choose one in chocies and fill the rectangle box plz~ Integr.nl Test: Drill Use

Question

choose one in chocies and fill the rectangle box plz~

Integr.nl Test: Drill Use the Integral Test to determine whether the series converges or diverges. The Integral Test - Reminder: Let f be a continuous, positive decreasing function on [N, infinity), where N is a nonnegative integer, and let an = f(n) for all non-negative integers n N. Then the series is convergent if and only if the improper integral is convergent. Choices: Divergent, Convergent. Remainder Estimate for Integral Test Approximate the sum of the series by using the sum of the first 6 terms. Estimate the error involved in this approximation. How many terms are required to ensure that the sum is accurate to within 0.0005 ? Remainder Estimate for the Integral Test: Suppose f(i) = ai, where f is continuous, positive, decreasing function for is convergent with Solution: In both part (a) and (b) we need to know With we have By approximating the sum by the first 6 terms we get According to the Remainder Estimate for the Integral Test we have So the size of the error is at most Tries 0/30 Accuracy to within 0.0005 means that we have to find the value of n such that Solving this inequality we get

Explanation / Answer

1)

We assume the condition applies, it's easy to verify anyway

a) int(1<x<+inf) xe^(-x)dx = [-xe^(-x)](1<x<+inf) + int(1<x<+inf) e^(-x)dx = 2/e , series converges

b) int(1<x<+inf) dx/(2x^2+1)= [ 1/sqrt(2) atan(sqrt(2) x ) ](1<x<+inf) =(Pi/2-atan(sqrt(2))/sqrt(2), so series converges

c) int(1<x<+inf) dx/sqrt(x) = [2 sqrt(x)](1<x<+inf) = +infinity, so series diverges

d) int(2<x<+inf) dx/(xln(x)^2) = [-1/(ln(x))](2<x<+inf) = 1/ln(2), so series converges

e) int(1<x<+inf) dx/(2x+1) = [ 1/2ln(2x+1) ](1<x<+inf) = +infinity, so series diverges

2)

a)

The sum of 6 first term by computation gives 85701/8000 = 10.712625

Wen can say that the error is R_6 and we have :

R_6 <= int (6<x<+inf) 9/x^3 dx

int 9/x^3 = -9/(2x^2) so the choice is 9/(2n^2)

R_6 <= 9/(2*6^2) = 1/8

The error is at most 0.125

b)

Choice again 9/(2n^2), this is the same integral than before

We need 9/(2*n^2) <= 0.0005 <=>

n >= sqrt( 9/(2*0.0005) ) = 94.98, So we take n=95

It's the choice n

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