Study Companion: Programme F.13 — Integration

One idea runs through this whole programme: adding up many small pieces of a rate gives the total, and that total is found by running differentiation backwards. A speedometer tells you how fast; integration tells you how far. The same sum, drawn on a graph, is an area — and the deepest fact in the programme is that areas are found by undoing derivatives.

This file answers why. The rules, tables and drill are in the companion reference sheet; every rule there links back to the unit here that earns it.

How to use this companion

Each unit is built as an argument: a real situation and a question you cannot yet answer, the attempt a sensible person would make first, the picture that settles it, a check with actual numbers, and only then the rule. Read it once front to back. The Narration spine at the end of each unit summarises what the video shows.

As in Programme F.12, \delta t is a small but finite change — a real number like 0.1 — and \frac{d}{dt} is the derivative it leads to.

Unit 01: Working Backwards from a Rate

The problem

A drone lifts off from a flat roof and climbs. Its flight log records only its climb rate:

with t in seconds. How high is it after 2 s?

First attempt

"Height is rate times time": 3(2)^2\times2=24 m. But the rate at t=2 is only its final rate; for most of the two seconds it was climbing more slowly. That product overestimates, and it has another, deeper problem: it ignores where the drone started.

The picture

Differentiation turns height into climb rate. So ask the question backwards: which height function has derivative 3t^2? From the power rule, \frac{d}{dt}t^3=3t^2. So h=t^3 is one answer.

But not the only one. Draw h=t^3, and then the same curve shifted up by 5, and down by 2:

Sliding a graph up or down does not change its steepness anywhere: every tangent moves up with it. All three curves have slope 3t^2 at every t. The climb rate cannot tell them apart, because differentiating t^3+C turns the constant C into 0 — the information is simply gone.

So the rate fixes the shape of the height graph, and one extra fact fixes its position. Here it is the roof: the drone started 5 m above the ground, so h=5 when t=0, and C=5.

Check it with numbers

h=t^3+5. At t=0: 5 m, the roof. At t=2: 8+5=13 m.

And the derivative check: \frac{d}{dt}(t^3+5)=3t^2, the logged climb rate.

The rule it gives

Finding a function from its derivative is integration, the reverse of differentiation. It is written

read "the integral of 3t^2 with respect to t". The function being integrated, 3t^2, is the integrand; the dt names the variable. C is the constant of integration, and it must always be written: without it the answer claims to know something the rate cannot tell you. An integral with an unknown C is an indefinite integral; one extra condition, such as a starting value, fixes C.

The check on any integral is to differentiate the answer and get the integrand back.

Worked example

Find I=\int8x^3\,dx given that I=20 when x=1.

\frac{d}{dx}(2x^4)=8x^3, so I=2x^4+C. At x=1: 20=2+C, so C=18 and I=2x^4+18.

Where this shows up

Every measured rate — flow from a flow meter, current from an ammeter, speed from a wheel sensor — becomes an amount by integration, and every such amount needs a starting value. A water meter reading is the integral of flow plus the reading when it was installed; the constant of integration is the dial's starting position.

Narration spine. Show the drone lifting off a roof with its climb-rate log. Ask for the height at t=2 and strike out "rate times time". Plot t^3, then slide copies up and down with their tangents at one t moving in parallel. Name the family t^3+C, then drop the curve onto the roof's height to fix C=5.

Unit 02: Standard Integrals, and Polynomials

The problem

A capacitor is charged from empty by a current that varies as

Charge is the integral of current. How much charge has it stored after 2 s?

First attempt

Differentiation had a table of standard results. Integration has no new table to learn — every derivative, read backwards, is an integral. The only work is to turn each rule round so it starts from the integrand.

The picture

Take the power rule, \frac{d}{dx}x^{n+1}=(n+1)x^n. It produces x^n with an unwanted factor (n+1) in front. Divide it out:

"Raise the power by one, and divide by the new power" — the power rule run in reverse. It fails for exactly one power: n=-1, where the new power is 0 and the division is impossible. That gap is filled by a derivative we already know, \frac{d}{dx}\ln x=\frac1x (F.12, Unit 09):

The rest of the table comes the same way, each line a derivative from F.12 turned round:

Because differentiation works term by term and leaves constant factors alone, so does integration. The separate constants from each term add up to a single C.

Check it with numbers

\int(6t^2-4t+3)\,dt=2t^3-2t^2+3t+C. Differentiate to check: 6t^2-4t+3.

The capacitor starts empty, so q=0 at t=0, giving C=0. At t=2:

The rule it gives

Worked example

Find I=\int(6x^2-4x+3)\,dx given I=10 when x=1, then evaluate I at x=2.

I=2x^3-2x^2+3x+C. At x=1: 2-2+3+C=10, so C=7. At x=2: 16-8+6+7=21.

For a long polynomial, evaluate in nested form: 2x^3-2x^2+3x+7=\big((2x-2)x+3\big)x+7; at x=2: 2\to(4-2)=2; 2\times2+3=7; 7\times2+7=21.

Where this shows up

The capacitor: 14 mC after two seconds. The same integral, with different numbers, gives the energy drawn by a load from its power trace, the volume pumped from a flow trace, or the heat delivered from a heat-flow trace — each a rate turned into an amount.

Narration spine. Write the power rule for derivatives, and slide it backwards into the integral: raise the power, divide by it. Show why n=-1 breaks the division and fill the gap with \ln x. Build the table by reversing each F.12 derivative. Integrate the capacitor current term by term and fix C from the empty start.

Unit 03: Functions of a Linear Function

The problem

A capacitor discharging through a resistor delivers a current

How much charge has flowed after 2 s? The table has \int e^x\,dx, but here the exponent is -\frac t2, not t.

First attempt

Copy the table: \int e^{-t/2}\,dt=e^{-t/2}? Check by differentiating: \frac{d}{dt}e^{-t/2}=-\frac12e^{-t/2} by the chain rule. That is off by a factor of -\frac12.

The picture

The chain rule says what went wrong. Replacing x by ax+b squeezes the graph horizontally by a factor a: everything happens a times faster, so every slope is a times steeper.

So differentiating F(ax+b) gives a\,F'(ax+b) — an extra factor a. To undo it, integrate as usual with ax+b in place of x, and divide by a:

Check it with numbers

\int(2x+5)^3\,dx=\frac{(2x+5)^4}{4}\cdot\frac12+C=\frac{(2x+5)^4}{8}+C.

Differentiate: \frac{4(2x+5)^3\times2}{8}=(2x+5)^3. At x=0 the slope of \frac{(2x+5)^4}8 is 5^3=125; numerically, with \delta x=0.001, 125.08.

The rule it gives

Replace x in the standard integral by the linear expression, and divide by the coefficient of x:

The same result comes from the substitution u=ax+b, du=a\,dx: \int f(u)\frac{du}{a}=\frac1aF(u).

Worked example

\int\cos3x\,dx=\frac{\sin3x}3+C. Check: \frac{d}{dx}\frac{\sin3x}3=\frac{3\cos3x}3=\cos3x.

Where this shows up

For the discharging capacitor, with a=-\frac12:

No charge has flowed at t=0, so 0=-0.012+C and C=0.012. After 2 s,

about 63\,\% of the 0.012 C it will eventually deliver — the "one time constant, 63\,\%" rule every electronics engineer quotes, arriving straight out of the integral.

Narration spine. Integrate e^{-t/2} by copying the table and check it by differentiating: off by -\frac12. Squeeze a graph horizontally by 3 and show its tangent steepen by 3. Conclude "divide by the coefficient", then integrate the capacitor current and fix C from zero charge.

Unit 04: Partial Fractions

The problem

In a chemical reactor the product concentration x (mol/L) grows at

with t in seconds. How long does it take to go from 0 to 1 mol/L?

Turning it round, \frac{dt}{dx}=\frac{10}{(2-x)(3-x)}, so the time is an integral of that fraction. It is not in the table.

First attempt

Integrate the top and bottom separately? There is no such rule — the integral of a quotient is not a quotient of integrals.

The idea that works

The table does have \int\frac{dx}{ax+b}. So split the fraction into a sum of fractions with linear denominators, each of which the table handles. From Programme F.8:

Put x=2: 1=A. Put x=3: 1=-B, so B=-1. Therefore

This is algebra, not a picture: the split is chosen because it lands every piece in the table.

Check it with numbers

At x=1: \frac{1}{1\times2}=0.5, and \frac11-\frac12=0.5.

The rule it gives

1. If the numerator's degree is not lower than the denominator's, divide first.
2. Factorise the denominator and split into partial fractions.
3. Integrate each piece with \int\frac{A}{ax+b}\,dx=\frac{A}{a}\ln(ax+b)+C.

Example with division first. \frac{x^2+3x+4}{x+1}: dividing, x^2+3x+4=(x+1)(x+2)+2, so

Check at x=1: the derivative is 1+2+1=4, and the integrand is \frac{8}{2}=4.

Worked example

5x+3=A(x+3)+B(x+1). At x=-1: -2=2A, A=-1. At x=-3: -12=-2B, B=6.

Check at x=1: the derivative is -\frac12+\frac64=1; the integrand is \frac{8}{2\times4}=1.

Where this shows up

For the reactor, integrating each piece (with a=-1 in both):

At x=0, t=0: C=-10\ln1.5. At x=1:

Narration spine. Show the reactor's rate equation and the fraction to integrate. Try "integrate top and bottom" and strike it. Split the fraction into two simple pieces, check the split at x=1, integrate each into a logarithm, and fix C from the empty start to get 2.88 s.

Unit 05: Distance from a Speedometer — Adding Up Strips

The problem

A car's speedometer is recorded for six seconds as it accelerates away and brakes to a stop:

How far did it travel? This time, answer it without antiderivatives — using only what the speedometer shows.

First attempt

If the speed were constant, distance would be speed times time: at 5 m/s for 6 s, 30 m. On a velocity–time graph that is the area of a rectangle 6 s wide and 5 m/s high — and the units agree: \frac{\text{m}}{\text{s}}\times\text{s}=\text{m}.

But this speed is never constant for an instant, so there is no single rectangle.

The picture

Pretend it is constant over short intervals. Chop the six seconds into strips of width \delta t, and on each strip use the speed at its start:

Each strip is a rectangle whose area is (speed) × (time) = the distance covered in that strip, if the speed held steady. The strips undershoot where the car speeds up and overshoot where it slows, but as the strips get thinner the staircase hugs the curve more and more closely. In the limit, the total of the strips is exactly the area under the velocity graph — and that area is the distance.

Follow one strip to see what it means: at t=2 s the speed is 8 m/s, so a strip 0.5 s wide is 8\times0.5=4 m of travel.

Check it with numbers

Total of the strips, speed taken at the start of each:

The totals settle on 36 m.

For one function the limit can be computed exactly. The area under y=x^2 from 0 to 3, with n strips of width \frac3n and heights at the right-hand ends:

using \sum r^2=\frac{n(n+1)(2n+1)}6 from Programme F.7. For n=3,6,30,300 this is 14, 11.375, 9.455, 9.045; as n\to\infty it tends to \frac{9\times2}{2}=9.

The rule it gives

The area under y=f(x) between x=a and x=b is the limit of the strip sum:

The sign \int is an elongated S, for sum; a and b are the limits, and an integral with limits is a definite integral. The dx is what is left of the strip width \delta x.

Worked example

Estimate \int_0^2x^3\,dx with four strips of width 0.5, heights at the right-hand ends: 0.5^3+1^3+1.5^3+2^3=0.125+1+3.375+8=12.5, times 0.5 gives 6.25. (Right-hand heights overestimate a rising curve; Unit 06 finds the exact value, 4.)

Where this shows up

The car travelled 36 m. A trip computer does exactly this sum, many times a second, from the wheel-speed sensor — it never knows a formula for v, only the strips.

Narration spine. Show the constant-speed rectangle and read its area as distance, units and all. Replace the real curve with a staircase of constant-speed strips; follow one strip (8 m/s × 0.5 s = 4 m). Sum them, write \sum v\,\delta t, and morph \sum into \int. Thin the strips and watch the staircase close on the curve while the total settles on 36.

Unit 06: Why Areas Are Undone Derivatives

The problem

Unit 05 found the car's distance, 36 m, by adding ever more strips. Unit 01 found heights by undoing a derivative. These look like two different operations. Are they?

First attempt

Check with numbers: an antiderivative of v=6t-t^2 is 3t^2-\frac{t^3}3. At t=6: 108-72=36. The same 36. A coincidence would be strange; we need a reason.

The picture

Let the right-hand end of the area move. Call it T, and call the area under the graph from 0 up to T by the name A(T) — the distance travelled by time T.

Nudge T by \delta T. The area grows by a thin sliver:

The sliver is almost a rectangle: height v(T), width \delta T. So

The error is the little curved cap on top of the rectangle, which shrinks faster than the sliver itself — the same kind of vanishing corner as in the growing square of F.12, Unit 02. In the limit,

The rate at which the area grows is the height of the graph at its edge. So the area function is an antiderivative of v. That is the connection between the two operations: summing strips and undoing derivatives give the same thing.

Check it with numbers

At T=3, with \delta T=0.1: the exact extra area is A(3.1)-A(3)=0.89967 m, and the rectangle estimate is v(3)\times0.1=9\times0.1=0.9 m. Their ratio to \delta T is 8.997 against v(3)=9.

The rule it gives

Because A and any antiderivative F of v have the same derivative, they differ only by a constant: A(T)=F(T)+C. Since A(0)=0, C=-F(0), and

This is the Fundamental Theorem of Calculus. To evaluate a definite integral: find any antiderivative, evaluate it at the upper limit, subtract its value at the lower limit. The constant of integration always cancels in the subtraction, so it is left out.

The assumption doing the work is that f is continuous, so the sliver really is nearly a rectangle.

Worked example

And the estimate from Unit 05: \int_0^2x^3\,dx=\left[\frac{x^4}4\right]_0^2=4-0=4.

Where this shows up

The car: \int_0^6(6t-t^2)\,dt=\left[3t^2-\frac{t^3}3\right]_0^6=108-72=36 m, in one line instead of six hundred strips. And the distance covered between the third and fourth seconds is A(4)-A(3)=26.67-18=8.67 m — any interval's distance is a subtraction.

Narration spine. Put the area and the antiderivative side by side and note they both give 36. Make the area's right edge a movable marker T and sweep it. Nudge T by \delta T, shade the sliver, and compare it with the rectangle v(T)\,\delta T as \delta T shrinks. Write \frac{dA}{dT}=v(T), then F(b)-F(a), and evaluate the car's distance in one line.

Unit 07: Negative Areas

The problem

After stopping at t=6 s, the car reverses. Its velocity is still v=6t-t^2, now negative. What does \int_0^8v\,dt measure — and how far did the car actually drive?

First attempt

\int_0^8(6t-t^2)\,dt=\left[3t^2-\frac{t^3}3\right]_0^8=192-170.67=21.33 m. But the car drove 36 m forwards before it even started reversing. 21.33 m cannot be the distance driven.

The picture

Go back to the strips. Below the axis, every strip has a positive width \delta t but a negative height v, so its product v\,\delta t is negative — a stretch travelled backwards. The integral adds the forward stretches and subtracts the backward ones:

So the integral is the net change — the displacement, where the car ends up relative to where it started. The distance driven counts both parts as positive.

Check it with numbers

From 6 to 8 s: \left[3t^2-\frac{t^3}3\right]_6^8=21.33-36=-14.67 m — a negative area, 14.67 m of reversing.

• Displacement: 36+(-14.67)=21.33 m.
• Distance driven: 36+14.67=50.67 m.

The rule it gives

\int_a^bf(x)\,dx counts area above the axis as positive and area below as negative. To find a physical area (or total distance), find where the graph crosses the axis, integrate each part separately, and add their sizes. Always sketch first.

Worked example

Area between y=x^2-4 and the x-axis from x=0 to x=3. It crosses at x=2.

• \int_0^2(x^2-4)\,dx=\left[\frac{x^3}3-4x\right]_0^2=\frac83-8=-\frac{16}3: an area of \frac{16}3 below.
• \int_2^3(x^2-4)\,dx=\left(9-12\right)-\left(\frac83-8\right)=-3+\frac{16}3=\frac73: an area of \frac73 above.

Total area \frac{16}3+\frac73=\frac{23}3=7.67; integrating straight through would give -\frac{16}3+\frac73=-3.

Where this shows up

A GPS tracker that integrates velocity reports displacement; an odometer reports distance. They agree only while the vehicle never reverses. The same distinction separates net charge from total charge in an alternating current: the integral of a pure sine current over a whole cycle is zero, even though charge moved the whole time.

Narration spine. Extend the velocity graph past t=6 so it dips below the axis while the car reverses. Show a strip below the axis with positive width and negative height. Colour the two areas +36 and -14.67, compute the net 21.33, and contrast it with the 50.67 m driven.

Unit 08: The Area Between Two Curves

The problem

A steel plate is cut with one curved edge, y=4-x^2, and one straight edge, y=x+2 (both in decimetres). The plate is 5 mm thick. What is its mass?

First attempt

Area under the top edge minus area under the bottom edge — but between which x-values, and what if part of it dips below the axis?

The picture

Use strips again, but now each strip runs from the lower edge up to the upper edge:

A strip at x has height (top − bottom) and width \delta x, so it has area \big[(4-x^2)-(x+2)\big]\delta x. Add up the strips from where the edges meet on the left to where they meet on the right. The axis never enters: only the difference in height matters, so the result holds wherever the plate sits.

The edges meet where 4-x^2=x+2, that is x^2+x-2=0, (x+2)(x-1)=0: at x=-2 and x=1.

Check it with numbers

So A=4.5 dm² =0.045 m².

The rule it gives

where a and b are the x-values at which the curves meet.

Worked example

The area between y=x and y=x^2: they meet at x=0 and x=1, with x on top. \int_0^1(x-x^2)\,dx=\frac12-\frac13=\frac16.

Where this shows up

The plate's mass is its area times its thickness times the density of steel, 7850 kg/m³:

The same strip-between-curves integral gives the cross-sectional area of a channel between its bed and the water surface, and the area of a cam profile — any region bounded top and bottom by known curves.

Narration spine. Draw the plate between the parabola and the line, find their crossings, and sweep a single vertical strip across it with its height top minus bottom. Fill the plate with thinning strips, integrate, and turn the 4.5 dm² into 1.77 kg of steel.