I need some help with this, I have understood problems like this much better with help from this forum, here is the link, its problem A16 : http://www.baruch.cuny.edu/sacc/documents/MTH2205_FinalExamv8.pdf
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∫(4x - 3(x)^(1/2)dx limit 0-4
= [2(x)^2 - 2(x)^(3/2)]
putting in 4 and 0
[2(4)^2 - 2(4)^(3/2)] - [0]
= 16
= [2(x)^2 - 2(x)^(3/2)]
putting in 4 and 0
[2(4)^2 - 2(4)^(3/2)] - [0]
= 16
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ya
am when you integrate ∫x^n => (1/n+1)x(n+1) well it's the same with fractions, you have say (making it different to the question first) ∫x^(1/3)dx => using the rule (1/n+1)x(n+1) so first adding 1 to 1/3 => 3/3 + 1/3 = 4/3 so you have to divide by this number so you have 1/(4/3) which (cont)
am when you integrate ∫x^n => (1/n+1)x(n+1) well it's the same with fractions, you have say (making it different to the question first) ∫x^(1/3)dx => using the rule (1/n+1)x(n+1) so first adding 1 to 1/3 => 3/3 + 1/3 = 4/3 so you have to divide by this number so you have 1/(4/3) which (cont)
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which is the same as multiplying by (3/4) so the integral ends up being (3/4)x^(4/3)
So for the question you have 1/2 + 1 = 3/2 => so again divide by this number so you have 3/(3/2) so again invert and multiply you have 3(2/3) => 3 cancel so you're left with 2x^3/2 ...hope this helps
So for the question you have 1/2 + 1 = 3/2 => so again divide by this number so you have 3/(3/2) so again invert and multiply you have 3(2/3) => 3 cancel so you're left with 2x^3/2 ...hope this helps
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