Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produced by your calculator. (Round your answer to six decimal places.) y = xe^−x + 3, 0 ≤ x ≤ 5

= 1/15 ( 0 + 7.2761738 + 4(1.4911851 +

Step-by-step explanation:

y = ln x , 1 <= x <= 3, about x axis and n = 10, dy/dx = 1/ x

S = (b a) ∫ 2π y √( 1 + (dy/dx) ^2) dx

so our f(x) is 2π y √( 1 + (dy/dx) ^2)

(b - a) / n = / 3 = (3-1) / 30 = 1/15

x0 = 1 , x1 = 1.2, x2 = 1.4, x3 = 1.6 x(10) = 3

So we have , using Simpsons rule:-

S10 = (1/15) ( f(x0) + 4 f(x1) + 2 f)x2) + + f(x10) )

= (1/15) f(1) + f(3) + 4(f(1.2) + f(1.6) + f(2) + f(2.4) + f(2.8)) + 2(f(1.4) + f(1.8) + f(2.2) + f(2.6) )

( Note f(1) = 2 * π * ln 1 * √(1 + (1/1)^2) = 0 and f(3) = 2π ln3√(1+(1/3^2) = 7,276)

so we have S(10)

= 1/15 ( 0 + 7.2761738 + 4(1.4911851 +

the expected value of the random variable : 0, 1 and 2

step-by-step explanation:

answer: my brother can figure it out, imma go get him so i can answer : ) edit: my brother is sleeping, sorry man

step-by-step explanation: