# Integration AS P1

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#1

Some help doing part one of the question please. I think i can do part 2 and part 3 on my own but I can't get the expression right. Thanks
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3 years ago
#2
(Original post by Jian17)

Some help doing part one of the question please. I think i can do part 2 and part 3 on my own but I can't get the expression right. Thanks
You can integrate by substitution, if you so must.
0
3 years ago
#3
(Original post by Jian17)

Some help doing part one of the question please. I think i can do part 2 and part 3 on my own but I can't get the expression right. Thanks
x1/2 is the same as sqr x.
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#4
How do i find the second expression ? I don't understand that sign and the y after it.
0
3 years ago
#5
(Original post by Jian17)
How do i find the second expression ? I don't understand that sign and the y after it.
What sign? Explain clearly what's confusing you.
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#6
These https://gyazo.com/5a6d329b3f4b0c3fac34b74d0e1322ae

I got this answer by the way:

8x - ((4-x)^3/2 / 3/2)
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#7
https://gyazo.com/2336a6bb53201cbfeafaaa3c882b4013

These are the answers, why does he multiply and divide by -1 ?
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3 years ago
#8
(Original post by Jian17)
These https://gyazo.com/5a6d329b3f4b0c3fac34b74d0e1322ae

I got this answer by the way:

8x - ((4-x)^3/2 / 3/2)
Not sure how you don't understand the basic notation for differentiation and integration...

means the derivative of w.r.t . So differentiate w.r.t

means integral of w.r.t . So integrate w.r.t

1
3 years ago
#9
(Original post by RDKGames)
You can integrate by substitution, if you so must.
There's no point in complicating things by using U substitution in this case. Just use the reverse chain rule, it's an easy expression.

Add one to the power, then divide the whole function by the new power and also the derivative of the inner function.
0
3 years ago
#10
(Original post by Acc Dos)
There's no point in complicating things by using U substitution in this case. Just use the reverse chain rule, it's an easy expression.

Add one to the power, then divide the whole function by the new power and also the derivative of the inner function.
Reverse chain rule comes FROM substitution for linear expressions in the first place. It is not overxomplicating anything whatsoever.
If OP doesn’t understand this first, then whats the point in just blindly applying reverse chain rule if they dont know where it comes from/why it works.
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3 years ago
#11
I know that it does, that's why I specified 'U substitution' didn't really know how to word that, my bad. But I guess you're right about OP needing to understand some fundamentals beforehand.I was actually taught the reverse chain rule before I learnt how to do that kind of substitution, strange now that I think about it.
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