aqa fp3 improper integrals

how come that tan to the minus 1 of infinity approaches pi/2 but tan to the minus 1 of minus infinity doesn't

I'm doing question 1F a) in the FP3 booklet on the AQA website.

Reply 1

Hasufel

\arctan(-x)= - \arctan(x)

Reply 2

maggiehodgson

Original post by Hasufel

\arctan(-x)= - \arctan(x)

Thanks for your input but it hasn't made things clearer for me.

If I drew a tan curve, I can see that as x -> pi/2 y -> infinity but also, I view, as x -> pi/2 from the other side of the asymptote y -> - infinity.

What vis my misthink?

Reply 3

Hasufel

the x values go from (-infinity, infinity) , y values don`t go beyond +/- Pi/2,

So,

Lim_{x-> - \infty} arctan(x)= - Lim_{x-> \infty} arctan(x) = - \frac{\pi}{2}

because

Lim_{x-> - \infty}....

has now become

- Lim_{x-> + \infty}

(since in the limit, you move from -x approaching -infinity, to x approaches infinity of the negative function.

(edited 9 years ago)

Reply 4

Hasufel

also, to confirm your suspicions,you can take an identity:

\displaystyle \arctan(x) = \arcsin \left(\frac{x}{\sqrt{1+x^{2}}} \right) , (- \infty <x< \infty)

for large enough x this behaves like:

\displaystyle \arcsin \left(\frac{x}{|x|} \right) = \arcsin(-1) = - \frac{\pi}{2}

i`m sorry if this might be too advanced an explanation - i`m familiar with the scottish system, but not English or Welsh (again, forgive my ignorance, please!)

(edited 9 years ago)

Reply 5

maggiehodgson

Original post by Hasufel

also, to confirm your suspicions,you can take an identity:

\displaystyle \arctan(x) = \arcsin \left(\frac{x}{\sqrt{1+x^{2}}} \right) , (- \infty <x< \infty)

for large enough x this behaves like:

\displaystyle \arcsin \left(\frac{x}{|x|} \right) = \arcsin(-1) = - \frac{\pi}{2}

i`m sorry if this might be too advanced an explanation - i`m familiar with the scottish system, but not English or Welsh (again, forgive my ignorance, please!)

You're right. It is too advanced for me. I'll try to figure it out though.

Thanks

Reply 6

TeeEm

Original post by maggiehodgson

how come that tan to the minus 1 of infinity approaches pi/2 but tan to the minus 1 of minus infinity doesn't

I'm doing question 1F a) in the FP3 booklet on the AQA website.

arctan(x) is by definition the inverse branch of tan(x) for -pi/2 to pi/2

so its graph

tends to pi/2 as x tends to +infinity (asymptote at y = pi/2)
tends to -pi/2 as x tends to -infinity (asymptote at y =-pi/2)

is this any good?

Reply 7

maggiehodgson

I think I'd better post the complete question

Explain why each of the following integrals is improper

\int_{-}^{1}\frac{1}{1+x^{2}}\boldsymbol{dx}

The lower limit on the integral is -infinity - couldn't find the infinity symbol.

It integrates to

[tan^{-1}x]_{-n}^{1} = [tan^{-1}1]-[tan^{-1}-n]

=0.785... - -pi/2

So why is the answer "the interval of integration is infinite".

Hope this helps to explain my problem better.

(edited 9 years ago)

Reply 8

TeeEm

Original post by maggiehodgson

I think I'd better post the complete question

Explain why each of the following integrals is improper

\int_{-}^{1}\frac{1}{1+x^{2}}\boldsymbol{dx}[\latex]

The lower limit on the integral is -infinity - couldn't find the infinity symbol.

It integrates to

[tan^{-1}x]_{-n}^{1} = [tan^{-1}1]-[tan^{-1}-n][\latex]

=0.785... - -pi/2

So why is the answer "the interval of integration is infinite".

Hope this helps to explain my problem better.

code error

Reply 9

maggiehodgson