Calculus · Chapter 8 — Further Applications of Integration
0%
Calculus · Chapter 8

Further Applications
of Integration

Beyond areas, volumes, and work, the definite integral measures a much wider range of quantities. This chapter applies the same build-and-sum-then-take-a-limit strategy to geometry, physics, engineering, economics, biology, and statistics — measuring the length of a curve, the area of a surface of revolution, the force of water against a dam, the location of a center of mass, the flow of blood through an artery, and the spread of a random variable.

Every result in this chapter comes from the same four moves: slice the quantity into many small pieces, approximate each piece, add the approximations into a Riemann sum, and pass to the integral. If you can recite those four moves for arc length, you can re-derive every formula here. Don't memorize five formulas — memorize one method and the geometry that feeds it.

8.1
Arc Length

How long is a curve? For a polygon the answer is easy — add the lengths of its straight segments. A general curve is handled by inscribing a polygon in it, summing the segment lengths, and letting the number of segments grow without bound. The arc length is the limit of those polygonal lengths, exactly mirroring how area and volume were defined.

Arc length as a limit of polygons

Approximate a curve by a polygon whose vertices lie on the curve, then refine. As the segments shrink and multiply, the total polygonal length approaches a fixed value, which we define to be the length of the curve. This is the same circle-from-inscribed-polygons idea that gives a circle's circumference.

Smooth curve

A function is called smooth when its derivative is continuous. Geometrically, a small change in the input produces a small change in the slope, so the curve has no corners or sudden jumps in direction. Smoothness is exactly the condition that converts the limit definition into a clean integral formula.

Length as a limit of inscribed polygons
\[ L=\lim_{n\to\infty}\sum_{i=1}^{n}\bigl|P_{i-1}P_i\bigr| \]
The Arc Length Formula

If \(f'\) is continuous on \([a,b]\), then the length of the curve \(y=f(x)\), \(a\le x\le b\), is

\[ L=\int_a^b \sqrt{1+\bigl[f'(x)\bigr]^2}\,dx. \]

The derivation applies the Mean Value Theorem to each subinterval, replacing each segment length \(\sqrt{\Delta x^2+\Delta y^2}\) with \(\sqrt{1+[f'(x_i^*)]^2}\,\Delta x\); the resulting Riemann sum converges to the integral because the integrand is continuous.

Leibniz form (curve given as y of x)
\[ L=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\;dx \]
Curve given as x of y
\[ L=\int_c^d \sqrt{1+\left(\frac{dx}{dy}\right)^{2}}\;dy \]
Key idea

Choose the integration variable that makes \(f'\) easy. If the curve is a tidy function of \(x\), integrate in \(x\); if it is tidier as \(x=g(y)\), integrate in \(y\). The two formulas describe the same length.

x y P₀ P₁ P₂ Pᵢ Pₙ y = f(x)
Figure 8.1.1An inscribed polygon (steel) approximates the smooth curve (red). As the number of segments grows, the total polygonal length approaches the true arc length.
EXAMPLE 1 Length of a semicubical parabola
Find the length of the arc of \(y^2=x^3\) between \((1,1)\) and \((4,8)\).
Solution
On the upper branch \(y=x^{3/2}\), so \(\dfrac{dy}{dx}=\tfrac{3}{2}x^{1/2}\).
Apply the formula: \(\displaystyle L=\int_1^4\sqrt{1+\tfrac{9}{4}x}\;dx.\)
Substitute \(u=1+\tfrac{9}{4}x\), so \(du=\tfrac{9}{4}dx\); limits become \(u=\tfrac{13}{4}\) to \(u=10\).
\(\displaystyle L=\frac{4}{9}\cdot\frac{2}{3}\Bigl[u^{3/2}\Bigr]_{13/4}^{10}=\frac{1}{27}\bigl(80\sqrt{10}-13\sqrt{13}\bigr).\)
\( L=\tfrac{1}{27}\bigl(80\sqrt{10}-13\sqrt{13}\bigr)\approx 7.6337.\) (Slightly longer than the straight chord \(\sqrt{58}\approx 7.6158\), as expected.)
EXAMPLE 2 An arc that needs a trig substitution
Find the length of the arc of the parabola \(y^2=x\) from \((0,0)\) to \((1,1)\).
Solution
Treat \(x\) as a function of \(y\): \(x=y^2\), so \(\dfrac{dx}{dy}=2y\).
\(\displaystyle L=\int_0^1\sqrt{1+4y^2}\;dy.\)
Let \(2y=\tan\theta\); this turns the integrand into \(\tfrac{1}{2}\sec^3\theta\), a standard secant-cubed integral.
Evaluating gives \(\displaystyle L=\frac{\sqrt5}{2}+\frac{\ln\bigl(\sqrt5+2\bigr)}{4}.\)
\( L=\dfrac{\sqrt5}{2}+\dfrac{\ln(\sqrt5+2)}{4}\approx 1.4789.\)
EXAMPLE 3 When the integral resists — estimate it
Set up the arc length of the hyperbola \(xy=1\) from \((1,1)\) to \(\bigl(2,\tfrac12\bigr)\), then estimate it with Simpson's Rule, \(n=10\).
Solution
With \(y=1/x\), \(\dfrac{dy}{dx}=-\dfrac{1}{x^2}\), so the length is \(\displaystyle L=\int_1^2\sqrt{1+\frac{1}{x^4}}\;dx=\int_1^2\frac{\sqrt{x^4+1}}{x^2}\,dx.\)
The square root has no elementary antiderivative, so apply a numerical rule. With \(\Delta x=0.1\), Simpson's Rule samples the integrand at \(x=1,1.1,\dots,2\).
\( L\approx 1.1321\) — accurate to four decimal places.

The arc length function

Fixing a starting point and letting the endpoint vary turns arc length into a function of position along the curve.

Arc length function

For a smooth curve \(y=f(x)\), the distance measured along the curve from a fixed initial point \(a\) to a variable point \(x\) defines \(s(x)\). It is monotonically increasing — moving along the curve always adds length.

Arc length function
\[ s(x)=\int_a^x \sqrt{1+\bigl[f'(t)\bigr]^2}\;dt \]
Rate of change (Fundamental Theorem of Calculus)
\[ \frac{ds}{dx}=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\;\ge 1 \]
Differential of arc length (symmetric form)
\[ ds=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx \qquad\Longleftrightarrow\qquad (ds)^2=(dx)^2+(dy)^2 \]
Why ds ≥ dx

The arc length differential equals \(dx\) only where the slope is zero; everywhere else the curve climbs or falls, so a step of \(dx\) horizontally always traces more than \(dx\) of curve. The symmetric form \((ds)^2=(dx)^2+(dy)^2\) is just the Pythagorean theorem on an infinitesimal triangle and is the mnemonic for both arc length formulas.

EXAMPLE 4 Building an arc length function
Find the arc length function for \(y=x^2-\tfrac{1}{8}\ln x\), taking \((1,1)\) as the starting point.
Solution
\(f'(x)=2x-\dfrac{1}{8x}\). The expression \(1+[f'(x)]^2\) collapses to the perfect square \(\Bigl(2x+\dfrac{1}{8x}\Bigr)^2\).
So \(\sqrt{1+[f'(x)]^2}=2x+\dfrac{1}{8x}\), and integrating from \(1\) to \(x\):
\(\displaystyle s(x)=\int_1^x\Bigl(2t+\tfrac{1}{8t}\Bigr)dt=x^2-1+\tfrac{1}{8}\ln x.\)
\( s(x)=x^2-1+\tfrac{1}{8}\ln x.\) For instance, from \((1,1)\) to \((3,f(3))\): \(s(3)=8+\tfrac{1}{8}\ln 3\approx 8.1373.\)

The "perfect square under the root" in Example 4 is no accident — practice arc-length problems are reverse-engineered so that \(1+[f']^2\) factors cleanly. On a real part profile or cam contour it almost never will, and you go straight to Simpson's Rule (Example 3). In practice I tell students: try the formula once, and if the radical doesn't simplify in two lines, stop and integrate numerically. Don't burn an exam hour hunting for a substitution that the problem may not have.

8.2
Area of a Surface of Revolution

Rotate a curve about a line and it sweeps out a surface — the outer skin of a solid of revolution. Its area should equal the amount of paint needed to coat a flat region of the same size. Building from the lateral areas of cylinders and cones, the same slice-approximate-sum-limit strategy yields a clean integral.

Surface of revolution

The surface formed when a plane curve is rotated about an axis. It is the lateral boundary of the corresponding solid of revolution and is approximated by bands, each a frustum (a sliced-off cone) generated by rotating one short segment of the curve.

Lateral area of a cone (slant height ℓ, base radius r)
\[ A=\pi r\ell \]
Area of a frustum band (average radius r, slant height ℓ)
\[ A=2\pi r\,\ell \]

From band to integral

Each band has circumference \(2\pi(\text{radius})\) and width equal to the slant length \(ds\) of the curve segment. Summing \(2\pi(\text{radius})\,ds\) over all bands and taking the limit gives the surface area. The radius is just the distance from the curve to the axis of rotation.

Rotation about the x-axis (y = f(x))
\[ S=\int_a^b 2\pi f(x)\sqrt{1+\bigl[f'(x)\bigr]^2}\;dx \]
Rotation about the y-axis (x = g(y))
\[ S=\int_c^d 2\pi g(y)\sqrt{1+\bigl[g'(y)\bigr]^2}\;dy \]
Symbolic forms using the arc length differential ds
\[ S=\int 2\pi y\;ds \quad(\text{about } x\text{-axis}) \qquad S=\int 2\pi x\;ds \quad(\text{about } y\text{-axis}) \]
The one rule to remember

Surface area \(=\int 2\pi(\text{radius})\,ds\). The factor \(2\pi(\text{radius})\) is the circumference of the circle each point traces as it spins; \(ds\) is the slant length it carries. Get the radius right (distance to the axis) and pick \(ds\) in \(x\) or \(y\) — everything else follows.

x ds y (radius) y = f(x)
Figure 8.2.1A single band of the surface. Its circumference is \(2\pi y\) and its slant width is \(ds\), so its area is \(2\pi y\,ds\); summing over all bands gives the total surface area.
EXAMPLE 1 A zone of a sphere
The arc \(y=\sqrt{4-x^2}\), \(-1\le x\le 1\), part of a circle of radius 2, is rotated about the \(x\)-axis. Find the surface area.
Solution
\(\dfrac{dy}{dx}=\dfrac{-x}{\sqrt{4-x^2}}\), so \(1+\Bigl(\dfrac{dy}{dx}\Bigr)^2=\dfrac{4}{4-x^2}.\)
\(\displaystyle S=\int_{-1}^{1}2\pi\sqrt{4-x^2}\cdot\frac{2}{\sqrt{4-x^2}}\,dx=4\pi\int_{-1}^{1}dx.\)
The radical cancels — a hallmark of spherical zones.
\( S=4\pi(2)=8\pi.\)
EXAMPLE 2 Rotating a parabola about the y-axis
The arc of \(y=x^2\) from \((1,1)\) to \((2,4)\) is rotated about the \(y\)-axis. Find the surface area.
Solution
Use \(S=\int 2\pi x\,ds\) with \(\dfrac{dy}{dx}=2x\): \(\displaystyle S=\int_1^2 2\pi x\sqrt{1+4x^2}\,dx.\)
Substitute \(u=1+4x^2\), \(du=8x\,dx\); limits \(u=5\) to \(u=17\).
\(\displaystyle S=\frac{\pi}{4}\int_5^{17}\sqrt{u}\,du=\frac{\pi}{4}\cdot\frac{2}{3}\Bigl[u^{3/2}\Bigr]_5^{17}.\)
\( S=\dfrac{\pi}{6}\bigl(17\sqrt{17}-5\sqrt5\bigr)\approx 30.85.\)
EXAMPLE 3 An exponential profile
Find the surface area generated by rotating \(y=e^x\), \(0\le x\le 1\), about the \(x\)-axis.
Solution
\(\dfrac{dy}{dx}=e^x\), so \(\displaystyle S=\int_0^1 2\pi e^x\sqrt{1+e^{2x}}\,dx.\)
Substitute \(u=e^x\): \(\displaystyle S=2\pi\int_1^{e}\sqrt{1+u^2}\,du\), a secant-cubed integral after \(u=\tan\theta\).
\( S=\pi\Bigl[e\sqrt{1+e^2}+\ln\bigl(e+\sqrt{1+e^2}\bigr)-\sqrt2-\ln(\sqrt2+1)\Bigr].\)

Watch the radius like a hawk — it is the single most common place students lose marks. Rotating about the \(x\)-axis, the radius is \(y\); about the \(y\)-axis, it is \(x\). If the problem rotates about a shifted line such as \(y=c\), the radius becomes the distance \(|f(x)-c|\), not \(f(x)\) itself. Sketch the band and physically measure the gap from the curve to the spin axis before you write the \(2\pi(\text{radius})\) factor.

8.3
Applications to Physics & Engineering

Two classic engineering problems yield to integration: the hydrostatic force a fluid exerts on a submerged wall, and the center of mass (centroid) of a thin plate. The recurring difficulty is that pressure is not constant — it grows with depth — so we slice the wall into thin horizontal strips on which the pressure is nearly uniform.

Hydrostatic force and pressure

Hydrostatic pressure

Pressure is force per unit area. At depth \(d\) in a fluid the weight of the column above produces a pressure that is the same in every direction (a diver feels equal pressure on nose and both ears). It depends only on depth and fluid density, not on orientation.

Pressure at depth d (ρ = mass density, g = gravity)
\[ P=\rho g\,d=\delta d \]
Force on a horizontal plate of area A
\[ F=\rho g A d \]
Units

In SI, pressure is the pascal, \(1\text{ Pa}=1\text{ N/m}^2\); water has \(\rho=1000\text{ kg/m}^3\). In US customary units engineers use the weight density \(\delta=\rho g\) directly — for water \(\delta=62.5\text{ lb/ft}^3\) — and write \(P=\delta d\), skipping the explicit \(g\).

Force on a vertical wall

Because pressure rises with depth, a vertical wall cannot be treated as one block. Slice it into thin horizontal strips of width equal to the wall's local width and height \(\Delta x\); on each strip the pressure is nearly constant, so its force is \(\text{pressure}\times\text{area}\). Summing and taking the limit gives the total force as an integral over depth.

Hydrostatic force on a vertical plate (width w(x) at depth x)
\[ F=\int_a^b \rho g\,x\,w(x)\;dx \]
EXAMPLE 1 Force on a trapezoidal dam
A dam is a trapezoid 20 m tall, 50 m wide at the top and 30 m at the bottom. The water level sits 4 m below the top. Find the hydrostatic force.
Solution
Put the origin at the water surface, \(x\) measured downward; the water depth spans \(0\le x\le 16\).
By similar triangles the strip width is \(w=2(15+a)=46-x\) at depth \(x\).
Strip force \(\approx \rho g\,x\,(46-x)\,\Delta x\); summing gives \(\displaystyle F=\int_0^{16}1000\,g\,x(46-x)\,dx.\)
\(\displaystyle F=9800\Bigl[23x^2-\tfrac{x^3}{3}\Bigr]_0^{16}.\)
\( F\approx 4.43\times 10^{7}\text{ N}.\)
EXAMPLE 2 End of a submerged drum
Find the hydrostatic force on one circular end (radius 3 ft) of a drum submerged with its center 7 ft below the surface (water 10 ft deep above the bottom of a 10-ft setup).
Solution
Center the circle at the origin: \(x^2+y^2=9\). A horizontal strip at height \(y\) has width \(2\sqrt{9-y^2}\) and lies at depth \(7-y\).
\(\displaystyle F=\int_{-3}^{3}62.5\,(7-y)\,2\sqrt{9-y^2}\,dy.\)
Split into two integrals: the \(y\sqrt{9-y^2}\) term is odd and integrates to 0; the constant term gives \(7\) times the area of a semicircular disk of radius 3.
\(\displaystyle F=875\int_{-3}^{3}\sqrt{9-y^2}\,dy=875\cdot\tfrac{9\pi}{2}.\)
\( F=\dfrac{7875\pi}{2}\approx 12{,}370\text{ lb}.\)

Moments and centers of mass

Center of mass

The single point at which a body balances. For a system of point masses it is the mass-weighted average of their positions; for a thin plate it is the point where the plate would balance on a pin.

Moment

The product of a mass and its distance from an axis — a measure of the tendency to rotate about that axis. The moment of a whole system about an axis is the sum of the individual moments, and the center of mass is the total moment divided by the total mass.

Center of mass of point masses on a line
\[ \bar{x}=\frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i}=\frac{M}{m} \]
Moments of a planar system
\[ M_y=\sum_{i=1}^{n} m_i x_i,\qquad M_x=\sum_{i=1}^{n} m_i y_i,\qquad \bar{x}=\frac{M_y}{m},\ \ \bar{y}=\frac{M_x}{m} \]

Symmetry principle

If a region is symmetric about a line, its centroid lies on that line. This often eliminates one of the two integrals — for any left–right symmetric plate, \(\bar{x}\) is read straight off the axis of symmetry.

Centroid of a region under y = f(x)
\[ \bar{x}=\frac{1}{A}\int_a^b x\,f(x)\,dx,\qquad \bar{y}=\frac{1}{A}\int_a^b \tfrac{1}{2}\bigl[f(x)\bigr]^2\,dx \]
Centroid of a region between f(x) and g(x), f ≥ g
\[ \bar{x}=\frac{1}{A}\int_a^b x\bigl[f(x)-g(x)\bigr]dx,\quad \bar{y}=\frac{1}{A}\int_a^b \tfrac{1}{2}\Bigl[f(x)^2-g(x)^2\Bigr]dx \]
Density cancels

For a uniform plate the density appears in both the moment and the mass, so it divides out. The centroid is a purely geometric property of the region — its location does not depend on what the plate is made of.

EXAMPLE 3 Center of mass of three point masses
Masses 3, 4, 8 sit at \((-1,1)\), \((2,-1)\), \((3,2)\). Find the moments and center of mass.
Solution
\(M_y=3(-1)+4(2)+8(3)=29\); \(M_x=3(1)+4(-1)+8(2)=15\).
Total mass \(m=3+4+8=15\).
\(\bar{x}=\dfrac{29}{15}\), \(\bar{y}=\dfrac{15}{15}=1\).
Center of mass \(\bigl(\tfrac{29}{15},\,1\bigr)\).
EXAMPLE 4 Centroid of a semicircular plate
Find the center of mass of a semicircular plate of radius \(r\).
Solution
Place the flat edge on the \(x\)-axis. By symmetry \(\bar{x}=0\).
Area \(A=\tfrac{1}{2}\pi r^2\). Then \(\displaystyle \bar{y}=\frac{1}{A}\int_{-r}^{r}\tfrac{1}{2}\bigl(\sqrt{r^2-x^2}\bigr)^2 dx=\frac{1}{\pi r^2}\int_{-r}^{r}(r^2-x^2)\,dx.\)
Evaluating: \(\displaystyle \bar{y}=\frac{1}{\pi r^2}\cdot\frac{4r^3}{3}.\)
Centroid \(\Bigl(0,\,\dfrac{4r}{3\pi}\Bigr)\).
EXAMPLE 6 Centroid between a line and a parabola
Find the centroid of the region bounded by \(y=x\) and \(y=x^2\).
Solution
The curves meet at \(x=0\) and \(x=1\), with \(x\ge x^2\) between them. Area \(\displaystyle A=\int_0^1 (x-x^2)\,dx=\tfrac{1}{6}.\)
\(\displaystyle \bar{x}=6\int_0^1 x(x-x^2)\,dx=6\Bigl[\tfrac{x^3}{3}-\tfrac{x^4}{4}\Bigr]_0^1=\tfrac{1}{2}.\)
\(\displaystyle \bar{y}=6\int_0^1 \tfrac{1}{2}(x^2-x^4)\,dx=\tfrac{2}{5}.\)
Centroid \(\bigl(\tfrac{1}{2},\,\tfrac{2}{5}\bigr)\).
Theorem of Pappus

Let a plane region \(R\) lie entirely on one side of a line \(\ell\) in its plane. When \(R\) is rotated about \(\ell\), the volume of the resulting solid equals the area of \(R\) times the distance travelled by its centroid:

\[ V=A\,d, \qquad d=2\pi\,(\text{centroid distance to }\ell). \]
EXAMPLE 7 Volume of a torus by Pappus
A circle of radius \(r\) is rotated about a line a distance \(R\;(R>r)\) from its center. Find the volume of the torus.
Solution
The disk has area \(A=\pi r^2\). Its centroid is the center, which travels \(d=2\pi R\) in one revolution.
By Pappus, \(V=A\,d=(\pi r^2)(2\pi R)\).
\( V=2\pi^2 r^2 R.\)

Two engineering habits worth drilling. First, hydrostatic problems are about depth, not height — always set up your variable so it increases downward from the free surface, then read each strip's depth directly. Second, the Theorem of Pappus is a genuine shortcut on the job: if you already know a cross-section's area and centroid (standard for rolled steel shapes, weld beads, machined fillets), you get the swept volume in one multiplication, with no integral at all. I expect you to reach for Pappus before setting up shells whenever the centroid is already in a table.

8.4
Applications to Economics & Biology

The integral measures accumulated quantities far outside geometry. Here it captures the money consumers save versus what they would have paid (consumer surplus), the volume of blood crossing an artery per unit time (Poiseuille's Law), and the rate the heart pumps blood (cardiac output).

Consumer surplus

Demand function

The price \(p(x)\) a seller must set to sell \(x\) units. It typically decreases — moving more units requires a lower price — so its graph, the demand curve, slopes downward.

Consumer surplus

The total amount buyers save by paying a single market price \(P\) rather than the higher prices many of them would have been willing to pay. Geometrically it is the area between the demand curve and the horizontal line \(p=P\), from \(0\) to the quantity sold \(X\).

Consumer surplus at sales level X, price P = p(X)
\[ \text{CS}=\int_0^X \bigl[p(x)-P\bigr]\,dx \]
EXAMPLE 1 Surplus for a quadratic demand curve
Demand is \(p=1200-0.2x-0.0001x^2\) dollars. Find the consumer surplus when 500 units are sold.
Solution
Market price: \(P=p(500)=1200-0.2(500)-0.0001(500)^2=1075.\)
\(\displaystyle \text{CS}=\int_0^{500}\bigl(1200-0.2x-0.0001x^2-1075\bigr)dx=\int_0^{500}\bigl(125-0.2x-0.0001x^2\bigr)dx.\)
\(\displaystyle =\Bigl[125x-0.1x^2-\tfrac{0.0001}{3}x^3\Bigr]_0^{500}.\)
\( \text{CS}\approx \$33{,}333.33.\)

Blood flow

Law of laminar flow

Blood velocity in a vessel is largest on the central axis and falls to zero at the wall, varying with the square of the distance \(r\) from the center. The driving pressure difference and the blood's viscosity set the scale of the velocity profile.

Velocity profile (radius R, length l, pressure difference P, viscosity η)
\[ v(r)=\frac{P}{4\eta l}\bigl(R^2-r^2\bigr) \]

Flux (rate of flow)

To get the volume of blood crossing a cross-section per unit time, sum the flow through thin concentric rings: ring area \(2\pi r\,\Delta r\) times local velocity. The limit is an integral, and it shows flow scales with the fourth power of the radius — Poiseuille's Law.

Poiseuille's Law
\[ F=\int_0^R 2\pi r\,v(r)\,dr=\frac{\pi P R^{4}}{8\eta l} \]
Why the fourth power matters

Because flow depends on \(R^4\), narrowing an artery has a dramatic effect. Reduce the radius to three-quarters of normal and, to keep the same flow, the required pressure more than triples — the quantitative root of high blood pressure from constricted arteries.

Cardiac output

Cardiac output & the dye dilution method

Cardiac output is the volume of blood the heart pumps into the aorta per unit time. A known amount of dye is injected, and its changing concentration downstream is recorded over time. The flow rate is the injected amount divided by the time-integral of the concentration.

Cardiac output (A = amount of dye, c(t) = concentration)
\[ F=\frac{A}{\displaystyle\int_0^T c(t)\,dt} \]
EXAMPLE 2 Estimating cardiac output
A 5-mg bolus of dye is injected; concentration is measured each second for 10 s. Estimate the cardiac output.
Solution
t (s)012345678910
c(t)00.42.86.59.88.96.14.02.31.10
Approximate \(\displaystyle\int_0^{10}c(t)\,dt\) by Simpson's Rule with \(\Delta t=1\): \(\approx 41.87.\)
\(\displaystyle F=\frac{A}{\int_0^{10}c(t)\,dt}=\frac{5}{41.87}.\)
\( F\approx 0.12\text{ L/s}\approx 7.2\text{ L/min}.\)

Notice the shared shape of all three results: a measured quantity divided or accumulated against an integral. Consumer surplus, flux, and cardiac output look unrelated, but each is "integrate the density of something over its domain." When you meet a new applied problem, the first question is always the same — what small piece am I summing, and over what variable? Answer that and the integral writes itself. The \(R^4\) dependence in Poiseuille's Law is also worth carrying into any pipe- or duct-flow course; it is the same physics.

8.5
Probability

When a measured quantity ranges over a continuum — a battery's lifetime, an adult's height, the wait time on a support call — its behavior is captured by a probability density function. Probabilities become areas, and the integral becomes the central tool for describing random outcomes.

Continuous random variable

A quantity whose values fill an interval of real numbers rather than a discrete list. We ask not for the probability of an exact value but for the probability that the value falls within a range.

Probability density function (pdf)

A function \(f\) such that the probability of landing between \(a\) and \(b\) equals the area under \(f\) from \(a\) to \(b\). It is never negative, and the total area beneath it is exactly 1, reflecting that some outcome is certain.

Probability as area
\[ P(a\le X\le b)=\int_a^b f(x)\,dx \]
Defining conditions of a pdf
\[ f(x)\ge 0 \quad\text{for all }x, \qquad \int_{-\infty}^{\infty} f(x)\,dx=1 \]
EXAMPLE 1 Verifying a density and finding a probability
Let \(f(x)=0.006\,x(10-x)\) for \(0\le x\le 10\), and \(0\) elsewhere. Verify \(f\) is a pdf and find \(P(4\le X\le 8)\).
Solution
On \([0,10]\), \(x(10-x)\ge 0\), so \(f\ge 0\). Total area: \(\displaystyle\int_0^{10}0.006(10x-x^2)\,dx=0.006\bigl(500-\tfrac{1000}{3}\bigr)=1.\) ✓
\(\displaystyle P(4\le X\le 8)=0.006\int_4^8(10x-x^2)\,dx=0.006\Bigl[5x^2-\tfrac{x^3}{3}\Bigr]_4^8.\)
\( P(4\le X\le 8)=0.544.\)

Exponential density (waiting / failure times)

Waiting times and equipment lifetimes are commonly modelled by an exponentially decreasing density: zero before time 0, then \(c\,e^{-ct}\) afterward. The constant \(c\) is fixed by requiring the total area to equal 1.

Exponential density function
\[ f(t)=\begin{cases}0 & t<0\\[2pt] c\,e^{-ct} & t\ge 0\end{cases} \]

Mean and median

Mean (expected value)

The long-run average value of the random variable, found by integrating \(x\,f(x)\). It is also the balance point of the density: a thin plate cut to the shape under \(f\) balances on the vertical line through the mean — exactly the centroid idea from §8.3.

Mean of a probability density function
\[ \mu=\int_{-\infty}^{\infty} x\,f(x)\,dx \]
Mean of the exponential density
\[ \mu=\frac{1}{c}\quad\Longrightarrow\quad f(t)=\frac{1}{\mu}e^{-t/\mu}\ \ (t\ge 0) \]

Median

The value \(m\) that splits the distribution into two equal halves — half the area lies on each side. For skewed distributions like the exponential, the median is smaller than the mean, because a few very long waits drag the average upward.

Median m
\[ \int_{m}^{\infty} f(x)\,dx=\frac{1}{2} \]
EXAMPLE 4 Waiting on hold
The average wait for a call to be answered is 5 minutes (exponential). Find (a) the probability of being answered within the first minute, and (b) the probability of waiting more than 5 minutes.
Solution
Mean \(\mu=5\) gives \(f(t)=0.2\,e^{-t/5}\) for \(t\ge 0\).
(a) \(\displaystyle P(0\le T\le 1)=\int_0^1 0.2e^{-t/5}\,dt=1-e^{-1/5}\approx 0.181.\)
(b) \(\displaystyle P(T>5)=\int_5^{\infty}0.2e^{-t/5}\,dt=e^{-1}\approx 0.368.\)
(a) ≈ 18% answered in the first minute. (b) ≈ 37% wait longer than 5 minutes — even though the mean is 5, since long waits skew the average up.

Normal distributions

Normal distribution

The bell-shaped family modelling test scores, heights, weights, and rainfall. It is governed by two parameters: the mean \(\mu\) (the center) and the standard deviation \(\sigma\) (the spread). Small \(\sigma\) clusters values tightly about the mean; large \(\sigma\) spreads them out.

Normal density function
\[ f(x)=\frac{1}{\sigma\sqrt{2\pi}}\;e^{-(x-\mu)^2/(2\sigma^2)} \]
No elementary antiderivative

The normal density cannot be integrated in closed form, so normal probabilities are always computed numerically — by calculator, Simpson's Rule, or tables. The familiar landmark: about 68% of the population lies within one standard deviation of the mean.

EXAMPLE 5 IQ scores
IQ scores are normal with \(\mu=100\), \(\sigma=15\). Find (a) the percentage scoring between 85 and 115, and (b) the percentage scoring above 140.
Solution
(a) 85 and 115 are exactly one standard deviation either side of 100, so \(\displaystyle P(85\le X\le 115)=\int_{85}^{115}\frac{1}{15\sqrt{2\pi}}e^{-(x-100)^2/450}\,dx\approx 0.68.\)
(b) \(\displaystyle P(X>140)\approx\int_{140}^{200}\frac{1}{15\sqrt{2\pi}}e^{-(x-100)^2/450}\,dx\approx 0.0038\) (evaluated numerically).
(a) ≈ 68% score between 85 and 115. (b) ≈ 0.4% score above 140.

Two takeaways I want you to keep past the exam. First, the mean–median gap in the exponential (Example 4) is the whole reason "average wait time" can be misleading in reliability and queueing work — report the median too when a distribution is skewed. Second, almost nothing in real engineering statistics integrates in closed form; the normal is the prime example. Get comfortable with Simpson's Rule and your calculator's numerical integrator now, because that is how these probabilities actually get computed on the job.

This topic touches sensitive applications (lifetimes, failure, medical data). Treated here purely as mathematics — if any of it raises a personal concern, talk it through with the right professional rather than relying only on a course.