Remove the modulus on the actual interval

Without the use of a calculator, show that

∫−22∣x−2∣x2−2x+4dx=pln⁡3+qπ,\int_{-2}^{2}\frac{|x-2|}{x^2-2x+4}dx=p\ln3+q\pi,

where pp and qq are exact real constants to be determined. marks

[2023/HCI/Prelim/I/4]

Solution

Constants

Absolute value and completing the square

Without a calculator, show that ∫−22∣x−2∣x2−2x+4 dx=pln⁡3+qπ\int_{-2}^{2}\dfrac{|x-2|}{x^2-2x+4}\,dx=p\ln3+q\pi and find pp and qq.

Explore this idea

Hint

  1. Remove the modulus using the sign of x−2x-2 on the interval.

  2. Split into f′f\frac{f'}f and a standard tan⁡−1\tan^{-1} form.

  3. Evaluate between the limits.

Answer · Constants

p=1/2,q=1/(23)p=1/2,q=1/(2\sqrt3); integral =12ln⁡3+π23=\frac12\ln3+\frac{\pi}{2\sqrt3}.

Explore the idea

The area from −2 to 2

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 = 2.00
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
1.456
Below-axis contribution
0
Signed integral
1.456

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

Signed integral from -2.00 to x
1.456
antiderivative − F(−2)
1.456
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 antiderivative formula holds only on x≤2x\le2, where the modulus can be dropped; the area up to x=2x=2 is 12ln⁡3+π23\frac12\ln3+\frac{\pi}{2\sqrt3}.

Ideas to try
Drag xx to 2: the readout should be 12ln⁡3+π23≈1.456\frac12\ln3+\frac\pi{2\sqrt3}\approx1.456.