I need solution to number 5,7 and 8 Approximate the following integral using n =
ID: 2863533 • Letter: I
Question
I need solution to number 5,7 and 8 Approximate the following integral using n = 4 subintervals by the Midpoint and Trapezoid Rules (4 D.P.) integral^pi/4_0 8sin(2x)dx (Exact = 4, M_4 = 4.0258. E_M = 0.0258. T_4 = 3.9485, E_T = 0.0515) 6. For the following series find a formula for the n^th term of the sequence of partial sums {S_n). Then evaluate limnrightarrowinfinity S_n to obtain the value of the series or state that the series diverges. a. sigma^infinity_k=3 3. 4^k/7^k (geometric series, S_n = 3. 4^3/7^3 (1 - (4/7)^n+2)/(1 - 4/7) = 64/49(1 -(4/7)6n+2),S = 64/49) b. sigma^infinity_k=1 2/4k^2 - 1 (Telescoping, S_n = 1 - 1/2n + 1,S = 1) 7. Find the radius of convergence, test at both endpoints to determine the interval of convergence of the series. sigma^infinity_k=0 (-1)^k(x - 2)^k/4^k (R = 4, the series diverge at both endpoints, the interval of convergence is (-2,6)) 8. Find a power series representation centered at 0 (Maclaurin series) for the function f(x) = 2/2 + x using 1/1 - x = 1 + x + x^2 +... = sigma^infintiy_k=0 x^k. Then, give the interval of convergence for the resulting series. (sigma^infinity_k=0(-x/2)^k, R = 2, the interval of convergence is(-2,2))Explanation / Answer
5)[0 to /4] 8sin(2x) dx
=[0 to /4] -4cos(2x)
=(-4cos(2/4))-(-4cos(2/4))
=0+4
=4
f(x)=8sin(2x)
a=0,b=/4 , n=4
x=(b-a)/n
x=[/4 -0]/4=/16
mid point rule:
[0 to /4] 8sin(2x) dx=x[f(/32)+f(3/32)+f(5/32)+f(7/32)]
[0 to /4] 8sin(2x) dx=(/16)[8sin(/16)+8sin(3/16)+8sin(5/16)+8sin(7/16)]
[0 to /4] 8sin(2x) dx=4.0258
error =|4.0258-4|=0.0258
trapezoidal rule :
[0 to /4] 8sin(2x) dx=(x/2)[f(0)+ 2*[f(/16)+f(2/16)+f(3/16)]+f(4/16)]
[0 to /4] 8sin(2x) dx=(x/2)[f(0)+ 2*[f(/16)+f(/8)+f(3/16)]+f(/4)]
[0 to /4] 8sin(2x) dx=(/32)[8sin(0)+ 2*[8sin(/8)+8sin(/4)+8sin(3/8)]+8sin(/2)]
[0 to /4] 8sin(2x) dx=3.9485
error =|3.9485-4|=0.0515