Asked by Ernie
1. Express the given integral as the limit of a Riemann sum but do not evaluate:
integral[0 to 3]((x^3 - 6x)dx)
2.Use the Fundamental Theorem to evaluate integral[0 to 3]((x^3 - 6x)dx).(Your answer must include the antiderivative.)
integral[0 to 3]((x^3 - 6x)dx)
2.Use the Fundamental Theorem to evaluate integral[0 to 3]((x^3 - 6x)dx).(Your answer must include the antiderivative.)
Answers
Answered by
Steve
If there are n rectangles and we use right-side values, then
the width of each rectangle is 3/n
xi=(3/n)*i
so the sum is the limit as n->∞ of
n
∑ f(xi)*∆x = ∑[((3/n)i)^3-6((3/n)i)]*(3/n) = (27/n^4)∑(3i^4-2i^2)
i=1
You can evaluate that sum using the formulas
n
∑ i^2 = (2n^3+3n^2+n)/6
i=1
n
∑ i^4 = (6n^5+15n^4+10n^3-n)/30
i=1
But it's a lot less work to use the integral
∫[0,3] x^3-6x dx = x^4/4-3x^2 [0,3] = 81/4-3*9 = -27/4
the width of each rectangle is 3/n
xi=(3/n)*i
so the sum is the limit as n->∞ of
n
∑ f(xi)*∆x = ∑[((3/n)i)^3-6((3/n)i)]*(3/n) = (27/n^4)∑(3i^4-2i^2)
i=1
You can evaluate that sum using the formulas
n
∑ i^2 = (2n^3+3n^2+n)/6
i=1
n
∑ i^4 = (6n^5+15n^4+10n^3-n)/30
i=1
But it's a lot less work to use the integral
∫[0,3] x^3-6x dx = x^4/4-3x^2 [0,3] = 81/4-3*9 = -27/4
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