How Does the Limit Sum Integer Method Solve Integral Calculations?

[Orion1]

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

\int_{2}^{10} x^6 \; dx

 

[lurflurf]

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

\int_{2}^{10} x^6 \; dx

Limit Sum Integer method is an odd name.
I assume you mean as a Reiman sum. Usually the Reiman sum is calculated with even spacing.
\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}
Thus all that is needed to work through to the end is the ability to do sums of polynomials.

 

[HallsofIvy]

Of course, to anyone who actually knows how to do an integral,
\int_2^{10}x^6dx= \frac{1}{7}x^7 evaluated between 2 and 10. You can use that to check your work.

 

[Orion1]

Riemann Sum...

\lim_{n\rightarrow\infty}\sum_{i=1}^n (2+i(10-2))^6\frac{(10-2)}{n}

Should not this Riemann sum actually be:
\lim_{n\rightarrow\infty}\sum_{i=1}^n \left( 2 + \frac{i(10-2)}{n} \right)^6 \frac{(10-2)}{n}

This was my approach:
\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}

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