UsukiDoll
  • UsukiDoll
Converting to a double integral Evaluate the integral... see attachment
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
UsukiDoll
  • UsukiDoll
1 Attachment
UsukiDoll
  • UsukiDoll
hint: write the integrand as an integral
UsukiDoll
  • UsukiDoll
where do I even begin? @wio

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anonymous
  • anonymous
Hmmm, not completely sure actually.
UsukiDoll
  • UsukiDoll
I wasn't even taught how to do it but I have to do it for hw wt hexagon
anonymous
  • anonymous
However: \[ f(x) = \int^x_{f^{-1}(0)} f'(t)dt \]
UsukiDoll
  • UsukiDoll
??????????
anonymous
  • anonymous
That is how you'd turn a function into an integral.
anonymous
  • anonymous
You take the derivative and put it inside an integral.
UsukiDoll
  • UsukiDoll
so take the derivative of tan^-1pix - tan^-1x?
anonymous
  • anonymous
Yeah.
anonymous
  • anonymous
You also need to find the roots.
anonymous
  • anonymous
any root will do.
UsukiDoll
  • UsukiDoll
how do I find the roots??? so take the derivative and plug it into the integral and then find the roots?
anonymous
  • anonymous
Well you only need to find one.
anonymous
  • anonymous
Actually you don't necessarily have to find the roots.
anonymous
  • anonymous
\[ f(x) = \int^x_{f^{-1}(0)} f'(t)dt = \int^x_{0} f'(t)dt+f(0) \]
anonymous
  • anonymous
Just do that... that will be easier.
anonymous
  • anonymous
Did you find \(f'(t)\) and \(f(0)\)?
UsukiDoll
  • UsukiDoll
sorry maybe after I eat dinner hungry atm
UsukiDoll
  • UsukiDoll
back
UsukiDoll
  • UsukiDoll
|dw:1378794480198:dw|
UsukiDoll
  • UsukiDoll
|dw:1378794579010:dw|
UsukiDoll
  • UsukiDoll
isn't tan 90 degrees 0?
UsukiDoll
  • UsukiDoll
@wio I'm back
anonymous
  • anonymous
OK
anonymous
  • anonymous
Hmm well \[ \tan(0) = \frac 01=0 \]
anonymous
  • anonymous
What did you get for the derivative?
UsukiDoll
  • UsukiDoll
probably because tan 90 doesn't exist...I mean I still remember the graph of it.
anonymous
  • anonymous
Anyway \(f(0) = 0\)
UsukiDoll
  • UsukiDoll
|dw:1378794996127:dw|
anonymous
  • anonymous
What happens to the \(x\)?
UsukiDoll
  • UsukiDoll
it gets bigger?
anonymous
  • anonymous
No I mean in the first term.
anonymous
  • anonymous
The first term should have an \(x\).
anonymous
  • anonymous
Then we can try to add them I guess?
UsukiDoll
  • UsukiDoll
but the derivative of pix is pi and since it was in tan^-1x format I had to put it as pi/1+(pi)^2
anonymous
  • anonymous
Chain rule: \[ (f\circ g)' = (f'\circ g)\cdot g' \]
UsukiDoll
  • UsukiDoll
-___-
UsukiDoll
  • UsukiDoll
so leave the tan inverses alone and deal with the dxx? huh?
anonymous
  • anonymous
since \(g=\pi x\) I should see it somwehre.
anonymous
  • anonymous
Anyway, is it possible to simplify them?
anonymous
  • anonymous
\[ f'(t) = \frac{\pi}{1+(\pi t)^2}-\frac 1{1-t^2} \]
anonymous
  • anonymous
\[ f'(t) = \frac{\pi}{1+(\pi t)^2}-\frac 1{1+t^2} \]
anonymous
  • anonymous
I suppose this could be simplified.
UsukiDoll
  • UsukiDoll
-___- loooks like something I wrote earlier.
anonymous
  • anonymous
Yeah, but like I said the first term needs a variable.
UsukiDoll
  • UsukiDoll
so how does it get simplified lcd?
anonymous
  • anonymous
yeah
UsukiDoll
  • UsukiDoll
1 Attachment
anonymous
  • anonymous
Okay, I'm wondering if you aren't suppose to be checking the limits of integration first, actually...
anonymous
  • anonymous
Look at the limits of integration: \[ \int_0^2\int_0^xdtdx \]If we switch order of integration it might help us.
anonymous
  • anonymous
|dw:1378797162816:dw|
anonymous
  • anonymous
|dw:1378797231738:dw|
anonymous
  • anonymous
Meaning we could change our bounds to: \[ \int_0^2\int_t^2dxdt \]
anonymous
  • anonymous
I have no idea if this will work. I honestly don't now how to do this problem. But this is what they hinted at.
UsukiDoll
  • UsukiDoll
I'm being honest. I don't get it either
UsukiDoll
  • UsukiDoll
prof didn't even go thorough the steps or anyhting just assigned this for hw. sorry typos sleepy behind the pc screen
anonymous
  • anonymous
And I don't get the whole \(\tan^{-1}(\pi x)\). You don't take inverse trig functions on angles.
UsukiDoll
  • UsukiDoll
that's what it was displayed in my text
anonymous
  • anonymous
even wolfram alpha doesn't give anything that looks nice.

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