What integrals compute

An integral was defined to compute an area, and area is close to the least interesting thing it does.

The pattern behind every application in this lesson is the same three-step move: cut the quantity into slices, approximate each slice by something elementary enough to write down, and recognise the total as a Riemann sum whose limit is an integral. Learning that move is worth more than memorising any of the formulas it produces, because the formulas are numerous and the move is one thing.

Area between two curves

If $f \geq g$ on $[a, b]$ , a vertical strip at position $x$ has height $f (x) - g (x)$ and width $d x$ , so the area between the curves is

\int_{a}^{b} [f (x) - g (x)] d x

The subtraction handles signs automatically: the formula is correct even where both curves are below the axis, because only their difference enters.

Example. Find the area enclosed between $y = x + 2$ and $y = x^{2}$ .

First find where they meet: $x^{2} = x + 2$ gives $x^{2} - x - 2 = 0$ , so $x = - 1$ and $x = 2$ . Between those points the line is above the parabola, checked at $x = 0$ where the line gives $2$ and the parabola gives $0$ . So the area is

\int_{- 1}^{2} (x + 2 - x^{2}) d x = {[\frac{x^{2}}{2} + 2 x - \frac{x^{3}}{3}]}_{- 1}^{2} = 3.3333 - (- 1.16667) = 4.5

Now you. Find the area enclosed between $y = 2 x$ and $y = x^{2}$ .

Answer

They meet where $x^{2} = 2 x$ , at $x = 0$ and $x = 2$ , and the line is above in between. The area is $\int_{0}^{2} (2 x - x^{2}) d x = [x^{2} - x^{3} / 3]_{0}^{2} = 4 - 2.6667 = 1.3333$ .

Volumes by slicing

A solid can be cut into thin slabs perpendicular to an axis. If the slab at position $x$ has cross-sectional area $A (x)$ and thickness $d x$ , its volume is $A (x) d x$ , and the total is $\int A (x) d x$ . When the solid is made by revolving a curve about the $x$ -axis, each slab is a disc of radius $f (x)$ , so

V = \int_{a}^{b} π [f (x)]^{2} d x

Example. Derive the volume of a sphere of radius $R$ .

A sphere is the solid of revolution of the semicircle $y = \sqrt{R^{2} - x^{2}}$ from $- R$ to $R$ . The disc at position $x$ has radius $y$ , so its area is $π (R^{2} - x^{2})$ , and

V = \int_{- R}^{R} π (R^{2} - x^{2}) d x = π {[R^{2} x - \frac{x^{3}}{3}]}_{- R}^{R}

Working the bracket out: at $x = R$ it is $R^{3} - R^{3} / 3 = 2 R^{3} / 3$ , and at $x = - R$ it is $- 2 R^{3} / 3$ , so the difference is $4 R^{3} / 3$ and $V = \frac{43}{π} R^{3}$ . The formula that school geometry asserts without justification falls out of one integral. Archimedes obtained it by exhaustion around 225 BC and was proud enough of the result to have the figure carved on his tomb.

Now you. Find the volume generated by revolving $y = \sqrt{x}$ from $x = 0$ to $x = 4$ about the $x$ -axis.

Answer

The disc radius is $\sqrt{x}$ , so the area is $π x$ and the volume is $\int_{0}^{4} π x d x = π [x^{2} / 2]_{0}^{4} = 8 π = 25.13$ .

The same slicing gives the cone. Revolving the line $y = (R / h) x$ from $0$ to $h$ gives $V = π (R / h)^{2} \int_{0}^{h} x^{2} d x = π (R / h)^{2} h^{3} / 3 = \frac{13}{π} R^{2} h$ , which is where the mysterious one third comes from: it is the $x^{3} / 3$ of the power rule and nothing more.

Work against a varying force

Elementary mechanics defines work as force times distance, which is valid only when the force is constant. When it varies, cut the displacement into pieces small enough that the force is nearly constant on each, multiply, and sum:

W = \int_{a}^{b} F (x) d x

For a spring obeying Hooke's law, $F = k x$ , so stretching from $0$ to $d$ takes $W = \frac{12}{k} d^{2}$ . That familiar formula is an integral, and its quadratic form is the reason a spring stretched twice as far stores four times the energy.

Example. A cylindrical tank of radius $2$ m contains water to a depth of $3$ m. How much work is needed to pump it all out over the rim, $5$ m above the base?

Slice the water into horizontal layers. The layer at height $y$ above the base has volume $π (2)^{2} d y = 4 π d y$ cubic metres, so its mass is $4000 π d y$ kilograms and its weight is $4000 π g d y$ newtons. That layer must be lifted a distance $5 - y$ . So

W = \int_{0}^{3} 4000 π g (5 - y) d y = 39200 π {[5 y - \frac{y^{2}}{2}]}_{0}^{3} = 39200 π (15 - 4.5)

which is $411600 π = 1.293 \times 10^{6}$ J, using $g = 9.8$ m s⁻². The check is that the total mass is $4 π \times 3 \times 1000 = 37700$ kg and its centre of mass sits at $y = 1.5$ , needing a lift of $3.5$ m, giving $37700 \times 9.8 \times 3.5 = 1.293 \times 10^{6}$ J. The integral and the centre-of-mass shortcut agree, as they must.

Now you. A spring with stiffness $k = 200$ N/m is stretched from its natural length to $0.3$ m. How much work is done?

Answer

$W = \int_{0}^{0.3} 200 x d x = 100 (0.09) = 9$ J.

Escape velocity, and integrals over infinite ranges

Gravity weakens with distance, so lifting a mass away from a planet is a work integral with a variable force, and the interesting case has no upper limit at all.

An improper integral is defined as a limit: $\int_{1}^{\infty} f (x) d x$ means $lim_{b \to \infty} \int_{1}^{b} f (x) d x$ , and the integral converges if that limit exists. The behaviour depends delicately on the integrand. For $1 / x^{2}$ the partial integral is $1 - 1 / b$ , which converges to $1$ . For $1 / x$ it is $\ln b$ , which grows without bound, so that integral diverges. The tail of $1 / x$ is too fat to have a finite total, which is the same fact as the divergence of the harmonic series in the previous course.

Now the escape problem. The force on a mass $m$ at distance $r$ from a planet of mass $M$ is $G M m / r^{2}$ , so the work needed to move it from the surface at $R$ to infinity is

W = \int_{R}^{\infty} \frac{G M m}{r^{2}} d r = G M m {[- \frac{1}{r}]}_{R}^{\infty} = \frac{G M m}{R}

finite, because the inverse square law falls off fast enough. Escape is possible with a finite energy budget, and if gravity fell off like $1 / r$ instead it would not be.

Setting the initial kinetic energy equal to that work gives $\frac{12}{m} v^{2} = G M m / R$ , so

v = \sqrt{\frac{2 G M}{R}}

independent of the escaping mass. For Earth, with $G = 6.674 \times 10^{- 11}$ m³ kg⁻¹ s⁻², $M = 5.972 \times 10^{24}$ kg and $R = 6.371 \times 10^{6}$ m, this gives $11186$ m/s, the familiar $11.2$ km/s. The number is a direct consequence of an improper integral converging.

Improper integrals also produce genuine curiosities. Revolving $y = 1 / x$ for $x \geq 1$ about the axis gives a horn whose volume is $π \int_{1}^{\infty} d x / x^{2} = π$ , finite, while its surface area involves $\int_{1}^{\infty} d x / x$ and is infinite. The solid can be filled with a finite amount of paint and its surface cannot be painted, which is a statement about the definitions rather than about paint: a real coat of paint has a thickness, and the horn narrows below it.

Averages, and the difference between two of them

The average value of $f$ over $[a, b]$ was defined two lessons ago as $\frac{1}{b - a} \int_{a}^{b} f$ , and it settles a question that confuses people about alternating current.

The average of $\sin x$ over a half cycle from $0$ to $π$ is $\frac{1}{π} \int_{0}^{π} \sin x d x = 2 / π = 0.6366$ . The root mean square value, computed in the previous lesson from the average of $\sin^{2}$ , is $1 / \sqrt{2} = 0.7071$ . Both are honest averages of the same wave and they differ by ten per cent, because one averages the voltage and the other averages the power. A meter reading the first and a meter reading the second will disagree, and which one is correct depends entirely on what the number is for.

Centre of mass

A rod along $[a, b]$ with linear density $ρ (x)$ has mass $\int_{a}^{b} ρ (x) d x$ , since the piece at $x$ has mass $ρ (x) d x$ . Its centre of mass is the point where a support would balance it, and balancing means the total moment about that point vanishes:

\overline{x} = \frac{\int_{a}^{b} x ρ (x) d x}{\int_{a}^{b} ρ (x) d x}

which is a weighted average of position, weighted by mass. For a rod of length $2$ m with density $ρ (x) = 1 + x$ kg/m, the mass is $\int_{0}^{2} (1 + x) d x = 2 + 2 = 4$ kg and the moment is $\int_{0}^{2} x (1 + x) d x = 2 + 8 / 3 = 4.667$ , so $\overline{x} = 1.167$ m, right of the geometric centre because the rod is denser at that end. The same construction in two and three dimensions gives centroids of plates and solids, and the same integral with $x^{2}$ in place of $x$ gives the moment of inertia, which is what rotational dynamics runs on.

Arc length, and a warning

The slicing pattern applies to length too. A short piece of curve is nearly straight, with horizontal run $d x$ and rise $f^{'} (x) d x$ , so by Pythagoras its length is $\sqrt{1 + [f^{'} (x)]^{2}} d x$ and

L = \int_{a}^{b} \sqrt{1 + [f^{'} (x)]^{2}} d x

The derivation is easy and the integrals are usually impossible. The integrand contains a square root of a sum of squares, which resists every technique of the previous lesson except in contrived cases. The perimeter of an ellipse is the standard example: it produces an integral that Legendre and others studied for decades and which has no elementary antiderivative at all. The functions invented to express it are called elliptic integrals, and they are named after this problem.

That is a fair summary of the state of the subject at this point. Setting up the integral is a skill that generalises, the answer exists whenever the integrand is continuous, and evaluating it in closed form is a matter of luck. Where luck runs out, either a numerical method is used, or the function is approximated by something easier before integrating.

That last option is more powerful than it sounds, and it is the subject of the final lesson: replacing a function by a polynomial that matches it to whatever accuracy is required, which is how physics gets usable answers out of expressions that cannot be evaluated at all.