# Determine a region whose area is equal to the given limit. Do not evaluate the limit. limit as n approaches infinity of sum_(i=1)^n (2/n)((5+(2i)/n)^10)

You need to use the given Riemann sum to evaluate the definite integral such that:

`lim_(n->oo) sum_(i=1)^n (2/n)(5 + i(2/n))^10 = lim_(n->oo) sum_(i=1)^n Delta x_i*f(x_i)`

Notice that `f(x_i) = (5 + i(2/n))^10 , hence x_i = 5 + i(2/n) =>`  `Delta x_i = 2/n = (7-5)/n`

`lim_(n->oo) sum_(i=1)^n (2/n)(5 + i(2/n))^10 =...

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You need to use the given Riemann sum to evaluate the definite integral such that:

`lim_(n->oo) sum_(i=1)^n (2/n)(5 + i(2/n))^10 = lim_(n->oo) sum_(i=1)^n Delta x_i*f(x_i)`

Notice that `f(x_i) = (5 + i(2/n))^10 , hence x_i = 5 + i(2/n) =>`  `Delta x_i = 2/n = (7-5)/n`

`lim_(n->oo) sum_(i=1)^n (2/n)(5 + i(2/n))^10 = int_5^7 x^10 dx = x^11/11|_5^7`

`int_5^7 x^10 dx = (7^11 - 5^11)/11`

Hence, evaluating the definite integral, using the given Riemann sum, yields `int_5^7 x^10 dx = (7^11 - 5^11)/11.`

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