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Signed integrals, geometric area and motion

Start a set of 8
Step 16 of 27
Given
\(\displaystyle v(t)=3,\quad0\le t\le4\)

A particle has constant velocity 3 m/s for 4 seconds. Find its displacement in metres

Year 12, VCE Units 3 & 4, Chapter 8, 4 difficulty levels, No calculatorBuild a worksheet

Signed totals and sums of magnitudes

Preserve direction for net change; add every region’s size for area or distance

\[I=\int_a^bf(x)\,dx,\qquad A=\int_a^b|f(x)|\,dx\]
  1. Above the axis The signed integral equals area when f is nonnegative throughout
  2. Below the axis Negate an entirely negative contribution to obtain its geometric area
  3. Across a crossing Split at each interior sign change. Add signed contributions for I; add separate magnitudes for A.
  4. Motion Velocity integrates to displacement. Its absolute value integrates to distance. Keep the time domain and units

Worked example

Find the signed integral and total geometric area for the function and interval shown below

\[f(x)=x-1,\quad0\le x\le3\]
  1. \(\displaystyle f(x)=0\Rightarrow x=1\)This root is inside the interval; f is negative to its left and positive to its right
  2. \(\displaystyle F(x)=\frac{x^2}{2}-x\)Differentiate to check the primitive
  3. \(\displaystyle F(0)=0,\quad F(1)=-\frac12,\quad F(3)=\frac32;\qquad \int_0^1f=F(1)-F(0)=-\frac12,\quad \int_1^3f=F(3)-F(1)=2\)Use right primitive value minus left primitive value on each interval
  4. \(\displaystyle I=\frac32,\qquad A=\frac52\)The signed integral adds contributions; total area adds their separate magnitudes

Common mistakes

Taking the absolute value too late
✕\(\displaystyle A=\left|-\frac12+2\right|=\frac32\)✓\(\displaystyle A=\left|-\frac12\right|+|2|=\frac52\)
Cancellation loses region size before the final absolute value
Adding signed changes for distance
✕\(\displaystyle D=-1+4=3\)✓\(\displaystyle D=|-1|+|4|=5\)
The signed total is displacement; distance counts motion in either direction positively
Assuming zero integral means no area
✕\(\displaystyle -4+4=0\Rightarrow A=0\)✓\(\displaystyle A=|-4|+|4|=8\)
Two nonzero regions can have opposite contributions

Difficulty levels

introRelate a positive line or constant velocity to an exact total
standardUse an interval sign certificate for a one-sided polynomial region or positive velocity
challengeSplit at one constructed crossing and distinguish signed contributions from their magnitudes
examChoose the integral setup, then calculate the signed total and total area or distance

Questions about this topic

Must every zero change the sign?+

No. Test the sign on each side. A squared factor can touch zero without changing sign

Can displacement be zero when distance is positive?+

Yes. Equal travel in opposite directions can return a particle to its starting position

Can I read an exact area from the graph’s pixels?+

Use the printed function and exact interval bounds to calculate. The graph helps you identify the regions and signs

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 8: Integration

In this chapter

Approximate area by an endpoint or trapezium ruleEvaluate and transform a definite integralElementary primitives and exact definite integralsFind complete families of polynomial anti-derivativesThe area function and the fundamental theorem of calculusSigned integrals, geometric area and motionFind the area between two curvesAverage value of a functionTotal change and kinematics from a rateDetermine a polynomial from its derivative and a boundary conditionRecover motion from acceleration