understood the first sentence but I'm not quite sure about the second. Could you please evaluate?

6. Re: A quick question about integrals

My point was that $\pm\sqrt{3}$ means "both $\sqrt{3}$ and $-\sqrt{3}$". If just $\sqrt{3}$ referred to both numbers, we would not need the " $\pm$".

7. Re: A quick question about integrals

I see. But technically the graph of $(2x-4)^\frac{1}{2}$ can have a range of from 0 to negative infinity right? I understand that it has to be a function, but is there a rule where the range can only be the positive numbers? Thanks

8. Re: A quick question about integrals

no ...

$y=-(2x-4)^{1/2}$ has the range $y \le 0$

9. Re: A quick question about integrals

When you plug the numbers into the formula do you get one answer or two?

10. Re: A quick question about integrals

Originally Posted by skeeter
no ...

$y=-(2x-4)^{1/2}$ has the range $y \le 0$
Oh I get it now! Thanks for helping