Module 1: Mathematical Foundations · Practice · assumes Foundation 9
Six worked examples on integration — what it is, the two techniques worth knowing, and the identity that removes time from a problem.
What this set is for
Foundation 9 introduced substitution and integration by parts once each, and Foundation 11 called a dx = v dv the identity most worth memorising. None of them had been practised. This set does that.
It opens with what an integral actually is — because the first example is one you can already do two ways — and ends with two problems that cannot be solved by the usual route at all. Same three-part structure: The Physics, The Mathematics, The Contemplation.
On this page
1. The integral is the area
Problem. A body’s velocity increases uniformly from 2 m/s to 10 m/s over 4 s, so that v(t) = 2 + 2t. Find the displacement twice — once by geometry, once by integration. Then find the displacement of a second body with v(t) = t² over 0 ≤ t ≤ 3, and say why only one method survives.
The physics
Displacement is what accumulates when velocity acts over time, and on a velocity–time graph that accumulation is the area beneath the curve. Foundation 7 computed such an area with school geometry. The claim now is that integration is not a rival method giving a similar answer — it is the same quantity, obtained differently, and it should agree exactly.
Same quantity, two methods — until the graph bends.
The mathematics
By geometry. The region is a trapezium with parallel sides 2 and 10 and width 4:
Δx = ½(2 + 10)(4) = 24 m
By integration.
Δx = ∫04 (2 + 2t) dt = [2t + t²]04
= (8 + 16) − 0 = 24 m
Identical, as it must be. Notice that the integration did more work than the geometry here, and gained nothing — which is worth knowing, because it means the integral is not automatically the better tool.
The curved case. Now v = t². There is no trapezium, no triangle, no combination of straight-sided pieces that fits under a parabola. Geometry has nothing to offer. The integral does not notice the difficulty:
Δx = ∫03 t² dt = [t³/3]03 = 9 m
The contemplation
Students often hold the trapezium rule and the definite integral as two separate techniques learned in different years. They are one thing. The integral does not approximate the area or represent it; it is a way of computing exactly the quantity the trapezium computed, and the agreement in the first case is not a coincidence to be checked but a consequence of what the symbol means. What integration buys is not accuracy — it is that the method stops caring whether the boundary is straight.
2. Where the journey doubles back
Problem. A particle moves along a line with v(t) = 6 sin(πt) m/s. Find the displacement and the total distance travelled over 0 ≤ t ≤ 2 s.
The physics
The sine is positive for the first second and negative for the second, so the particle goes out and comes back. Before computing anything: the motion over the second half mirrors the first, so the particle should return exactly to where it started, and the displacement should be zero. The distance, however, counts both legs.
The mathematics
Displacement is the plain integral, signs and all:
Δx = ∫02 6 sin(πt) dt = [−(6/π) cos(πt)]02
= −(6/π)(cos 2π − cos 0) = −(6/π)(1 − 1) = 0 m
Distance requires the speed, |v|, which means splitting the interval where the velocity changes sign — at t = 1, since sin(πt) is positive on [0, 1] and negative on [1, 2]:
∫01 6 sin(πt) dt = −(6/π)(cos π − cos 0) = −(6/π)(−2) = 12/π
∫12 −6 sin(πt) dt = (6/π)(cos 2π − cos π) = (6/π)(2) = 12/π
S = 24/π ≈ 7.64 m
The particle travelled 7.64 m and finished where it began.
The contemplation
The work that mattered here was not the integration — both integrals are one line. It was finding t = 1. A student who integrates |v| across the whole interval without locating the sign change is not making an arithmetic error; the expression ∫|6 sin πt| dt simply cannot be evaluated by antidifferentiation, because the modulus is not a function you have an antiderivative for. The splitting is not tidiness. It is what converts an unanswerable expression into two answerable ones.
3. Substitution, and the limits that must move with it
Problem. A particle moves along the x-axis from x = 0 to x = √3 m under a force F(x) = x/√(x² + 1) N. Find the work done.
The physics
The force varies with position, so W = Fd is unavailable and the integral is compulsory. Before computing, note the size of the force: it starts at zero and rises toward 1 N as x grows, so over a distance of about 1.7 m the work should come out somewhere near 1 J. That estimate is worth having, because it will catch a slip.
The mathematics
The awkward part is x² + 1 inside a square root — and its derivative, 2x, is sitting outside. That is the signal for substitution. Put u = x² + 1, so du = 2x dx, giving x dx = du/2.
Now move the limits. The integral was over x from 0 to √3; in the new variable it runs over u:
x = 0 → u = 1 x = √3 → u = 3 + 1 = 4
W = ∫14 u−1/2 (du/2) = ½[2√u]14
= √4 − √1 = 1 J
Which matches the estimate. If instead you prefer to substitute back to x before evaluating, that is equally valid — but then the limits stay as 0 and √3, and mixing the two conventions is where marks are lost.
The contemplation
The most common error in this topic is leaving the limits at 0 and √3 while changing the integrand to u. It produces a wrong answer that looks entirely reasonable — no negative under a root, no division by zero, nothing to alert you. The reason it fails is that the limits are not decorations on the integral sign; they are values of the variable being integrated, and changing the variable changes them. Substitution is not a trick applied to the integrand alone. It is a change of what you are accumulating over.
4. By parts, over an infinite tail
Problem. During an impact, a bumper experiences a force F(t) = F0t e−βt for t ≥ 0, with F0 = 100 N/s and β = 2 s−1. Find the total impulse delivered.
The physics
Impulse is the accumulation of force over time, and equals the momentum transferred. The force is zero at impact, rises as the bumper compresses, then decays away — and it never quite reaches zero, so the integral runs to infinity. That should give pause: an infinite duration might suggest an infinite impulse. It does not, because the exponential decay outpaces the linear growth, and the total momentum transfer is finite.
The mathematics
The integrand is a product of two functions of different kinds, neither of which is the derivative of the other. That is the signal for integration by parts. Choose u = t, because differentiating it makes it simpler:
u = t, du = dt dv = e−βt dt, v = −(1/β)e−βt
∫ te−βt dt = −(t/β)e−βt + (1/β)∫ e−βt dt
= −e−βt(t/β + 1/β²)
Evaluating the limits. As t → ∞, the exponential decays faster than the bracket grows, so the whole expression tends to zero. At t = 0 it equals −1/β². Hence:
J = F0[0 − (−1/β²)] = F0/β²
= 100/4 = 25 N s
The bumper receives 25 N s of impulse, and a body of mass 5 kg struck this way would change speed by 5 m/s.
The contemplation
The choice of u is the whole method, and it is a choice rather than a rule. Had we taken u = e−βt instead, the remaining integral would have contained t² and been worse than the one we started with — the method would have run backwards, correctly, into a harder problem. Integration by parts does not solve an integral so much as trade it for another one, and the skill is entirely in making sure the trade is favourable.
5. When acceleration depends on position
Problem. A particle moving along the x-axis has acceleration a = −2x m/s². At x = 0 its speed is 6 m/s in the positive direction. Find its speed at x = 4 m, and how far it travels before momentarily stopping.
The physics
The acceleration is given as a function of position, not of time — and no time appears anywhere in the question. The usual route, integrating acceleration to get velocity, needs a(t) and is unavailable. Attempting it would require first solving for x(t), which is a differential equation and far more work than the question deserves. Something has to let us relate speed to position directly, without t ever entering.
The mathematics
The chain rule supplies it. Since a = dv/dt and v = dx/dt:
a = dv/dt = (dv/dx)(dx/dt) = v dv/dx
so a dx = v dv
Time has gone. Both sides can now be integrated — the left over position, the right over velocity:
∫6v v′ dv′ = ∫0x (−2x′) dx′
v²/2 − 18 = −x²
v² = 36 − 2x²
At x = 4: v² = 36 − 32 = 4, so v = 2 m/s.
It stops where v = 0, that is 2x² = 36, giving x = 3√2 ≈ 4.24 m.
Notice what the relation v² = 36 − 2x² is. Multiply through by m/2 and it says ½mv² + mx² is constant — kinetic energy plus a potential energy quadratic in x. We have derived conservation of energy for this force without invoking it.
The contemplation
The identity a dx = v dv is one line of chain rule and it removes an entire variable from the problem. That is worth noticing as a general move: when a question mentions no time, and you find yourself introducing time in order to answer it, you have made the problem harder than it was asked. And the energy relation that fell out at the end is not a coincidence — the work–energy theorem is this identity, integrated and multiplied by mass. Foundation 9 said displacement, work and impulse are one operation applied to different pairs. Here is a fourth face of it.
6. When acceleration depends on speed
Problem. A boat moving at 20 m/s cuts its engine. The water resists its motion so that the deceleration is proportional to speed, a = −0.5v m/s². How far does it travel before coming to rest?
The physics
Again the question asks for a distance and mentions no time. But this case has a wrinkle worth seeing in advance: as the boat slows, the retarding force weakens too, so the deceleration fades away and the boat never quite stops. That suggests the time taken is infinite — and yet the distance may still be finite, in the same way the impulse in example 4 was finite over an infinite tail.
The mathematics
Use the same identity, and notice what happens:
v dv/dx = −0.5v
dv/dx = −0.5 (dividing by v, valid while the boat still moves)
A striking result: velocity falls off linearly with distance, though not with time. Integrating from the start:
v = 20 − 0.5x
Setting v = 0: x = 40 m
The boat covers 40 m. And it does so while taking, strictly, forever — the speed approaches zero without reaching it, so no finite time answers the question “when does it stop”, while “how far does it go” has the perfectly definite answer 40 m.
The contemplation
A finite distance covered in infinite time is not a paradox, but it takes a moment to accept. It happens because the two questions are integrals over different variables, and one converges while the other does not. Notice also the modelling assumption doing the work: a = −0.5v is a reasonable description of drag at moderate speeds, and it predicts a boat that drifts forever. A real boat stops, because at very low speeds other resistances dominate and the linear law fails. The infinity is in the model, not in the water.
Choosing the method
Six problems, and in most of them the integration itself was a line or two. The difficulty was elsewhere: recognising that a modulus needs the interval split, that the limits move under substitution, that one choice of u improves the integral and the other ruins it, and that a question with no time in it should be answered without introducing any.
A useful habit follows. Before integrating, ask which variable the quantity is given as a function of, and which variable the answer is wanted in. If those two differ, the setup matters more than the technique — and getting the setup right usually leaves an integral simple enough to do in your head.
Assumes: Foundation 9 · Type: Worked examples · Reading: 25 min · Difficulty: Intermediate to Advanced