- #1

Orion1

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How is this problem solved using the Limit Sum Integer method?

[tex]\int_{2}^{10} x^6 \; dx[/tex]

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- Thread starter Orion1
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- #1

Orion1

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How is this problem solved using the Limit Sum Integer method?

[tex]\int_{2}^{10} x^6 \; dx[/tex]

- #2

lurflurf

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Limit Sum Integer method is an odd name.Orion1 said:

How is this problem solved using the Limit Sum Integer method?

[tex]\int_{2}^{10} x^6 \; dx[/tex]

I assume you mean as a Reiman sum. Usually the Reiman sum is calculated with even spacing.

[tex]\int_{2}^{10} x^6 \; dx=\lim_{n\rightarrow\infty}\sum_{i=1}^n (x^*_i)^6{\Delta}x_i=\lim_{n\rightarrow\infty}\sum_{i=1}^n (2+i(10-2))^6\frac{(10-2)}{n}[/tex]

Thus all that is needed to work through to the end is the ability to do sums of polynomials.

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HallsofIvy

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Of course, to anyone who actually knows how to do an integral,

[tex]\int_2^{10}x^6dx= \frac{1}{7}x^7[/tex] evaluated between 2 and 10. You can use that to check your work.

[tex]\int_2^{10}x^6dx= \frac{1}{7}x^7[/tex] evaluated between 2 and 10. You can use that to check your work.

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- #4

Orion1

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lurflurf said:[tex]\lim_{n\rightarrow\infty}\sum_{i=1}^n (2+i(10-2))^6\frac{(10-2)}{n}[/tex]

Should not this Riemann sum actually be:

[tex]\lim_{n\rightarrow\infty}\sum_{i=1}^n \left( 2 + \frac{i(10-2)}{n} \right)^6 \frac{(10-2)}{n}[/tex]

This was my approach:

[tex]\int_2^{10} x^6 dx = \lim_{n \rightarrow \infty} \frac{8}{n} \sum_{i = 1}^n x^6 = \lim_{n \rightarrow \infty} \sum_{i = 1}^n \left( 2 + \frac{8i}{n} \right)^6 \cdot \frac{8}{n}[/tex]

This Riemann sum must be expanded before one can sum the polynomial?

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