Unit 4: Further calculus and statistical inference

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 OpusReviewed by Better Tuition Academy9 min answer

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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).

Common errors

Mixing displacement and distance. Displacement is the signed integral; distance is the absolute-value integral or the split-and-sum.

Forgetting to split at zeros of vv for distance. A particle that changes direction has distance greater than displacement. The split-and-sum is mandatory.

Top-bottom backwards. Picking the wrong "top" gives a negative area. Test a value in each sub-interval before integrating.

Forgetting the constant of integration in position-from-velocity. x(t)x(t) has CC, which must be determined from x(t0)=x0x(t_0) = x_0.

Average velocity vs average speed. Average velocity = displacement / time. Average speed = distance / time. They differ when the particle reverses direction.

Forgetting to divide by interval length for average value. fˉ=1baabfdx\bar f = \frac{1}{b-a} \int_a^b f \, dx; the 1ba\frac{1}{b-a} is essential.

In one sentence

The definite integral applies to compute area between curves (top minus bottom on each sub-interval between intersections), the average value of a function (integral divided by interval length), and kinematics quantities (signed displacement from vdt\int v \, dt, total distance from vdt\int |v| \, dt via split-and-sum, position function from vdt\int v \, dt plus an initial condition); the PSMT and EA examine these applications most heavily in real-world contexts involving rate functions and accumulated change.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

2024 QCAA-style P25 marksA particle moves along a straight line with velocity $v(t) = t^2 - 4t + 3$ m/s for $0 \leq t \leq 5$ s. (a) Find the displacement from $t = 0$ to $t = 5$. (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 $y = x^3$ and $y = x$.
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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