Calculus · Chapter 6 Applications of Integration
Calculus · Chapter 6

Applications of Integration

The definite integral is more than an antiderivative bookkeeping tool — it is a machine for summing infinitely many infinitesimal contributions. Every application in this chapter follows one recipe: cut a quantity into thin slices, approximate each slice by a simple shape, sum with a Riemann sum, and pass to the limit to recover an integral.

Read 6.1–6.3 as one continuous skill, not three. The whole game is choosing the slicing direction that makes the height (or radius) easy to write. If solving for x in terms of y is ugly, slice the other way or switch to shells before you commit to an integral.

For 6.4 and 6.5, watch your units and your reference point. Most "wrong" work answers come from measuring distance from the wrong end, not from a calculus error.

6.1

Areas Between Curves

In Chapter 5, area meant the region under a single curve. Here we measure the region trapped between two curves by treating the gap between them as the height of a thin approximating rectangle.

Region between two curves

Given continuous functions with \(f(x)\ge g(x)\) on \([a,b]\), the region \(S\) lies above \(y=g(x)\), below \(y=f(x)\), and between the vertical lines \(x=a\) and \(x=b\). Each thin vertical strip has width \(\Delta x\) and height equal to the vertical gap \(f(x_i^*)-g(x_i^*)\).

From Riemann sum to integral

Summing the rectangle areas gives \(\sum_{i=1}^{n}\bigl[f(x_i^*)-g(x_i^*)\bigr]\Delta x\), an approximation that sharpens as \(n\to\infty\). The limit is exactly the definite integral of the difference of the two functions.

Area between two curves (vertical strips)
\[ A=\int_{a}^{b}\bigl[\,f(x)-g(x)\,\bigr]\,dx,\qquad f(x)\ge g(x)\text{ on }[a,b] \]
a b y=f(x) y=g(x) Δx
Figure 6.1 A typical vertical strip of width \(\Delta x\) spans the vertical gap between the upper curve \(f\) and the lower curve \(g\). (Original schematic.)
EXAMPLE 1 Strip with fixed vertical sides
Find the area bounded above by \(y=e^{x}\), below by \(y=x\), and on the sides by \(x=0\) and \(x=1\).
Solution
The top curve is \(f(x)=e^{x}\) and the bottom curve is \(g(x)=x\); the side limits are given as \(a=0\), \(b=1\).
\[ A=\int_{0}^{1}\bigl(e^{x}-x\bigr)\,dx=\Bigl[\,e^{x}-\tfrac{1}{2}x^{2}\,\Bigr]_{0}^{1} \]
\[ A=e-\tfrac{1}{2}-1=e-1.5\approx 1.18 \]
EXAMPLE 2 Sides found from intersections
Find the area enclosed by the parabolas \(y=x^{2}\) and \(y=2x-x^{2}\).
Solution
Set the curves equal: \(2x-x^{2}=x^{2}\Rightarrow 2x^{2}-2x=0\Rightarrow x=0\) or \(x=1\). The curves meet at \((0,0)\) and \((1,1)\).
On this interval the top is \(y_T=2x-x^{2}\) and the bottom is \(y_B=x^{2}\), so the strip height is \(y_T-y_B=2x-2x^{2}\).
\[ A=\int_{0}^{1}\bigl(2x-2x^{2}\bigr)\,dx=2\int_{0}^{1}\bigl(x-x^{2}\bigr)\,dx=2\Bigl[\tfrac{x^{2}}{2}-\tfrac{x^{3}}{3}\Bigr]_{0}^{1} \]
\[ A=\tfrac{1}{3} \]
Setup heuristic: sketch first, label the top curve \(y_T\) and bottom curve \(y_B\), then the area is simply \(\displaystyle\int (y_T-y_B)\,dx\) — the integral just adds up all the strip areas.
EXAMPLE 3 Intersections found numerically
Approximate the area bounded by \(y=\dfrac{x}{\sqrt{x^{2}+1}}\) and \(y=x^{4}-x\).
Solution
The intersection equation cannot be solved in closed form. A graph shows one intersection at the origin and another near \(x\approx 1.18\).
\[ A\approx\int_{0}^{1.18}\!\left(\frac{x}{\sqrt{x^{2}+1}}-x^{4}+x\right)dx \]
Substituting \(u=x^{2}+1\) for the first term and integrating the rest yields the numerical value.
\[ A\approx 0.785 \]
EXAMPLE 4 Area as distance between two cars
Two cars A and B start side by side. Given their velocity curves, what does the area between the curves mean, and estimate it with the Midpoint Rule over the first 16 seconds.
Solution
Area under a velocity curve is distance traveled, so the area between the curves is the gap in distance — how far ahead car A is after 16 s.
With \(n=4\) and \(\Delta t=4\), evaluate \(v_A-v_B\) at the midpoints \(t=2,6,10,14\), giving values \(13,23,28,29\).
\[ \int_{0}^{16}(v_A-v_B)\,dt\approx 4\,(13+23+28+29) \]
\[ \approx 372\ \text{ft} \]

When the upper curve switches

If neither curve stays on top across the whole interval, split the region where the curves cross and integrate the absolute gap. In practice: integrate \(\bigl|f(x)-g(x)\bigr|\), breaking the integral at each intersection.

Area when the top curve changes
\[ A=\int_{a}^{b}\bigl|\,f(x)-g(x)\,\bigr|\,dx \]
EXAMPLE 5 Splitting at a crossing point
Find the area of the region bounded by \(y=\sin x\), \(y=\cos x\), \(x=0\), and \(x=\pi/2\).
Solution
The curves cross where \(\sin x=\cos x\), i.e. \(x=\pi/4\). On \([0,\pi/4]\), \(\cos x\ge\sin x\); on \([\pi/4,\pi/2]\), \(\sin x\ge\cos x\).
\[ A=\int_{0}^{\pi/4}(\cos x-\sin x)\,dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,dx \]
By symmetry about \(x=\pi/4\), this equals \(2\displaystyle\int_{0}^{\pi/4}(\cos x-\sin x)\,dx\).
\[ A=2\sqrt{2}-2 \]

Integrating with respect to y

Some regions are far simpler when \(x\) is treated as a function of \(y\). Horizontal strips of height \(\Delta y\) span from a left boundary \(x_L\) to a right boundary \(x_R\).

Area between two curves (horizontal strips)
\[ A=\int_{c}^{d}\bigl[\,x_R-x_L\,\bigr]\,dy \]
EXAMPLE 6 Slicing horizontally to avoid splitting
Find the area enclosed by the line \(y=x-1\) and the parabola \(y^{2}=2x+6\).
Solution
Intersections are \((-1,-2)\) and \((5,4)\). Solving for \(x\): the right boundary is \(x_R=y+1\) and the left boundary is \(x_L=\tfrac{1}{2}y^{2}-3\).
\[ A=\int_{-2}^{4}\Bigl[(y+1)-\bigl(\tfrac{1}{2}y^{2}-3\bigr)\Bigr]dy=\int_{-2}^{4}\Bigl(-\tfrac{1}{2}y^{2}+y+4\Bigr)dy \]
\[ A=18 \]
Integrating in \(x\) would have forced splitting the region in two — the horizontal slice is much cleaner.

Decision rule I want you to internalize: choose the slicing variable that lets you write each boundary as a single function. If both top and bottom are clean functions of \(x\), slice vertically. If left and right are clean functions of \(y\), slice horizontally. Example 6 is the poster child — vertical slicing there needs two integrals, horizontal needs one.

6.2

Volumes

To find a volume, slice the solid into thin slabs perpendicular to an axis. Each slab is approximately a cylinder whose volume is cross-sectional area times thickness. Summing and taking the limit turns the sum into an integral of the cross-sectional area.

Cylinder volume (the building block)

A cylinder is a region (the base) swept straight to a congruent region in a parallel plane. Its volume is base area times height, \(V=Ah\). Special cases: a circular cylinder gives \(V=\pi r^{2}h\), and a rectangular box gives \(V=lwh\).

Cross-sectional area function

Slice a general solid \(S\) by a plane perpendicular to the \(x\)-axis at position \(x\). The area of that slice is \(A(x)\). As \(x\) runs from \(a\) to \(b\), \(A(x)\) traces how the solid grows and shrinks.

Definition of Volume

Let \(S\) lie between \(x=a\) and \(x=b\). If the cross-sectional area perpendicular to the \(x\)-axis is \(A(x)\), with \(A\) continuous, then

\[ V=\lim_{n\to\infty}\sum_{i=1}^{n}A(x_i^{*})\,\Delta x=\int_{a}^{b}A(x)\,dx. \]

A(x) a x b
Figure 6.2 A moving cross-section of area \(A(x)\) sweeps the solid from \(a\) to \(b\); volume is the integral of that area. (Original schematic.)
EXAMPLE 1 Volume of a sphere
Show that a sphere of radius \(r\) has volume \(V=\tfrac{4}{3}\pi r^{3}\).
Solution
Center the sphere at the origin. A slice at \(x\) is a disk of radius \(\sqrt{r^{2}-x^{2}}\), so \(A(x)=\pi(r^{2}-x^{2})\).
\[ V=\int_{-r}^{r}\pi(r^{2}-x^{2})\,dx=2\pi\int_{0}^{r}(r^{2}-x^{2})\,dx \]
\[ V=2\pi\Bigl(r^{3}-\tfrac{r^{3}}{3}\Bigr)=\tfrac{4}{3}\pi r^{3} \]

Solids of revolution: disks

Rotating a region about an axis produces a solid whose cross-sections are disks. The disk radius is the distance from the axis to the curve.

Disk method
\[ A=\pi(\text{radius})^{2},\qquad V=\int_{a}^{b}\pi\,[\,R(x)\,]^{2}\,dx \]
EXAMPLE 2 Rotating about the x-axis
Rotate the region under \(y=\sqrt{x}\) from \(0\) to \(1\) about the \(x\)-axis. Find the volume.
Solution
A slice at \(x\) is a disk of radius \(\sqrt{x}\), so \(A(x)=\pi(\sqrt{x})^{2}=\pi x\).
\[ V=\int_{0}^{1}\pi x\,dx=\pi\Bigl[\tfrac{x^{2}}{2}\Bigr]_{0}^{1} \]
\[ V=\tfrac{\pi}{2} \]
EXAMPLE 3 Rotating about the y-axis
Rotate the region bounded by \(y=x^{3}\), \(y=8\), and \(x=0\) about the \(y\)-axis.
Solution
Slice perpendicular to the \(y\)-axis. At height \(y\), the disk radius is \(x=\sqrt[3]{y}\), so \(A(y)=\pi y^{2/3}\).
\[ V=\int_{0}^{8}\pi y^{2/3}\,dy=\pi\Bigl[\tfrac{3}{5}y^{5/3}\Bigr]_{0}^{8} \]
\[ V=\tfrac{96\pi}{5} \]

Solids of revolution: washers

When the region does not touch the axis, the cross-section is a washer (an annulus). Subtract the inner disk area from the outer disk area.

Washer method
\[ A=\pi\Bigl[\,(\text{outer radius})^{2}-(\text{inner radius})^{2}\,\Bigr] \]
r_out r_in
Figure 6.3 A washer cross-section: outer radius from the axis to the far curve, inner radius to the near curve. (Original schematic.)
EXAMPLE 4 Washer about the x-axis
Rotate the region enclosed by \(y=x\) and \(y=x^{2}\) about the \(x\)-axis.
Solution
The curves meet at \((0,0)\) and \((1,1)\). At position \(x\), outer radius is \(x\), inner radius is \(x^{2}\).
\[ V=\int_{0}^{1}\pi\bigl(x^{2}-x^{4}\bigr)\,dx=\pi\Bigl[\tfrac{x^{3}}{3}-\tfrac{x^{5}}{5}\Bigr]_{0}^{1} \]
\[ V=\tfrac{2\pi}{15} \]
EXAMPLE 5 Rotating about a shifted horizontal line
Rotate the region of Example 4 about the line \(y=2\).
Solution
Distances are now measured from \(y=2\). Outer radius is \(2-x^{2}\), inner radius is \(2-x\).
\[ V=\int_{0}^{1}\pi\Bigl[(2-x^{2})^{2}-(2-x)^{2}\Bigr]dx=\pi\int_{0}^{1}\bigl(x^{4}-5x^{2}+4x\bigr)\,dx \]
\[ V=\tfrac{8\pi}{15} \]
EXAMPLE 6 Rotating about a shifted vertical line
Rotate the region of Example 4 about the line \(x=-1\).
Solution
Slice horizontally. Distances are measured from \(x=-1\): outer radius \(1+\sqrt{y}\), inner radius \(1+y\).
\[ V=\pi\int_{0}^{1}\Bigl[(1+\sqrt{y})^{2}-(1+y)^{2}\Bigr]dy=\pi\int_{0}^{1}\bigl(2\sqrt{y}-y-y^{2}\bigr)\,dy \]
\[ V=\tfrac{\pi}{2} \]

Solids that are not solids of revolution

The slicing definition works for any solid with a known cross-section, not just revolutions.

EXAMPLE 7 Equilateral-triangle cross-sections
A solid has a circular base of radius 1. Cross-sections perpendicular to the base are equilateral triangles. Find the volume.
Solution
Take the base circle as \(x^{2}+y^{2}=1\). At position \(x\), the chord length (triangle base) is \(2\sqrt{1-x^{2}}\), and the equilateral height is \(\sqrt{3}\,\sqrt{1-x^{2}}\).
So \(A(x)=\tfrac12\cdot 2\sqrt{1-x^{2}}\cdot\sqrt{3}\sqrt{1-x^{2}}=\sqrt{3}\,(1-x^{2})\).
\[ V=\int_{-1}^{1}\sqrt{3}\,(1-x^{2})\,dx=2\sqrt{3}\Bigl[x-\tfrac{x^{3}}{3}\Bigr]_{0}^{1} \]
\[ V=\tfrac{4\sqrt{3}}{3} \]
EXAMPLE 8 Volume of a pyramid
Find the volume of a pyramid with square base of side \(L\) and height \(h\).
Solution
Place the vertex at the origin with the \(x\)-axis down the central axis. By similar triangles, the cross-section at \(x\) is a square of side \(s=\dfrac{Lx}{h}\).
\[ A(x)=s^{2}=\frac{L^{2}}{h^{2}}x^{2} \]
\[ V=\int_{0}^{h}\frac{L^{2}}{h^{2}}x^{2}\,dx=\frac{L^{2}}{h^{2}}\Bigl[\tfrac{x^{3}}{3}\Bigr]_{0}^{h} \]
\[ V=\tfrac{1}{3}L^{2}h \]
EXAMPLE 9 A wedge cut from a cylinder
A wedge is cut from a cylinder of radius 4 by a plane through a diameter at \(30^{\circ}\) to the base. Find the volume.
Solution
Place the \(x\)-axis along the diameter where the planes meet. A cross-section at \(x\) is a right triangle with base \(\sqrt{16-x^{2}}\) and height \(\dfrac{\sqrt{16-x^{2}}}{\sqrt{3}}\) (from \(\tan 30^{\circ}\)).
\[ A(x)=\frac{16-x^{2}}{2\sqrt{3}} \]
\[ V=\int_{-4}^{4}\frac{16-x^{2}}{2\sqrt{3}}\,dx=\frac{1}{\sqrt{3}}\int_{0}^{4}(16-x^{2})\,dx \]
\[ V=\frac{128}{3\sqrt{3}} \]

Disk vs. washer is not a separate formula to memorize — a disk is just a washer with inner radius zero. Always ask "does the region touch the axis of rotation?" If yes → disk. If a gap → washer. And when you rotate about a shifted line, write each radius as (the line) minus (the curve), not the bare curve value. That single habit kills most sign errors in Examples 5 and 6.

6.3

Volumes by Cylindrical Shells

Some rotations are awkward with washers because solving the curve for the other variable is hard. The shell method slices parallel to the axis of rotation, building the solid from nested thin cylindrical shells.

Volume of one shell

A thin cylindrical shell of average radius \(r\), height \(h\), and thickness \(\Delta r\) unrolls into a flat sheet of length \(2\pi r\) (its circumference), height \(h\), and thickness \(\Delta r\). Its volume is therefore \(2\pi r h\,\Delta r\) — circumference times height times thickness.

Volume of a cylindrical shell
\[ V=2\pi r h\,\Delta r=[\text{circumference}]\,[\text{height}]\,[\text{thickness}] \]
shell → 2πx Δx
Figure 6.4 A shell of radius \(x\) and thickness \(\Delta x\) unrolls into a sheet of length \(2\pi x\). (Original schematic.)
Shell method (rotation about the y-axis)
\[ V=\int_{a}^{b}2\pi x\,f(x)\,dx,\qquad 0\le a
EXAMPLE 1 When washers would need a cubic
Rotate the region bounded by \(y=2x^{2}-x^{3}\) and \(y=0\) about the \(y\)-axis.
Solution
A shell at \(x\) has radius \(x\), circumference \(2\pi x\), and height \(f(x)=2x^{2}-x^{3}\).
\[ V=\int_{0}^{2}2\pi x\,(2x^{2}-x^{3})\,dx=2\pi\int_{0}^{2}\bigl(2x^{3}-x^{4}\bigr)\,dx \]
\[ V=\tfrac{16\pi}{5} \]
No need to solve the cubic for \(x\) in terms of \(y\) — that is exactly what makes shells easier here.
EXAMPLE 2 Shell height as a difference
Rotate the region between \(y=x\) and \(y=x^{2}\) about the \(y\)-axis.
Solution
A shell at \(x\) has radius \(x\) and height \(x-x^{2}\) (top curve minus bottom curve).
\[ V=\int_{0}^{1}2\pi x\,(x-x^{2})\,dx=2\pi\int_{0}^{1}\bigl(x^{2}-x^{3}\bigr)\,dx \]
\[ V=\tfrac{\pi}{6} \]
EXAMPLE 3 Shells about the x-axis
Use shells to find the volume from rotating the region under \(y=\sqrt{x}\), \(0\) to \(1\), about the \(x\)-axis.
Solution
Relabel as \(x=y^{2}\). For rotation about the \(x\)-axis, a shell at height \(y\) has radius \(y\), circumference \(2\pi y\), and height \(1-y^{2}\).
\[ V=\int_{0}^{1}2\pi y\,(1-y^{2})\,dy=2\pi\int_{0}^{1}\bigl(y-y^{3}\bigr)\,dy \]
\[ V=\tfrac{\pi}{2} \]
Same answer as the disk solution in 6.2, Example 2 — here disks were actually simpler.
EXAMPLE 4 Shells about a shifted line
Rotate the region bounded by \(y=x-x^{2}\) and \(y=0\) about the line \(x=2\).
Solution
A shell at \(x\) now has radius \(2-x\) (distance to the line) and height \(x-x^{2}\).
\[ V=\int_{0}^{1}2\pi(2-x)(x-x^{2})\,dx=2\pi\int_{0}^{1}\bigl(x^{3}-3x^{2}+2x\bigr)\,dx \]
\[ V=\tfrac{\pi}{2} \]
Disks/washers vs. shells: washers slice perpendicular to the axis of rotation; shells slice parallel to it. Pick whichever makes the integrand a clean function of a single variable without solving the curve for the other variable.

Quick mental check for the shell radius when rotating about a shifted line: it is always the perpendicular distance from the line to the strip, written as a positive quantity. About \(x=2\) with strips at \(x\le 2\), that is \(2-x\); about the \(y\)-axis it is just \(x\). Get the radius right and the rest is routine integration.

6.4

Work

Work measures effort: force acting through distance. When force is constant, work is just their product. When force varies along the path, we slice the path, treat force as constant on each tiny piece, and integrate.

Force and work (constant force)

By Newton's Second Law, force is mass times acceleration, \(F=ma\). When a constant force \(F\) moves an object a distance \(d\), the work is \(W=Fd\). SI units: force in newtons (N), distance in meters, work in joules (J). US units use foot-pounds (ft-lb).

Newton's Second Law of Motion

The force on an object equals its mass times its acceleration:

\[ F=m\,\frac{d^{2}s}{dt^{2}}=ma. \]

A force of 1 N gives a 1 kg mass an acceleration of \(1\ \text{m/s}^{2}\).

Work done by a variable force
\[ W=\int_{a}^{b}f(x)\,dx \]
EXAMPLE 1 Lifting against gravity
(a) Work to lift a 1.2 kg course packet to a desk 0.7 m high (\(g=9.8\)). (b) Work to lift a 20 lb weight 6 ft.
Solution
(a) Force opposes gravity: \(F=mg=1.2\times 9.8=11.76\) N. Then \(W=Fd=11.76\times 0.7\approx 8.2\) J.
(b) Weight is already a force: \(W=Fd=20\times 6=120\) ft-lb.
In US units the pound is a force, so no multiplication by \(g\) is needed.
EXAMPLE 2 A position-dependent force
A force of \(x^{2}+2x\) pounds acts at \(x\) feet from the origin. Find the work moving from \(x=1\) to \(x=3\).
Solution
\[ W=\int_{1}^{3}(x^{2}+2x)\,dx=\Bigl[\tfrac{x^{3}}{3}+x^{2}\Bigr]_{1}^{3} \]
\[ W=\tfrac{50}{3}\approx 16.7\ \text{ft-lb} \]
Hooke's Law

The force needed to hold a spring stretched \(x\) units beyond its natural length is proportional to that stretch:

\[ f(x)=kx, \]

where \(k\) is the spring constant. Valid while the stretch \(x\) is not too large.

EXAMPLE 3 Work to stretch a spring
A 40 N force holds a spring stretched from 10 cm to 15 cm. Find the work stretching it from 15 cm to 18 cm.
Solution
A 5 cm \(=0.05\) m stretch needs 40 N, so \(0.05k=40\Rightarrow k=800\), giving \(f(x)=800x\).
Stretching from 15 cm to 18 cm means \(x\) goes from \(0.05\) to \(0.08\) m.
\[ W=\int_{0.05}^{0.08}800x\,dx=800\Bigl[\tfrac{x^{2}}{2}\Bigr]_{0.05}^{0.08} \]
\[ W=400\,(0.08^{2}-0.05^{2})=1.56\ \text{J} \]
EXAMPLE 4 Lifting a hanging cable
A 200 lb cable is 100 ft long, hanging vertically. Find the work to lift it all to the top.
Solution
The cable weighs \(2\) lb/ft. A small piece at depth \(x\) of length \(\Delta x\) weighs \(2\,\Delta x\) and must rise about \(x\) ft, so it contributes \(2x\,\Delta x\) ft-lb.
\[ W=\int_{0}^{100}2x\,dx=\Bigl[x^{2}\Bigr]_{0}^{100} \]
\[ W=10{,}000\ \text{ft-lb} \]
Different pieces travel different distances, which is why an integral is required rather than \(Fd\).
EXAMPLE 5 Pumping water from a cone
An inverted cone (height 10 m, base radius 4 m) is filled with water to 8 m. Find the work to pump all the water to the top. (Water density \(1000\ \text{kg/m}^{3}\).)
Solution
Measure depth \(x\) from the top; water occupies \(x\in[2,10]\). By similar triangles, a thin layer at depth \(x\) has radius \(r=\tfrac{2}{5}(10-x)\).
Layer volume \(\approx\pi r^{2}\,\Delta x=\tfrac{4\pi}{25}(10-x)^{2}\Delta x\); its weight (force) is \(\approx 1570\,(10-x)^{2}\Delta x\) N.
Each layer rises distance \(x\), so the work for that layer is \(\approx 1570\,x\,(10-x)^{2}\Delta x\).
\[ W=\int_{2}^{10}1570\,x\,(10-x)^{2}\,dx \]
\[ W\approx 3.4\times 10^{6}\ \text{J} \]

The pumping problems (Example 5 and its exercises) trip everyone up because two things vary: the amount of water in a layer (a volume/weight factor) and the distance it must travel. Build the layer's weight first, then multiply by the lift distance, then integrate. Decide once where your coordinate's zero sits and keep it consistent — measuring depth from the top vs. height from the bottom both work, but mixing them is the classic mistake.

6.5

Average Value of a Function

Averaging finitely many numbers is easy. To average a continuously varying quantity — like temperature over a day — we sample, average the samples, and take the limit, which produces an integral.

From discrete average to continuous average

The average of \(n\) sampled values \(f(x_1^{*}),\dots,f(x_n^{*})\) is their sum divided by \(n\). Writing \(n=(b-a)/\Delta x\) turns the average into \(\dfrac{1}{b-a}\sum f(x_i^{*})\,\Delta x\), whose limit is a definite integral.

Average value of a function
\[ f_{\text{ave}}=\frac{1}{b-a}\int_{a}^{b}f(x)\,dx \]
Geometric reading: \(f_{\text{ave}}\) is the height of a rectangle on \([a,b]\) whose area equals the area under \(f\). In words, area \(\div\) width \(=\) average height.
EXAMPLE 1 Computing an average value
Find the average value of \(f(x)=1+x^{2}\) on \([-1,2]\).
Solution
\[ f_{\text{ave}}=\frac{1}{2-(-1)}\int_{-1}^{2}(1+x^{2})\,dx=\frac{1}{3}\Bigl[x+\tfrac{x^{3}}{3}\Bigr]_{-1}^{2} \]
\[ f_{\text{ave}}=2 \]
The Mean Value Theorem for Integrals

If \(f\) is continuous on \([a,b]\), then there exists a number \(c\) in \([a,b]\) such that \(f(c)=f_{\text{ave}}\); equivalently,

\[ \int_{a}^{b}f(x)\,dx=f(c)\,(b-a). \]

Geometrically, some rectangle of height \(f(c)\) over \([a,b]\) has exactly the same area as the region under \(f\).

EXAMPLE 2 Finding c explicitly
For \(f(x)=1+x^{2}\) on \([-1,2]\), find a value \(c\) guaranteed by the Mean Value Theorem for Integrals.
Solution
From Example 1, \(f_{\text{ave}}=2\). Solve \(f(c)=2\): \(1+c^{2}=2\Rightarrow c^{2}=1\).
\[ c=\pm 1,\quad\text{and } c=1 \text{ lies in } [-1,2] \]
Here both \(c=1\) and \(c=-1\) satisfy the theorem on this interval.
EXAMPLE 3 Average velocity equals average of velocities
Show that a car's average velocity over \([t_1,t_2]\) equals the average value of its velocity function.
Solution
Average value of \(v\): \(\displaystyle v_{\text{ave}}=\frac{1}{t_2-t_1}\int_{t_1}^{t_2}v(t)\,dt\).
Since \(v=s'\), the Net Change Theorem gives \(\displaystyle\int_{t_1}^{t_2}v(t)\,dt=s(t_2)-s(t_1)\).
\[ v_{\text{ave}}=\frac{s(t_2)-s(t_1)}{t_2-t_1}=\text{average velocity} \]

Two student errors I see every term here. First, people compute \(\int f\,dx\) and forget to divide by the interval width — that gives area, not an average. Second, the Mean Value Theorem for Integrals only guarantees \(c\) exists; it does not say it is unique (Example 2 has two). Use the theorem to justify that an average is actually attained, not to "find the" single point.

MechForge · Calculus · Chapter 6 — Applications of Integration. Concepts, formulas, and worked examples distilled in original wording for structured study; figures are original schematics. Section exercises and chapter review problems are intentionally omitted.