Using Riemann sums, find integral representations of the following:?

(1) #lim_(nrarroo)sum_(i=1)^n 2e^sqrt(i/n)1/n #
(2) #lim_(nrarroo)sum_(i=1)^n (i/n)^27 * 1/n#

(1) #lim_(nrarroo)sum_(i=1)^n 2e^sqrt(i/n)1/n #
(2) #lim_(nrarroo)sum_(i=1)^n (i/n)^27 * 1/n#

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Jim H Share
Jan 26, 2018

Answer:

Please see below for (2).

Explanation:

Using #x_i = #right endpoints, we have

#int_a^b f(x) dx = lim_(nrarroo)sum_(i=1)^n [f(a+i(b-a)/n)(b-a)/n]#

# = lim_(nrarroo)sum_(i=1)^n i^27/n^27 * 1/n#

# = lim_(nrarroo)sum_(i=1)^n (i/n)^27 * 1/n#

#f(x) = x^27#, #a = 0# and #b = 1#, so

#int_0^1 x^27 dx#

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Write your answer here...
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Answer

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Explanation

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Jim H Share
Jan 26, 2018

Answer:

Please see below for (1).

Explanation:

Using #x_i = #right endpoints, we have

#int_a^b f(x) dx = lim_(nrarroo)sum_(i=1)^n [f(a+i(b-a)/n)(b-a)/n]#

So #lim_(nrarroo)sum_(i=1)^n 2e^sqrt(i/n)1/n#

has #f(a+i(b-a)/n) = 2e^sqrt(i/n)# and #(b-a)/n - 1/n#

So #(b-a)/n = 1/n# and #a+i(b-a)/n = i/n = i(1/n)#, so #a = 0#.

Observe that when #i = n#, we have #a+i(b-a)/n = b#.
In this case we get #b = n/n = 1#

We have

#int_a^b f(x) dx = lim_(nrarroo)sum_(i=1)^n [f(a+i(b-a)/n)(b-a)/n]#

# lim_(nrarroo)sum_(i=1)^n 2e^sqrt(0+i/n) 1/n#

So #f(x) = 2e^sqrtx#

and we get

#lim_(nrarroo)sum_(i=1)^n 2e^sqrt(i/n)1/n = int_0^1 2e^sqrtx dx#

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