Integration can leave the convergence range

(i) Using standard series from the List of Formulae (MF26), expand ln⁡(k+2x)n\ln(k+2x)^n in ascending powers of xx as far as the term in x3x^3, where kk and nn are real constants, k>0k>0, n<0n<0 and x>−k2x>-\frac k2. marks

(ii) State the range of values of xx for which the expansion in part (i) is valid. mark

(iii) Find the exact value of ∫01ln⁡(1+2x)−2 dx\int_0^1\ln(1+2x)^{-2}\,dx. marks

(iv) Use your series from part (i) to estimate ∫01ln⁡(1+2x)−2 dx\int_0^1\ln(1+2x)^{-2}\,dx. marks

(v) Calculate the percentage error of the estimate in part (iv), and comment on the accuracy of your approximation using your answer from part (ii). marks

[2020/NYJC/Prelim/I/9]

Total marks: 10

Solution

(i)

Logarithmic series

Expand ln⁡(k+2x)n\ln(k+2x)^n in ascending powers of xx up to x3x^3, where k>0k>0 and n<0n<0.

Hint

  1. Use ln⁡(k+2x)n=nln⁡k+nln⁡(1+2xk)\ln(k+2x)^n=n\ln k+n\ln\left(1+\frac{2x}k\right).

  2. Expand ln⁡(1+u)=u−u22+u33\ln(1+u)=u-\frac{u^2}2+\frac{u^3}3 with u=2xku=\frac{2x}k.

Answer · (i)

nln⁡k+2nkx−2nk2x2+8n3k3x3+⋯n\ln k+\frac{2n}kx-\frac{2n}{k^2}x^2+\frac{8n}{3k^3}x^3+\cdots.

(ii)

Validity

State the range of xx for which the expansion in (i) is valid.

Hint

  1. ln⁡(1+u)\ln(1+u) is valid for −1<u≤1-1<u\le1 with u=2xku=\frac{2x}k.

Answer · (ii)

−k/2<x≤k/2-k/2<x\le k/2.

(iii)

Integration by parts

Find the exact value of ∫01ln⁡(1+2x)−2 dx\int_0^1\ln(1+2x)^{-2}\,dx.

Explore this idea

Hint

  1. Rewrite as −2∫01ln⁡(1+2x) dx-2\int_0^1\ln(1+2x)\,dx.

  2. Parts with u=ln⁡(1+2x)u=\ln(1+2x) and dv=dxdv=dx.

Answer · (iii)

2−3ln⁡32-3\ln3.

Explore the idea

The exact area from 0 to 1

Drag x. The shading shows the selected interval; the lower curves compare the numerical signed integral with the change in the antiderivative.

Upper limit xx = 1.00
Try an example

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
0
Below-axis contribution
-1.296
Signed integral
-1.296

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from 0.00 to x
-1.296
exact antiderivative − F(0)
-1.296
agrees at displayed precision
Formula − integral
0
0.00% of the integral’s magnitude
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
  • F(x) − F(start), from the antiderivative
Key idea

The curve is below the axis, so the area is negative: 2−3ln⁡32-3\ln3.

Ideas to try
Drag xx to 1: the readout is 2−3ln⁡3≈−1.2962-3\ln3\approx-1.296.
(iv)

Integrating the series

Use the series from (i) with k=1k=1, n=−2n=-2 to estimate ∫01ln⁡(1+2x)−2 dx\int_0^1\ln(1+2x)^{-2}\,dx.

Explore this idea

Hint

  1. Substitute n=−2n=-2, k=1k=1: ln⁡(1+2x)−2≈−4x+4x2−163x3\ln(1+2x)^{-2}\approx-4x+4x^2-\frac{16}3x^3.

  2. Integrate from 0 to 1.

Answer · (iv)

−2-2.

Explore the idea

The series estimate against the true area

Drag x to compare the numerical signed integral with the series estimate. Their difference shows the approximation error.

Upper limit xx = 1.00
Try an example

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
0
Below-axis contribution
-1.296
Signed integral
-1.296

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from 0.00 to x
-1.296
series antiderivative − F(0)
-2
Estimate − integral
-0.704
54.34% of the integral’s magnitude
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
  • Series estimate of the signed integral; it need not agree with the true integral
Key idea

The estimate is accurate near 0 but drifts away as xx passes 12\frac12.

Ideas to try
Compare the dashed (series) value with the solid (true) area at x=1x=1 and at x=0.5x=0.5.
(v)

Percentage error

Calculate the percentage error of the estimate and comment on accuracy using (ii).

Hint

  1. Percentage error =∣estimate−exact∣∣exact∣×100=\frac{|\text{estimate}-\text{exact}|}{|\text{exact}|}\times100.

Answer · (v)

54.34%54.34\% error; unreliable because integration extends beyond the series interval.