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QLDMath MethodsSyllabus dot point

Topic 2: Trigonometric functions and integration applications

Apply the definite integral to compute the area between curves (including curves that change relative order), the average value of a function, and kinematics quantities (displacement, distance, position) from velocity and acceleration

A focused answer to the QCE Maths Methods Unit 4 dot point on the applications of integration. Area between curves, average value of a function, displacement and distance from velocity, position from acceleration with initial conditions, with worked PSMT-style examples.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Area under a curve
  3. Area between two curves
  4. Average value of a function
  5. Kinematics: position from velocity and acceleration
  6. Worked example. Constant-acceleration motion
  7. Worked example. Variable acceleration
  8. PSMT applications

What this dot point is asking

QCAA wants you to apply the definite integral to compute geometric area (between a curve and the xx-axis, or between two curves), the average value of a function on an interval, and kinematics quantities (displacement, distance, position from velocity). The dot point feeds the IA3 PSMT (which often models a real-world rate function) and the EA Paper 2 extended response.

Area under a curve

The signed area between y=f(x)y = f(x) and the xx-axis on [a,b][a, b] is:

abf(x)dx\int_{a}^{b} f(x) \, dx

For geometric area (always non-negative), split the interval at any zero of ff on (a,b)(a, b) and take absolute values:

Area=arfdx (or its absolute value)+rbfdx\text{Area} = \int_{a}^{r} f \, dx \text{ (or its absolute value)} + |\int_{r}^{b} f \, dx|

Area between two curves

If f(x)g(x)f(x) \geq g(x) on [a,b][a, b], the area between them is:

A=ab[f(x)g(x)]dxA = \int_{a}^{b} [f(x) - g(x)] \, dx

"Top minus bottom".

Procedure for area enclosed by two curves:

  1. Find intersection points. Solve f(x)=g(x)f(x) = g(x).
  2. Identify top and bottom in each sub-interval between intersections. Test a value inside.
  3. Integrate top minus bottom on each sub-interval.
  4. Sum the (positive) sub-integrals.

If the two curves cross within the interval of interest, the top-and-bottom roles swap and you must split.

Average value of a function

The average value of ff on [a,b][a, b] is:

fˉ=1baabf(x)dx\bar f = \frac{1}{b - a} \int_{a}^{b} f(x) \, dx

Geometrically: the height of the rectangle on [a,b][a, b] whose area equals the integral. In modelling: the typical value of a continuously varying quantity over an interval.

Example. Average velocity from t1t_1 to t2t_2 for a particle with velocity function v(t)v(t):

vˉ=1t2t1t1t2v(t)dt=displacementtime interval\bar v = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} v(t) \, dt = \frac{\text{displacement}}{\text{time interval}}

(Note: this is average velocity, not average speed; speed requires the integral of v|v|.)

Kinematics: position from velocity and acceleration

Displacement from velocity

For a particle with velocity v(t)v(t) from t=t1t = t_1 to t=t2t = t_2:

Displacement=t1t2v(t)dt\text{Displacement} = \int_{t_1}^{t_2} v(t) \, dt

Displacement is signed; positive if net motion is in the positive direction, negative otherwise.

Distance travelled from velocity

Total distance is the integral of speed v(t)|v(t)|:

Distance=t1t2v(t)dt\text{Distance} = \int_{t_1}^{t_2} |v(t)| \, dt

For a velocity that changes sign on the interval (the particle changes direction), find the zeros of vv, split the interval at each, and sum the absolute values of the sub-integrals.

Position function from velocity

If v(t)=dxdtv(t) = \frac{dx}{dt} and x(t0)=x0x(t_0) = x_0 is the initial position:

x(t)=x0+t0tv(s)dsx(t) = x_0 + \int_{t_0}^{t} v(s) \, ds

For most QCE Methods problems, antidifferentiate v(t)v(t) to get x(t)=vdt+Cx(t) = \int v \, dt + C, then use the initial condition to find CC.

Velocity and position from acceleration

If a(t)=dvdta(t) = \frac{dv}{dt}:

v(t)=v0+t0ta(s)dsv(t) = v_0 + \int_{t_0}^{t} a(s) \, ds

x(t)=x0+t0tv(s)dsx(t) = x_0 + \int_{t_0}^{t} v(s) \, ds

Two integrations with two initial conditions (v0v_0 and x0x_0) determine the position function from acceleration.

Worked example. Constant-acceleration motion

A particle has constant acceleration a=3a = 3 m/s2^2, starting from rest at x=0x = 0. Find v(t)v(t) and x(t)x(t).

v(t)=3dt=3t+C1v(t) = \int 3 \, dt = 3t + C_1. With v(0)=0v(0) = 0, C1=0C_1 = 0. So v(t)=3tv(t) = 3t m/s.

x(t)=3tdt=3t22+C2x(t) = \int 3 t \, dt = \frac{3 t^2}{2} + C_2. With x(0)=0x(0) = 0, C2=0C_2 = 0. So x(t)=3t22x(t) = \frac{3 t^2}{2} m.

This recovers the standard x=12at2x = \frac{1}{2} a t^2 formula from constant-acceleration kinematics.

Worked example. Variable acceleration

A particle has acceleration a(t)=62ta(t) = 6 - 2t m/s2^2, with v(0)=1v(0) = 1 m/s and x(0)=0x(0) = 0. Find v(t)v(t) and x(t)x(t).

v(t)=(62t)dt=6tt2+C1v(t) = \int (6 - 2t) \, dt = 6t - t^2 + C_1. v(0)=1    C1=1v(0) = 1 \implies C_1 = 1. So v(t)=6tt2+1v(t) = 6t - t^2 + 1.

x(t)=(6tt2+1)dt=3t2t33+t+C2x(t) = \int (6t - t^2 + 1) \, dt = 3 t^2 - \frac{t^3}{3} + t + C_2. x(0)=0    C2=0x(0) = 0 \implies C_2 = 0. So x(t)=3t2t33+tx(t) = 3 t^2 - \frac{t^3}{3} + t.

PSMT applications

The IA3 PSMT often presents a real-world rate function: water flow into a reservoir, drug clearance from blood, energy consumption over a day, traffic flow. Integration of the rate function over an interval gives the total accumulated quantity; division gives the average rate.

Typical PSMT moves:

  1. Identify the rate function and its domain.
  2. Integrate to get the total change.
  3. Apply initial / final conditions to determine constants.
  4. Compute averages, peak values, or time-to-reach thresholds.
  5. Discuss the model's limitations and refinements (boundary cases, real-world constraints not in the model).

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 QCAA-style P25 marksA particle moves along a straight line with velocity v(t)=t24t+3v(t) = t^2 - 4t + 3 m/s for 0t50 \leq t \leq 5 s. (a) Find the displacement from t=0t = 0 to t=5t = 5. (b) Find the total distance travelled.
Show worked answer →

(a) Displacement. Displacement is the signed integral of velocity.

Disp=05(t24t+3)dt=[t332t2+3t]05\text{Disp} = \int_{0}^{5} (t^2 - 4t + 3) \, dt = \left[ \frac{t^3}{3} - 2 t^2 + 3 t \right]_{0}^{5}.

Evaluate at t=5t = 5: 125350+15=125335=1251053=203\frac{125}{3} - 50 + 15 = \frac{125}{3} - 35 = \frac{125 - 105}{3} = \frac{20}{3}.

Evaluate at t=0t = 0: 00.

Displacement =203= \frac{20}{3} m (6.67\approx 6.67 m).

(b) Total distance. Find where v(t)=0v(t) = 0 on [0,5][0, 5].

t24t+3=0    (t1)(t3)=0    t=1t^2 - 4t + 3 = 0 \implies (t-1)(t-3) = 0 \implies t = 1 or t=3t = 3.

The particle changes direction at t=1t = 1 and t=3t = 3. Split the interval.

On [0,1][0, 1]: integral =[t332t2+3t]01=132+3=43= [\frac{t^3}{3} - 2t^2 + 3t]_{0}^{1} = \frac{1}{3} - 2 + 3 = \frac{4}{3} (positive, v>0v > 0 here).

On [1,3][1, 3]: integral =[]13=(27318+9)43=043=43= [\ldots]_{1}^{3} = (\frac{27}{3} - 18 + 9) - \frac{4}{3} = 0 - \frac{4}{3} = -\frac{4}{3} (negative, v<0v < 0).

On [3,5][3, 5]: integral =[]35=2030=203= [\ldots]_{3}^{5} = \frac{20}{3} - 0 = \frac{20}{3} (positive, v>0v > 0).

Total distance =43+43+203=4+4+203=283= \frac{4}{3} + |-\frac{4}{3}| + \frac{20}{3} = \frac{4 + 4 + 20}{3} = \frac{28}{3} m (9.33\approx 9.33 m).

Markers reward the displacement-vs-distance distinction, finding the zeros of vv, the interval split, and the absolute-value treatment.

2023 QCAA-style P24 marksFind the exact area enclosed between y=x3y = x^3 and y=xy = x.
Show worked answer →

Intersection points. x3=x    x3x=0    x(x1)(x+1)=0    x=1,0,1x^3 = x \implies x^3 - x = 0 \implies x(x-1)(x+1) = 0 \implies x = -1, 0, 1.

Test points to determine which is on top in each interval.

On (1,0)(-1, 0): at x=0.5x = -0.5, x3=0.125x^3 = -0.125 and x=0.5x = -0.5. So x3>xx^3 > x here (x3x^3 on top).

On (0,1)(0, 1): at x=0.5x = 0.5, x3=0.125x^3 = 0.125 and x=0.5x = 0.5. So x>x3x > x^3 here (xx on top).

Area =10(x3x)dx+01(xx3)dx= \int_{-1}^{0} (x^3 - x) \, dx + \int_{0}^{1} (x - x^3) \, dx.

First integral: [x44x22]10=0(1412)=14[\frac{x^4}{4} - \frac{x^2}{2}]_{-1}^{0} = 0 - (\frac{1}{4} - \frac{1}{2}) = \frac{1}{4}.

Second integral: [x22x44]01=(1214)0=14[\frac{x^2}{2} - \frac{x^4}{4}]_{0}^{1} = (\frac{1}{2} - \frac{1}{4}) - 0 = \frac{1}{4}.

Total area =14+14=12= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.

Markers reward finding all three intersection points, top-vs-bottom test in each sub-interval, and the symmetry observation that gives equal contributions from each.

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