WeBWorK: MAI 135 18 : C8 : 6 × © We Want To Find The Area Of X | + (D https://we

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WeBWorK: MAI 135 18 : C8 : 6 × © We Want To Find The Area Of X | + (D https://webwork.as.uky.edu/webwork2/MA1 13S1 8/C8/6/?effec B 80% asearch User Scttings Previous Problem Problem List Next Problem Grades Problems (1 point Problem 1 Problem2 Problem 3 Problem 4 Problem 5 Problem 5 Problem 7 Problem 8 Problem Problem 10 Problem 11 Problem 12 Problern 13 We want to find the area of a region S which lies below the graph of fa) 42 and above the interval a, b 4,9 on the z-axis. Thus To do this we begin by cdividing the interval 4,9 into N equal subintervals using the points4, x1,22,..-,x-9 The lengih of each sub-inis 5N Find a formula for the kth division point in terms of k and N help (formulas To approximate the area, we construct rectangles as pictured abovs. The base is the interval1 arca of lhis rclarigle on the r-axis and the height is f(). Find a formula for Ae, the Find an expression of the sum of the areas A, SN= 67 N-(42k/N°2) It will be help ul to use the formulas for the sums of powers of integers from the texthook Take the limit as N tends to infinity to find the area of S The area of S limN >00 SA- Note: You can earm Dartial credit O Type here to search 11:49 PM 3/17/2018

Explanation / Answer

length of each subinterval,x=(9-4)/N=5/N

xk=4+ k*(5/N)

=>xk=4+ (5k/N)

Ak= (4(4+ (5k/N))2)(5/N)

Ak= (320/N)+(800k/N2)+ (500k2/N3)

SN= [k=1 to N][(320/N)+(800k/N2)+ (500k2/N3)]

SN= (320N/N)+(800N(N+1)/2N2)+ (500N(N+1)(2N+1)/6N3)

SN= 320+(400(1+(1/N)))+ ((250/3)(1+(1/N))(2+(1/N)))

the area of S=lim[N-> ] SN

the area of S=lim[N-> ] 320+(400(1+(1/N)))+ ((250/3)(1+(1/N))(2+(1/N)))

the area of S= 320+(400(1+0))+ ((250/3)(1+0)(2+0))

the area of S= 320+400+ (500/3)

the area of S= 2660/3exactly = 886.67 approximately

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