calculus please help

Question

Use the Power Series to approximate the integral. (use 5 terms)∫

∫ln(x) dx from 0 to 1

Answer 1

For comparison, establish ∫_{(from 0 to 1)}[ln x]dx:

u = ln x-----------dv = dx

du = dx/x----------v = x

Guide on ∫[u]dv = uv − ∫[v]du or

------------∫[ln x]dx = x[ln x] − ∫[x](dx/x).

Rewrite this last as ∫[ln x]dx = x[ln x] − ∫dx or

∫[ln x] dx = x[ln x] − x + Constant.

Evaluate {x[ln x] − x + Constant|_{(from 0 to 1)}}:

{1[ln 1] − 1 + Constant} minus {0[ln 0] − 0 + Constant}

goes to 1× 0 − 1 + Constant minus 0 × lim_(x→0+) [ln x] + 0 − Constant.

Reduce the last expression above to -1. Note that lim_(x→0+) [ln x] is -∞.

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The series 1 + x + x² + ... + xⁿ + ... = 1/(1 − x) can be rewritten with -x in place

of x to obtain 1 − x + x² − x³ + ... + (-1)ⁿxⁿ + ... = 1/(1 + x). (Both series require |x| < 1.)

The second series can be integrated term-by-term to gain x − x²/2 + x³/3 − x⁴/4 + ... +

(-1)ⁿx⁽ⁿ⁺¹⁾/(n + 1) [for -1 < x ≤ 1], which will converge to ln (1 + x).

For x = -0.01, ln (1 + x) is ln 0.99; for x = -0.99, ln (1 + x) is ln 0.01. A loop program on

a calculator gives ln 0.99 to exact value in the summation of the first five terms as

-0.01005033585.

For ln 0.01, however, the program takes 1850 loops to give exact value as -4.605170186.

Then {0.99 × ln 0.99 − 0.99} minus {0.01 × ln 0.01 − 0.01} simplifies to -0.9438981306,

fairly comparable to -1 above.