Converting to a double integral
Evaluate the integral... see attachment

- UsukiDoll

Converting to a double integral
Evaluate the integral... see attachment

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- UsukiDoll

##### 1 Attachment

- UsukiDoll

hint: write the integrand as an integral

- UsukiDoll

where do I even begin? @wio

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## More answers

- anonymous

Hmmm, not completely sure actually.

- UsukiDoll

I wasn't even taught how to do it but I have to do it for hw wt hexagon

- anonymous

However: \[
f(x) = \int^x_{f^{-1}(0)} f'(t)dt
\]

- UsukiDoll

??????????

- anonymous

That is how you'd turn a function into an integral.

- anonymous

You take the derivative and put it inside an integral.

- UsukiDoll

so take the derivative of tan^-1pix - tan^-1x?

- anonymous

Yeah.

- anonymous

You also need to find the roots.

- anonymous

any root will do.

- UsukiDoll

how do I find the roots???
so take the derivative and plug it into the integral and then find the roots?

- anonymous

Well you only need to find one.

- anonymous

Actually you don't necessarily have to find the roots.

- anonymous

\[
f(x) = \int^x_{f^{-1}(0)} f'(t)dt = \int^x_{0} f'(t)dt+f(0)
\]

- anonymous

Just do that... that will be easier.

- anonymous

Did you find \(f'(t)\) and \(f(0)\)?

- UsukiDoll

sorry maybe after I eat dinner hungry atm

- UsukiDoll

back

- UsukiDoll

|dw:1378794480198:dw|

- UsukiDoll

|dw:1378794579010:dw|

- UsukiDoll

isn't tan 90 degrees 0?

- UsukiDoll

@wio I'm back

- anonymous

OK

- anonymous

Hmm well \[
\tan(0) = \frac 01=0
\]

- anonymous

What did you get for the derivative?

- UsukiDoll

probably because tan 90 doesn't exist...I mean I still remember the graph of it.

- anonymous

Anyway \(f(0) = 0\)

- UsukiDoll

|dw:1378794996127:dw|

- anonymous

What happens to the \(x\)?

- UsukiDoll

it gets bigger?

- anonymous

No I mean in the first term.

- anonymous

The first term should have an \(x\).

- anonymous

Then we can try to add them I guess?

- 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

Chain rule: \[
(f\circ g)' = (f'\circ g)\cdot g'
\]

- UsukiDoll

-___-

- UsukiDoll

so leave the tan inverses alone and deal with the dxx? huh?

- anonymous

since \(g=\pi x\) I should see it somwehre.

- anonymous

Anyway, is it possible to simplify them?

- anonymous

\[
f'(t) = \frac{\pi}{1+(\pi t)^2}-\frac 1{1-t^2}
\]

- anonymous

\[
f'(t) = \frac{\pi}{1+(\pi t)^2}-\frac 1{1+t^2}
\]

- anonymous

I suppose this could be simplified.

- UsukiDoll

-___- loooks like something I wrote earlier.

- anonymous

Yeah, but like I said the first term needs a variable.

- UsukiDoll

so how does it get simplified lcd?

- anonymous

yeah

- UsukiDoll

##### 1 Attachment

- anonymous

Okay, I'm wondering if you aren't suppose to be checking the limits of integration first, actually...

- anonymous

Look at the limits of integration: \[
\int_0^2\int_0^xdtdx
\]If we switch order of integration it might help us.

- anonymous

|dw:1378797162816:dw|

- anonymous

|dw:1378797231738:dw|

- anonymous

Meaning we could change our bounds to: \[
\int_0^2\int_t^2dxdt
\]

- 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

I'm being honest. I don't get it either

- UsukiDoll

prof didn't even go thorough the steps or anyhting just assigned this for hw. sorry typos sleepy behind the pc screen

- anonymous

And I don't get the whole \(\tan^{-1}(\pi x)\). You don't take inverse trig functions on angles.

- UsukiDoll

that's what it was displayed in my text

- anonymous

even wolfram alpha doesn't give anything that looks nice.

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