Question

Evaluate the Riemann sum for f ( x ) = 1 − 1 2 x , 2 ≤ x ≤ 14 , with six subintervals, taking the sample points to be left endpoints.

Answer 1

Answer:

Riemann sum for f ( x ) = 1 − 1 2 x , 2 ≤ x ≤ 14 , with six sub-intervals is -640

Step-by-step explanation:

f(x) = 1 − 12x

Plotting points from 2 to 14 as 2 ≤ x ≤ 14.

f(x) = 1 − 12x → f(2) = 1 - 12(2) = -23 → (2, -23)

f(x) = 1 − 12x → f(3) = 1 - 12(3) = -35 → (3, -35)

f(x) = 1 − 12x → f(4) = 1 - 12(4) = -47 → (4, -47)

f(x) = 1 − 12x → f(5) = 1 - 12(5) = -59 → (5, -59)

f(x) = 1 − 12x → f(6) = 1 - 12(6) = -73 → (6, -73)

f(x) = 1 − 12x → f(7) = 1 - 12(7) = -84 → (7, -83)

f(x) = 1 − 12x → f(8) = 1 - 12(8) = -96 → (8, -95)

f(x) = 1 − 12x → f(9) = 1 - 12(9) = -108 → (9, -107)

f(x) = 1 − 12x → f(10) = 1 - 12(10) = -119 → (10, -119)

f(x) = 1 − 12x → f(11) = 1 - 12(11) = -131 → (11, -131)

f(x) = 1 − 12x → f(12) = 1 - 12(12) = -143 → (12, -143)

f(x) = 1 − 12x → f(13) = 1 - 12(13) = -155 → (13, -155)

f(x) = 1 − 12x → f(14) = 1 - 12(14) = -168 → (14, -167)

Δx = 2 as domain difference is 2, because

(14 - 6)/6 = 2

So, the formula becomes:

${\displaystyle \textstyle \sum _{i}^{n}}f(x_{i})$Δx

${\displaystyle \textstyle \sum _{i}^{6}}(-23)(2)+(-35)(2)+(-47)(2)+(-59)(2)+(-73)(2)+(-83)(2)$

= (-46) + (-70) + (-94) + (-118)+ (-146) + (-166)

= -640

Hence, Riemann sum for f(x) = 1−12x , 2 ≤ x ≤ 14 , with

six sub-intervals = -640.

Keywords: Riemann sum, sum

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