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Random integration question.

I bump into this question was wondering if I got it right.

∫02∣x−1∣dx\displaystyle\int_0^2 |x-1|dx
Let uu = ∣x−1∣|x-1|
,Therefore dudu=∣dx∣|dx|
Since, ∣dx∣=dx|dx|=dx
∫udu\displaystyle\int u du
We will re-define the boundaries later.
Integrating we get:
u22\frac{u^2}{2}
Re-defining the boundaries;
Since, uu = ∣x−1∣|x-1|
Where x=0,u=1x=0,u=1
, x=2,u=1x=2,u=1
Hence, (u22)11(\frac{u^2}{2})_1^1
Substituting then subtracting, I need up with ZeroZero
Point out the mistake(s), If found.
(edited 6 years ago)
Original post by Darke1231
I bump into this question was wondering if I got it right.


Where is it?
Reply 2
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Darke1231
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Original post by RDKGames
Where is it?


Just posted it.
It would probably be easier to just sketch the graph and work out the areas from that - I think the answer is 1? Not sure about that though
Edit - from the graph below you can see it's 1
(edited 6 years ago)
IMG_5719.PNG
Original post by Darke1231
...


This is incorrect. (mostly your assumption on how to differentiate a modulus function)

Use the fact that ∣x−1∣=1−x|x-1|=1-x on 0<x<10<x<1 and ∣x−1∣=x−1|x-1| = x-1 on 1<x<21<x<2
Reply 6
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Darke1231
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Original post by RDKGames
This is incorrect. (mostly your assumption on how to differentiate a modulus function)

Use the fact that ∣x−1∣=1−x|x-1|=1-x on 0<x<10<x<1 and ∣x−1∣=x−1|x-1| = x-1 on 1<x<21<x<2


Knew I was missing something, I see. How will I continue then? Integrating them in separate?
Original post by Darke1231
Knew I was missing something, I see. How will I continue then? Integrating them in separate?


yes
Reply 8
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Darke1231
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Original post by PuffyPenguin
It would probably be easier to just sketch the graph and work out the areas from that - I think the answer is 1? Not sure about that though
Edit - from the graph below you can see it's 1

I see, Thanks.
Reply 9
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Darke1231
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Original post by RDKGames
yes

∫021−xdx\displaystyle \int_0^2 1-x dx + ∫02x−1dx\displaystyle \int_0^2 x-1 dx ???
(edited 6 years ago)
Original post by Darke1231
∫021−x\displaystyle \int_0^2 1-x + ∫02x−1\displaystyle \int_0^2 x-1 ???


No, the first one from 0 to 1 and the second from 1 to 2
Reply 11
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Darke1231
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Original post by RDKGames
No, the first one from 0 to 1 and the second from 1 to 2

Got it, cheers,mate.
Original post by Darke1231
I bump into this question was wondering if I got it right.



Puffy Penguin's approach is the better one here.

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