The interval [0,4] is partitioned into n equal subintervals, and a number x_i is arbitrarily chosen in the i^th subinterval for each i. Then....
lim_(n → ∞) ∑_(i = 1 → n)[2x_i+8)/n]=???
To find the limit of the given expression as n approaches infinity, we need to rewrite it as a definite integral.
The interval [0, 4] is partitioned into n equal subintervals, which means the width of each subinterval is (4-0)/n = 4/n.
Now, let's solve the expression:
∑(i = 1 to n)[2x_i+8)/n]
Multiply both the numerator and denominator by 4/n:
∑(i = 1 to n)[(8x_i+32)/n^2]
We can rewrite this expression as the summation of a series of rectangles, with the width being 4/n and the height being (8x_i+32)/n^2.
As n approaches infinity, the width of each rectangle approaches 0, resulting in an integral from 0 to 4:
∫(0 to 4) [(8x+32)/16] dx
Simplifying this integral:
(1/16) ∫(0 to 4) (8x+32) dx
= (1/16) [(4x^2 + 32x) / 2] evaluated from 0 to 4
= (1/16) [ (8(4)^2 + 32(4))/2 - (8(0)^2 + 32(0))/2 ]
= (1/16) [ 128 + 128 ]
= (1/16) [ 256 ]
= 16
Therefore, the limit as n approaches infinity of the given expression is 16.