Differentiate: 1/(2x-1)^(1/2) dx from x=5 to x=13

- anonymous

Differentiate: 1/(2x-1)^(1/2) dx from x=5 to x=13

- chestercat

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- anonymous

Use a u sub, and then F(13)-F(5)

- yuki

if you can differentiate
\[\int\limits 1/x dx\]
then you will know how to do it.
do you need more help ?

- anonymous

Yes I do, sorry.

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## More answers

- yuki

your integral looks awfully like 1/x, it's just shifted a little.
you you can guess that it is the family of 1/x.
can you integrate 1/x ?

- anonymous

log x +c?

- yuki

it's actually lnx + C

- anonymous

That's what I meant, lol

- yuki

okay just making sure

- anonymous

I just really don't understand definite integrals.

- yuki

now your goal is to make the integral look like that by letting
u = 2x+1

- yuki

oops, never mind about lnx, i did not see the ^1/2 part.
so it will be a family of \[\int\limits 1/\sqrt(x)\]

- anonymous

But then what?

- yuki

after letting u = 2x-1, \[\int\limits 1/\sqrt(u) dx\]

- yuki

is what you are going to get, but the problem is that
the integrand is now a function of u, which cannot be
integrated over x.
so you have to over come that by figuring out what "dx" is.
that technique is the so-called u-substitution.
do you know how to find dx ?

- anonymous

No

- yuki

okay, so u = 2x-1 right ?
if you differentiate both sides with respect to x,
d/dx(u) = d/dx(2x-1)
d/dx(u) = 2
do you get this so far ?

- anonymous

Yes

- yuki

so if you multiply dx on both sides and divide 2 on both sides,
you will get 1/2 du = dx
this is what is happening, but it is traditional and easier to say
u=2x-1
du = 2 dx
by differentiating both sides

- anonymous

Gotcha

- yuki

so once you substitute dx with the above equation,
your integral becomes\[1/2\int\limits 1/\sqrt(u) du\]

- yuki

now it is important to know that the limits of
the integration will change as well,
unless you want to know how to do that I won't
go into the detail.
to avoid having to change the limits, you will
find out the indefinite integral first, then substitute
2x-1 back into u.
that way you can plug in the limits again.

- anonymous

so the answer is 2?

- yuki

\[1/2\int\limits 1/\sqrt(u) du = 1/2 *1/2*(u^{1/2})+C\]

- yuki

1/4(2x-1) + C =F(x)
so F(13) -F(5)
is your answer.
just as a reminder, C will not matter so you can ignore it

- yuki

woops\[1/4 \sqrt(2x-1) +C\]
is what I meant

- yuki

it seems like the answer will be 3/4

- anonymous

Oh, ok, I see. Thanks!!! Can you help me with a couple word problems too?

- yuki

I need to go soon, so I can help you with one of them.
ask me the one that you think need most help

- anonymous

A manufacturer has been selling flashlights at $6 a piece, and at this price consumers have been buying 3000 flashlights per month. The manufacturer wishes to raise the price and estimates that for each $1 increase in price, 1000 fewer flashlights will be sold each month. The manufacturer can produce the flashlight a cost of $4 per flashlight. At what price should the manufacturer sell the flashlights to generate the greatest profit?

- yuki

okay so what do you not get from this problem ?

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