Maclaurin series/ Integration

Question

1.)

a-Evaluate the integral from 0 to 1/2 of ln(1+x)dx with an accuracy of 10^-4 using techniques of integration and a calculator.

b- Now approximate the integral from 0 to 1/2 of ln(1+x)dx to an accuracy of 10^-4 using the Maclaurin series of ln(1+x) and term-by-term integration. You will have to figure out how many terms of the series must be used. Are the answers to a) and b) very similar numbers?

Answer 1

a) Rather than integrate ln(1+x) from 0 to 1/2, I shall integrate ln(x) from 1 to 3/2

Using I[a,b] for the integral from a to b and E[a,b] for evaluation from a to b,

I[1,3/2] ln(x) dx = x ln x - x E[1, 3/2] = 3/2 ln 3/2 - 3/2 - (1 ln 1 - 1) = 3/2 ln(3/2) - 1/2 = .1082

Letting ∑ be the sum for n = 1 to infinity, the Maclaurin expansion of ln(1 + x) = ∑(-1)ⁿ⁺¹ xⁿ/n

As this is an alternating series, we only need to show that the next term is less than the accuracy.

Then, we integrate this from 0 to 1/2, so we get ∑(-1)ⁿ⁺¹ xⁿ⁺¹/(n*(n+1))

Then, as this is an alternating series, we have to find the first term less than the accuracy requested.

(1/2)ⁿ⁺¹/(n*(n+1)) < .0001, or

n(n+1)2ⁿ⁺¹ > 10000

6*7*2^7 = 5376 < 10000

7*8*2^8 = 14336 > 10000

Then, we add up to n=6

(1/2)²/(1*2) - (1/2)³/(2*3) + (1/2)⁴/(3*4) - (1/2)⁵/(4*5) + (1/2)⁶/(5*6) - (1/2)⁷/(6*7) = .1081

(The actual answer is less than .1082, so we are within .0001).