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Jorge
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#1
Report Thread starter 15 years ago
#1
How do you integrate the following:

sqrt(2x +1)

Thanks

jorge :hello:
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dvs
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#2
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∫ (2x+1)^(0.5) dx
= {1/[(0.5+1)*2]} (2x+1)^(0.5+1) + C
= (1/3) (2x+1)^(3/2) + C

∫ (ax+b)^n dx = [1/a(n+1)] (ax+b)^(n+1) + C
(Provided that n ≠ -1)
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Jorge
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#3
Report Thread starter 15 years ago
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(Original post by dvs)
∫ (2x+1)^(0.5) dx
= {1/[(0.5+1)*2]} (2x+1)^(0.5+1) + C
= (1/3) (2x+1)^(3/2) + C

∫ (ax+b)^n dx = [1/a(n+1)] (ax+b)^(n+1) + C
(Provided that n ≠ -1)

Thanks

Am I supposed to know this for C1-4 ??

The reason I ask is that I saw the expression being integrated using the Trapezium Rule, so I thought it was not possible with normal integration or was Einstein's stuff.

Jorge :hello:
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dvs
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#4
Report 15 years ago
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(Original post by Jorge)
Am I supposed to know this for C1-4 ??
I think it was on the P2 (or maybe P3) syllabus, so I guess it should be on C1-C4's syllabi.
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