Question

Find:

Answer 1

The derivative of \(\int_ 0 ^ 1 \dfrac {1 -\sqrt {x}}{1 + \sqrt{x}}\).

Answer 2

Suggestions:

1) Check your limits of integration. I'd expected to see at least an "x' as your upper limit.
2) Next, be sure to include the differential dx.

Answer 3

\(\int_ 0 ^ 1 \dfrac {1 -\sqrt {x}}{1 + \sqrt{x}}dx\)

Answer 4

iS there any chance that you meant\[\int\limits_ 0 ^ x \dfrac {1 -\sqrt {t}}{1 + \sqrt{t}}dt~ ?\]

Answer 5

There are no x?

Answer 6

It's with respect to x this time.

Answer 7

No x's in the limits i meant.

Answer 8

Note: here, I'm using "t" as a dummy variable. The lower limit of integration is 0 and the upper limit x.
OK. could you possibly mean:\[\frac{ d }{ dt }\int\limits_ 0 ^ t \dfrac {1 -\sqrt {x}}{1 + \sqrt{x}}dx~?\]

Answer 9

Here, x is the "dummy variable" and we are differentiating with respect to t.

Answer 10

If y es, then the result would look like this:\[ \dfrac {1 -\sqrt {t}}{1 + \sqrt{t}}\]

Answer 11

If you have the original problem on your computer display, why not take a screen shot and upload the screen shot to here?

Answer 12

\(\frac{ d }{ dx}\int\limits_ 0 ^ 1 \dfrac {1 -\sqrt {x}}{1 + \sqrt{x}}dx\)
This is how the question looks like in my book.

Answer 13

This is an unusual problem, then. Since the integral would be a constant, I suspect the derivative of this integral would simply be zero (since the derivative of a constant is zero). What do you think?

Answer 14

What i did first was find the anti-derivative of the function, and then used the first theorem of calculus. I also got 0, but the book got -1.

Answer 15

One thing I'd try would be to replace that limit of integration 1 with x. Then the derivative of your entire definite integral would be
\[\dfrac {1 -\sqrt {x}}{1 + \sqrt{x}}\] What would happen if we were now to replace that x by 1? I get 0/2 = 0. I don't see how the result could be -1, frankly.

Answer 16

Why would you do that?

Answer 17

Ok then.

Answer 18

What if the lower limit wasn't a one, like pi?

Answer 19

???? Your lower limit is currently 0, isn't it? and your upper limit 1, right?

Answer 20

I know. But I'm giving an example.

Answer 21

If you're going to change either limit of integration, you'll need to ensure that any such limit (or limits) is/are within the domain of the function in question. Note that both 0 and 1 are within the domain of your integrand here, and Pi would be also.

Hate to say it, but I don't think I know enough in this particular case to advise you what to do next. My suggestion is that you carefully write out a solution based upon our discussion and then show that solution (even if you think it's wrong) to your instructor and ask for guidance. If you don't see your instructor in person, can you share sketches/illustrations with that person as attachments to e-mail?

Answer 22

Teachers appreciate it when you do all you can towards finding a solution before asking them for help.

Answer 23

I haven't got any of my teachers e-mails.