Porisma

Worked solution

Write the modulus curve piecewise before integrating

Without using a calculator, find the area of the region bounded by f(x)=∣x2−4∣f(x)=|x^2-4|, the xx-axis and the lines x=0x=0 and x=3x=3.

CJC · Lecture · Example 9

Resolve the modulus

The source writes ∣x2−4∣=x2−4|x^2-4|=x^2-4 when x2−4≥0x^2-4\ge0 and −(x2−4)-(x^2-4) when x2−4<0x^2-4<0. Equivalently, ∣x2−4∣={x2−4,x≤−2 or x≥2,4−x2,−2<x<2.|x^2-4|=\begin{cases}x^2-4,&x\le-2\text{ or }x\ge2,\\4-x^2,&-2<x<2.\end{cases} The graph reflects the negative part of x2−4x^2-4 above the axis; it crosses at ±2\pm2 and has height 44 at 00.

How did this part go?

Split the required interval

On [0,2][0,2] use 4−x24-x^2; on [2,3][2,3] use x2−4x^2-4. Thus A=−∫02(x2−4) dx+∫23(x2−4) dxA=-\int_0^2(x^2-4)\,dx+\int_2^3(x^2-4)\,dx. Shade the entire region under the nonnegative modulus curve from 00 to 33.

How did this part go?

Evaluate exactly

A=−[x3/3−4x]02+[x3/3−4x]23=−(8/3−8)+[(27/3−12)−(8/3−8)]=23/3A=-[x^3/3-4x]_0^2+[x^3/3-4x]_2^3=-(8/3-8)+[(27/3-12)-(8/3-8)]=23/3 square units.

How did this part go?

Compare your sketch

Expand diagram

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Graph of absolute value of x squared minus four: the central section is reflected above the axis, with zeros at minus two and two and height four at zero. Shade from zero to three; split at two.
Graph of absolute value of x squared minus four: the central section is reflected above the axis, with zeros at minus two and two and height four at zero. Shade from zero to three; split at two.
Graph of absolute value of x squared minus four: the central section is reflected above the axis, with zeros at minus two and two and height four at zero. Shade from zero to three; split at two.

Common mistakes

Removing modulus bars without checking the sign of x squared minus 4 changes the function and the required area.

Read the hint