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Find the Riemann sum for f(x) = 3 sin x, 0 ≤ x ≤ 3π/2, with six terms, taking the...
Find the Riemann sum for f(x) = 3 sin x, 0 ≤ x ≤ 3π/2, with six terms, taking the sample points to be right endpoints.
(a) Find the Riemann sum for f(x) = 3 sin x, 0 ≤ x ≤ 3π/2, with six terms, taking the sample points to be right endpoints. (Round your answers to six decimal places.)
(b) Repeat part (a) with midpoints as the sample points.
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You need to find the width of interval, hence you may use the formula:
`Delta x = (3pi/2 - 0)/6 =gt Delta x = 3pi/12 =gt Delta x = pi/4`
You need to find the 6 right endpoints such that:
x_1 = pi/4 ; x_2 = pi/4 + pi/4 = pi/2
`x_3 = pi/2 + pi/4 = 3pi/4`
`x_4 = 3pi/4 + pi/4 = pi`
`x_5 = pi+pi/4 = 5pi/4`
`x_6 = 5pi/4 + pi/4 = 6pi/4 =gt x_6 = 3pi/2`
You need to remember the formula of Riemann's sum such that:
`R_6 = sum_(i=1)^6 f(x_i)Delta x`
`R_6 = (pi/4)*(f(x_1)+f(x_2)+f(x_3)+f(x_4)+f(x_5)+f(x_6))`
`R_6 = (3pi/4)*(sin (pi/4)+sin (pi/2)+sin (3pi/4)+sin (pi)+sin (5pi/4)+sin (3pi/2))`
`R_6 = (3pi/4)*(sqrt2/2 + 1 + sqrt2/2 + 0 - sqrt2/2 - 1)`
`R_6 = (3pi/4)*(sqrt2/2) =gt R_6 = 1.666081`
Hence, evaluating the Riemann's sum under given conditions yields `R_6 = 1.666081.`
b)You need to find the midpoints of intervals `[0,pi/4],[pi/4,pi/2],[pi/2,3pi/4],[3pi/4,pi],[pi,5pi/4],[5pi/4,3pi/2] ` such that:
`m_1 = pi/8`
`m_2 = (pi/4+pi/2)/2 = 3pi/8`
`m_3 = (pi/2+3pi/4)/2 = 5pi/8`
`m_4 = (3pi/4+pi)/2 = 7pi/8`
`m_5 = 9pi/8`
`m_6 = 11pi/8`
You should evaluate Riemann's sum substituting `m_1,m_2,m_3,m_4,m_5,m_6` in equation of function such that:
`M_6 = (3pi/4)*(sin (pi/8)+sin (3pi/8)+sin (5pi/8)+sin (7pi/8)+sin (9pi/8)+sin (11pi/8))`
`M_6 = (3pi/4)*(0.382683+0.923879+0.923879+0.382683-0.382683-0.923879)`
`M_6 = 3.078514`
Hence, evaluating the Riemann's sum under given conditions yields `M_6 = 3.078514.`
Posted by sciencesolve on April 19, 2012 at 4:48 PM (Answer #1)
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