The area A of the region S that lies under the graph of the continuous function
ID: 2884342 • Letter: T
Question
The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles A = lim n rightarrow infinity R_n = lim n rightarrow infinity [f(x_1) delta x + f(x_2) delta x + ... + f(x_n) delta x] Consider the function f(x) = x cos (x), 0 lessthanorequalto x lessthanorequalto pi/2. Using the above definition, determine which of the following expressions represents the area under the graph of f as a limit. A. lim n rightarrow infinity sigma^n_i = 1 pi/2n (pi i/2n cos (pi i/2n)) B. lim n rightarrow infinity sigma^n_i = 1 pi i/2n (pi i/2n cos (pi i/2n)) C. lim n rightarrow infinity sigma^n_i = 1 (pi i/2n cos (pi i/2n)) D. lim n rightarrow infinity sigma^n_i = 1 pi/2n (pi/2n cos (pi/2n)) E. lim n rightarrow infinity sigma^n_i = 1 pi/2n (cos (pi i/2n))Explanation / Answer
We have given f(x)=xcos(x),0<=x<=pi/2
we have a=0,b=pi/2 x=(b-a)/n
we divide the given interval into n subintervals with equal width x=(b-a)/n
x=(b-a)/n=(pi/2-0)/n=pi/(2n)
f(x1)=x1cos(x1)
A=lim n-->infinity Rn = lim n-->infinity[f(x1)x+f(x2)x+f(x3)x+...+f(xn)x]
=lim n-->infinity summation of (i=1 to n)[f(xi)]x
=lim n-->infinity summation of (i=1 to n)[(xi)*cos(xi)]x
=lim n-->infinity summation of (i=1 to n)[((pi*i)/2n)*cos((pi*i)/2n)*x] since xi is subinterval of i th= (pi*i)/2n
=lim n-->infinity summation of (i=1 to n)[((pi*i)/2n)*cos((pi*i)/2n)*(pi/2n)]
A=lim n-->infinity summation of (i=1 to n)[[(pi*i)/2n)*cos((pi*i)/2n)]*(pi/2n)]
answer is option A
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