please do it in full step Consider the definite integral x 1/3 dx What is the mi

Question

please do it in full step

Consider the definite integral x 1/3 dx What is the minimum number (N) of equally spaced subintervals needed such that the difference between the upper and lower Riemann sum estimates for this definite integral is less than 10? Find, giving reasons (and using summation notation), expressions for the upper Riemann sum (U64) and the lower Riemann sum (L64) for this definite integral using N = 64 equally spaced subintervals. Find the exact value of this definite integral. Use this value together with the equation L64

Explanation / Answer

a) lower sum = sum of (64/N*k)^(1/3) *64/N from k=0 to N-1
upper sum =sum of (64/N*k)^(1/3) *64/N from k=1 to N

so difference = (64/N*N)^(1/3)*64/N - (0)^(1/3)*64/N = 64^(1/3)*64/N = 256/N = 10

N = 256/10 = 25.6
so 26 intervals

b)
lower sum = sum of (64/N*k)^(1/3) *64/N from k=0 to N-1
upper sum =sum of (64/N*k)^(1/3) *64/N from k=1 to N

c)

integral of x^(1/3) = x^(4/3)/(4/3) = 3*(64)^(4/3)/4=192

so we when we plug N=64 into the sum the 64s cancels

sum of k^(1/3) from 0 to 63 < 192 < sum of k^(1/3) from 1 to 64

since 64^(1/3) = 4
192<sum ofk^(1/3) from 1 to 62< 196

2

a)let u = x^2-4x + 5
du = 2x - 4 dx
so
du/2 = (x-2) dx

integral = 1/u 1/2 du = 1/2 ln(x^2 - 4x + 5) = ln(5)/2

b) sin^2 = 1 - cos^2

so we have (1-cos^2)^2 sin dx
let u = cos x then du = - sin x

so integral = - integral of (1-u^2) du
= - integral of 1 - 2u^2 + u^4 du = -u + 2u^3/3 - u^5/5 = 53/480

c)
multiply top and bottom by e^x
so integral = integral of e^x/(e^x+1) dx
let u = e^x + 1 then du = e^x dx

= integral of du/u = ln|u|+ C = ln |e^x + 1| +c

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