Notes on Chapters 6–8: Definite Integrals, Differential Equations, and Series
6. Using Definite Integrals
- Purpose: represent geometric areas, lengths, volumes, and physical quantities by summing infinitesimal slices via definite integrals.
- Structure: sections 6.1–6.5 cover area & arc length, volume (disk/washer methods, horizontal/vertical slices), density/mass/center of mass, work/pressure, and improper integrals.
6.1 Using Definite Integrals to Find Area and Length
Core idea: area under a nonnegative function f on [a,b] is A = ∫_a^b f(x) dx. For area between two curves, use the upper curve minus the lower curve on the chosen interval.
Area between two curves (general principle):
- If y = g(x) is the upper function and y = f(x) is the lower, and a ≤ x ≤ b with g(x) ≥ f(x), then
A =
\int_{a}^{b} [g(x) - f(x)]\,dx.
- If y = g(x) is the upper function and y = f(x) is the lower, and a ≤ x ≤ b with g(x) ≥ f(x), then
Preview Activity 6.1.1 (example): area between f(x) = 5 − (x − 1)^2 and g(x) = 4 − x.
- Intersections found algebraically; interval between intersection points used for area integrals.
- Exact area results via definite integrals illustrated; area equivalence with Riemann sums.
Vertical vs horizontal slicing:
- Vertical slices (thickness dx) yield height g(x)−f(x) on [a,b]; area is ∫_a^b (g−f) dx.
- Horizontal slices (thickness dy) are used when the region is easier described as x = h(y) or x = f(y); area becomes ∫ (right − left) dy over y ∈ [c,d].
Example 6.1.2: area between f(x) = (x − 1)^2 + 1 and g(x) = x + 2 on [0,3].
- Intersections at x = 0 and x = 3; A = ∫0^3 g(x) dx − ∫0^3 f(x) dx = 21/2 − 6 = 9/2.
- Also expressed as A = ∫_0^3 [g(x) − f(x)] dx.
Equation (6.1.2): Area between two curves via difference of integrals:
A = \int{a}^{b} \big( {upper}(x) - {lower}(x) \big) \, dx = \int{a}^{b} (\,\text{upper}(x) - \text{lower}(x)\,)
dx.Area via representative slices (Riemann sum):
A \approx \sum{i=1}^{n} [\phi(xi) - \psi(x_i)] \Delta x \quad\text{with } \Delta x \to 0.The general principle (two-curve area): if y = G(x) and y = F(x) intersect at (a, G(a)) and (b, G(b)) and G(x) ≥ F(x) for a ≤ x ≤ b, then A = ∫_a^b [G(x) − F(x)] dx.
Activity 6.1.2: problems a–d
- Students determine intersections, sketch, draw representative slices, set up exact integrals, and evaluate.
6.1.2 Summary: key takeaways on area between curves and vertical vs horizontal slicing.
6.1.3 Finding area with horizontal slices:
- Example 6.1.5: area bounded by x = y^2 − 1 and y = x − 1.
- Intersections found by solving y^2 − y − 2 = 0 → y = −1, 2; intersection points (0,−1) and (3,2).
- Using horizontal slices (dy): width is [xright(y) − xleft(y)] with thickness dy.
- For y ∈ [−1,2], xleft(y) = y^2 − 1, xright(y) = y+1; area via A = ∫_{-1}^{2} [(y+1) − (y^2 − 1)] dy.
6.1.4 Summary: arc length formula and area principles summarized.
6.1.5 Exercises: problems on area between functions and arc length. (Various)
6.1.4 Summary and 6.1.5 Exercises provide practice applying the area-between-curves principle and arc length formula.
6.1.4 The Arc Length of a Curve
- Arc length of y = f(x) on [a,b] is
L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx. - Idea: partition the curve into small segments, approximate each by a straight line, and sum their lengths; in the limit, you obtain the integral above.
- Activity 6.1.4 covers arc length problems and the relationship to area/sum of slices.
6.2 Using Definite Integrals to Find Volume
Volume by cross-sections (washers/disks): slice perpendicular to axis of revolution; each slice is a disk or washer.
Disk method (single function): if revolving y = r(x) ≥ 0 about the x-axis on [a,b], the volume is
V = \int_{a}^{b} \pi [r(x)]^2 dx.Washer method (two curves): if revolving region between y = R(x) and y = r(x) (R ≥ r ≥ 0) about the x-axis, the volume is
V = \int_{a}^{b} \pi[ R(x)^2 - r(x)^2] \, dx.Example 6.2.2: solid from revolving R between y = 4 − x^2 and x-axis about the x-axis; radius = y = 4 − x^2; bounds x ∈ [−2, 2].
- V = ∫_{-2}^{2} π (4 − x^2)^2 dx; evaluation yields V = (512/15)π.
The Disk Method is the case of revolving a single function; the Washer Method handles annular washers when the axis is on one side.
Preview Activity 6.2.1 contrasts disk/washer thinking for a cone-like solid along x-axis.
6.2.1 The Volume of a Solid of Revolution: cone-volume via integral and comparison to V = (1/3)π r^2 h.
6.2.2 Revolving about the y-axis requires expressing the radius in terms of y and using horizontal slices (dy). Example 6.2.6 yields V = 7π/15 for a region between y = sqrt(x) and y = x^4 revolved about the y-axis.
6.2.3 Revolving about lines other than axes modifies radii by a constant shift; example of rotating about y = −1 uses radii R(x) = x+1, r(x) = x^2+1, giving V = ∫ π[(x+1)^2 − (x^2+1)^2] dx between x = 0 and x = 1, etc.
6.2.4 Summary and 6.2.5 Exercises: revolve around x- and y-axes and around other horizontal/vertical lines; disk/washer methods.
6.2.4 The Washer Method
- If revolving the region between y = R(x) and y = r(x) about the x-axis, volume slice is a washer with area π[R(x)^2 − r(x)^2].
- Example 6.2.4 demonstrates the washer method for two curves about the x-axis.
6.2.5 Revolving about the y-axis
- When revolving about the y-axis, slices parallel to the y-axis yield washers with radii in terms of x as a function of y; horizontal slicing is used (dy).
- Example 6.2.6 shows how to express the inner/outer radius r(y) and R(y) when rotating about the y-axis.
6.2.3 Revolving about horizontal/vertical lines other than the axis
- Radii are shifted by the distance from the axis; you must adjust radii accordingly in the integrand.
6.2.5 Exercises cover a variety of solids of revolution problems.
6.3 Density, Mass, and Center of Mass
Connection: density ρ(x) is mass per unit length along an axis; a thin slice of width Δx at x has mass ≈ ρ(x) Δx when cross-sectional area is constant.
Mass of a region: if density ρ(x) (units: mass per unit length) is distributed along [a,b] with constant cross-section, then the total mass is
M = \int_{a}^{b}
ho(x) \, dx.For a rod of length L with density ρ(x), total mass M = ∫_0^L ρ(x) dx.
Center of mass for a continuous mass distribution along a single axis: if density ρ(x) on [a,b], the center of mass is
x{ ext{cm}} = \frac{ \int{a}^{b} x \rho(x) \, dx }{ \int_{a}^{b} \rho(x) \, dx }.Discrete mass distribution (point masses) center of mass: x = (Σ xi mi) / (Σ m_i).
6.3.1 Density: intuitive interpretation; constant-cross-section rod; ρ(x) is mass per unit length; units: g/cm per length; slice mass ≈ ρ(x) Δx.
6.3.2 Activity: nonuniform density problems (rod, cone, etc.) to practice mass and center of mass computations.
6.3.3 Center of Mass (continuous) for a thin rod with density ρ(x): x_cm = [∫ x ρ(x) dx] / [∫ ρ(x) dx].
6.3.4 Summary: key formulas and relationships for mass and center of mass; discrete vs continuous.
6.3.5 Exercises: multiple problems on center of mass and mass with varying density.
6.4 Physics Applications: Work, Force, and Pressure
Work is the accumulation of force over distance; for constant force F acting through distance d: W = F d.
When force varies, use W = ∫ F(x) dx, by slicing into small segments: W_slice ≈ F(x) Δx; summing gives the integral.
Relationship to common formulas:
- Distance traveled: if velocity v(t) is the rate of change of position, distance = ∫ v(t) dt.
- Mass with variable density: M = ∫ ρ(x) dx when density varies along an axis.
Preview Activity 6.4.1 (bucket in a well): shows how to set up B(h), the weight at height h, and compute work to raise water by integrating B(h) over the height interval.
Example 6.4.2 (disk/pump in a tank) and Example 6.4.4 (hydrostatic pressure):
- For hydrostatic pressure on a dam: pressure at depth d is P = ρ g d; total force on a thin horizontal slice of thickness Δx is Fslice = P(x) Aslice; sum via integral: F = ∫ P(x) A(x) dx.
- In a trapezoidal dam, the linear side boundary leads to radii (or heights) that vary with x; integrate to get total force.
6.4.4 Summary: unify the three settings (work, force, pressure) under the integral framework.
6.4.5 Exercises: extensive practice including work to pump liquids, hydrostatic forces, and related problems.
6.5 Improper Integrals
Improper integrals arise when the interval is unbounded or the integrand has a vertical asymptote inside the interval.
Definitions and conventions:
- If f ≥ 0 on [a, ∞), then ∫a^∞ f(x) dx converges if the limit as B → ∞ of ∫a^B f(x) dx exists and is finite; otherwise it diverges.
- Example: ∫_1^∞ 1/x^p dx converges if p > 1, diverges if p ≤ 1 (harmonic-like behavior).
Improper integrals of unbounded integrands: e.g., ∫0^1 1/√x dx converges with value 2; ∫1^∞ 1/x dx diverges.
Techniques:
- Replace improper integral with a limit of proper integrals and evaluate the limit.
- The Divergence Test: if lim{k→∞} ak ≠ 0, then ∑ a_k diverges.
- The Integral Test: if f is positive, decreasing, and f(k) = ak, then ∫1^∞ f(x) dx and ∑ a_k either both converge or both diverge.
- Direct Comparison, Limit Comparison, Ratio Test, Root Test (summarized in notes).
6.5.5 Exercises: variety of improper integrals (finite and infinite, with comparisons).
6.5.5.6–6.5.5.7: additional exercises and applications.
7. Differential Equations
7.1 An Introduction to Differential Equations
Definition: a differential equation describes the derivative(s) of an unknown function; solutions are functions whose derivatives satisfy the equation.
Examples:
- ds/dt = 4t + 1 with initial condition s(0) = 7 leads to a specific s(t).
- A market example: dA/dt = 0.03 A corresponds to continuous compound interest; general form dA/dt = k A.
Initial Value Problems (IVPs): a DE plus an initial condition often yields a unique solution under well-behaved conditions.
Autonomous vs non-autonomous: a differential equation is autonomous if the independent variable (t) does not appear in the derivative; otherwise it is non-autonomous.
7.1.4 Summary: recap of differential equations and their solutions; 7.1.5 Exercises.
7.2 Qualitative behavior of solutions to DEs
Slope fields: graphical tool showing the slope dy/dt = f(t,y) at sample points; solutions are curves tangent to the slope field.
Autoonomous DEs: dy/dt = f(y); equilibrium (constant) solutions satisfy f(y) = 0. Stability depends on nearby solution behavior.
7.2.1–7.2.3: slope fields, equilibriums, and stability concepts; graphical methods for analyzing long-term behavior.
7.2.4 Summary and 7.2.4 Exercises.
7.3 Euler’s method
Numerical method to approximate solutions: build a sequence (ti, yi) with step Δt; use y{i+1} ≈ yi + f(ti, yi) Δt.
Example: dy/dt = t − y with y(0) = 1; illustrate steps and error considerations; comparison to exact solution.
Error behavior: local truncation error and global error; error roughly proportional to Δt; improved Euler method concepts introduced; table demonstrations.
7.3.3 Summary and 7.3.4 Exercises.
7.4 Separable differential equations
A separable DE has the form dy/dt = g(y) h(t) or equivalently g(y) dy = h(t) dt; solve by separation and integration.
Example: dy/dt = y^2? or dy/dt = t y^2 (as given) and how to handle y = 0 equilibrium separately.
7.4.2 Summary and 7.4.3 Exercises.
7.5 Modeling with differential equations
Real-world modeling approach: balance rates in/out; examples include lakes pollution, bank accounts, morphine absorption, etc.
7.5.1 Preview problems illustrate constructing differential equations from physical processes.
7.5.2 Summary; 7.5.3 Exercises.
7.6 Population Growth and the Logistic Equation
Two models:
- Exponential growth: dP/dt = kP (per-capita growth constant k).
- Logistic growth: dP/dt = k P (N − P) where carrying capacity N yields a saturating growth.
Earth population data example; per-capita growth rate as a function of P; empirical linear fit for per-capita rate leads to dP/dt = P(0.025 − 0.002P) with carrying capacity N = 12.5.
General logistic equation: dP/dt = k P (N − P). Equilibria P = 0 and P = N; P = N is stable (carrying capacity).
Solving logistic equation: via separation and partial fractions, obtain the explicit solution
P(t) = \frac{N}{1 + \left( \dfrac{N}{P_0} - 1 \right) e^{-k N t}}.- Alternative form shown in notes: P(t) = N (N − P0)/P0 e^{−kN t} + 1 (full expression in text); canonical form is above.
Long-term behavior: P(t) → N as t → ∞ for any P_0 in (0,N).
7.6.2 Activity: analyzing when rate of change is maximal; logistic curve properties; inflection at P = N/2; carrying capacity as N.
7.6.3 Summary and 7.6.4 Exercises.
8. Sequences and Series
8.1 Sequences
A sequence is a list {s_n}. It can be viewed as a function from positive integers to reals.
Convergence: {sn} → L means sn gets arbitrarily close to L for large n.
Preview Activity 8.1.1: compound interest sequence and limits; 8.1.2 Definitions; 8.1.3 Exercises.
8.2 Geometric Series
Finite geometric sum: S_n = a + ar + ar^2 + … + ar^{n-1} = a(1 − r^n)/(1 − r) for r ≠ 1.
Infinite geometric series converges if |r| < 1 with sum S = a/(1 − r).
Partial sums Sn and the limit as n → ∞; examples with Warfarin model; notation ∑_{k=0}^∞ a r^k.
8.2.2 and 8.2.3 Exercises; 8.2.4 Series starting at nonzero index; various geometric sums.
8.3 Series of Real Numbers
Infinite series ∑ ak with partial sums Sn = ∑{k=1}^n ak; convergence means S_n → S finite.
Divergence tests and major convergence tests:
- Divergence Test: if lim{n→∞} an ≠ 0, then ∑ a_n diverges.
- Integral Test: relates series to definite integrals for functions f(n) = a_n.
- Direct Comparison Test: compare termwise with a known convergent/divergent series.
- Limit Comparison Test: if lim ak/bk = c ∈ (0,∞), then ∑ ak and ∑ bk converge or diverge together.
- Ratio Test: uses lim |a{n+1}/an| to decide convergence (|r|
- Root Test: uses limsup sqrt[n]{|a_n|} to decide convergence.
8.3.4–8.3.7: harmonic series and p-series, Integral Test as a tool for p-series, comparison tests, and several exercises including direct comparisons and limit comparisons.
8.4 Alternating Series
Definition: an alternating series ∑ (-1)^k ak with ak ≥ 0.
Alternating Series Test: if a_k ↓ 0, then the alternating series converges.
Alternating Series Estimation Theorem: the error after n terms is less than a_{n+1} in magnitude.
Absolute vs conditional convergence: absolute convergence depends on ∑ |a_k|;
- Some alternating series converge conditionally (e.g., alternating harmonic series) but not absolutely.
8.4.5–8.4.7: additional properties, regrouping, and rearrangement phenomena for conditionally convergent series (Riemann’s theorem context).
8.5 Taylor Polynomials and Taylor Series
Taylor polynomial of f around a: Pn(x) = ∑{k=0}^n f^{(k)}(a) (x − a)^k / k!
If f^{(k)}(a) exists for all k, the Taylor series is Tf(x) = ∑_{k=0}^∞ f^{(k)}(a) (x − a)^k / k!; Maclaurin series is the case a = 0.
8.5.2 Degree vs order: the order n polynomial derivative matching up to n-th derivative at a; sometimes the polynomial degree is less than n.
8.5.3 Examples: e^x, cos x, sin x, and 1/(1 − x); Maclaurin series and intervals of convergence.
Interval of convergence via Ratio Test for Taylor series; endpoints require separate checking.
8.5.4 Error bounds: Lagrange form of the remainder, En(x) ≤ M|x−a|^{n+1}/(n+1)!, etc. (general form).
8.5.5 Applications to convergence questions and approximations with Taylor polynomials.
8.6 Power Series
Power series: f(x) = ∑{k=0}^∞ ck (x − a)^k with radius of convergence r (interval (a−r, a+r)), and e.g., f(x) = ∑ x^k/k! (expansion for e^x).
Theorem: on the interval of convergence, a power series is the Taylor series of its sum; differentiation/integration term-by-term inside the interval preserves convergence (no endpoints guaranteed).
Examples: f(x) = ∑ x^{k} / k^2, interval of convergence (−1,1), endpoints checked separately; x → 1 or x → −1 yield convergence under specific tests (e.g., alternating series for x = −1).
8.6.3–8.6.6: calculating derivatives/integrals of power series, manipulating series to find new ones, applications like arctan(x) series, and Airy-type expansions; applications to differential equations via power series.
8.6.2 Exercises: interval of convergence for various power series; 8.6.3–8.6.6: more manipulations and examples.
Appendices (B & C)
Appendix B: Answers to activities (practice problems and worked solutions).
Appendix C: Answers to selected exercises (non-WeBWorK problems with solutions).