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Hard maths question can you help?

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Reply 20
Avatar for Zacken
Zacken
22
Original post by keromedic
I made a mistake :tongue:


Happens to the best of us. :lol:
Original post by Zacken
Happens to the best of us. :lol:


Forgot that 1*-1=-1 for some reason :tongue:.

As for my initial point, it was that I felt that ∫−11f(x)dx=∫−10f(x)dx+∫01f(x)dx\displaystyle \int_{-1}^1 f(x)dx =\int_{-1}^0 f(x)dx+\int_{0}^1 f(x) dx. It then followed naturally in my reasoning at the time that the result would be 0 as they'd cancel. See my crude image below.

graph.png
(edited 8 years ago)
Reply 22
Avatar for Zacken
Zacken
22
Original post by keromedic
Forgot that 1*-1=-1 for some reason :tongue:.

As for my initial point, it was that I felt that ∫−11f(x)d=∫−10f(x)dx+∫01f(x)dx\displaystyle \int_{-1}^1 f(x)d =\int_{-1}^0 f(x)dx+\int_{0}^1 f(x) dx. It then followed naturally in my reasoning at the time that the result would be 0 as they'd cancel. See my crude image below.

graph.png


I understand your reasoning, it's just that I'm questioning why your sketch looks like −1x\frac{-1}{x}, 1x2\frac{1}{x^2} clearly lies above the x-axis all the way through.
Original post by Zacken
I understand your reasoning, it's just that I'm questioning why your sketch looks like −1x\frac{-1}{x}, 1x2\frac{1}{x^2} clearly lies above the x-axis all the way through.


Because I was wrong. It's been too long.
Reply 24
Avatar for Zacken
Zacken
22
Original post by keromedic
Because I was wrong. It's been too long.


Ah, okay. That's fine then. I once had a class test where I wrote x2=4⇒x=±4x^2 = 4 \Rightarrow x = \pm 4. :facepalm: - silly mistakes, the death of us all. xD

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