You need to find the interval width `Delta x` using the following formula such that:

`Delta x = (b-a)/n`

`Delta x = (2-0)/4 =gt Delta x = 2/4 =gt Delta x = 1/2` = 0.5

You need to evaluate the midpoints of intervals [0,0.5] , [0.5,1] , [1,1.5] , [1.5,...

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You need to find the interval width `Delta x` using the following formula such that:

`Delta x = (b-a)/n`

`Delta x = (2-0)/4 =gt Delta x = 2/4 =gt Delta x = 1/2` = 0.5

You need to evaluate the midpoints of intervals [0,0.5] , [0.5,1] , [1,1.5] , [1.5, 2] such that:

`x_1 = (0+0.5)/2 = 0.25`

`x_2 = (0.5+1)/2 = 0.75`

`x_3 = (1+1.5)/2 = 1.25`

`x_4 = (1.5+2)/2 = 1.75`

You need to use the formula of Riemann's sum such that:

`R_4 =` `sum_(i=1)^4``f(x_i)*Delta x`

`R_4 = (1/2)*(f(x_1) + f(x_2) + f(x_3) + f(x_4))`

`R_4 = (1/2)*(f(0.25) + f(0.75) + f(1.25) + f(1.75))`

`R_4 = (1/2)*(e^0.25 - 3 + e^0.75 - 3 + e^1.25 - 3 + e^1.75 - 3)` `R_4 = (1/2)*(2.2795 + 2.1063 + 3.4610 + 5.6870 - 12)` =>`R_4 = 0.7669`

**Hence, evaluating the Riemann's sum under given conditions yields `R_4 = 0.7669.` **