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syphonation
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I have posted a problem from my book below. I am having trouble with a homework problem similar to this one and can't, for the life of me, figure it out. The back of the book says the ansewer to this problem is 4/3. I can't figure how they're getting that. I've tried everything I can think of, and I'm not even close.
Could someone explain how to solve this problem? I hate to post it with no solution attempt, but I am lost.
Please excuse the sorry attempt at using LaTex..I wrote it out the best I could. Obviously, n is above Sigma and i = 1 is under it.
Use the form of the defenition of the integral given in Theorem 4 to evaluate the integral.
[tex]\int[/tex][tex]\stackrel{2}{0}[/tex] (2-x^2)dx
Theorem 4 states that:
"If f is integrable on [a,b], then the following is true:
[tex]\int[/tex][tex]\stackrel{b}{a}[/tex] f(x)dx = lim as n --> [tex]\infty[/tex] [tex]\sum[/tex][tex]\stackrel{n}{i=1}[/tex] f(x sub i)[tex]\Delta[/tex]x
where [tex]\Delta[/tex]x = (b-a)/n and x sub i = a + i[tex]\Delta[/tex]x
The answer is 4/3.
[tex]\Delta[/tex]x = (2-0)/n = 2/n
x sub i = a + i[tex]\Delta[/tex]x = 0 + (2/n)i
That's all I have that I know is right..
Could someone explain how to solve this problem? I hate to post it with no solution attempt, but I am lost.
Please excuse the sorry attempt at using LaTex..I wrote it out the best I could. Obviously, n is above Sigma and i = 1 is under it.
Homework Statement
Use the form of the defenition of the integral given in Theorem 4 to evaluate the integral.
[tex]\int[/tex][tex]\stackrel{2}{0}[/tex] (2-x^2)dx
Homework Equations
Theorem 4 states that:
"If f is integrable on [a,b], then the following is true:
[tex]\int[/tex][tex]\stackrel{b}{a}[/tex] f(x)dx = lim as n --> [tex]\infty[/tex] [tex]\sum[/tex][tex]\stackrel{n}{i=1}[/tex] f(x sub i)[tex]\Delta[/tex]x
where [tex]\Delta[/tex]x = (b-a)/n and x sub i = a + i[tex]\Delta[/tex]x
The Attempt at a Solution
The answer is 4/3.
[tex]\Delta[/tex]x = (2-0)/n = 2/n
x sub i = a + i[tex]\Delta[/tex]x = 0 + (2/n)i
That's all I have that I know is right..
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