Difficulty: Intermediate | Prerequisites: Integration techniques (substitution, by parts), basic differential equations concept, familiarity with limits at infinity.
Once you can evaluate integrals, the next question is: what can you do with them? This section covers three major applications. Differential equations solved by direct antidifferentiation model real growth and decay. Volumes of revolution let you compute the volume of a 3D solid from a 2D curve. Improper integrals extend the definite integral to infinite intervals and functions with vertical asymptotes, which is essential for convergence tests in series and for probability distributions.
Solve simple DEs by integrating both sides and using initial conditions. Compute volumes of revolution with the disk method (pi times the integral of f(x)^2). Improper integrals require limits: replace the infinity or the singularity with a variable, evaluate, then take the limit.
Initial Value Problem (IVP)
A differential equation paired with a specific condition at a known point, e.g. E(0) = 2000. The condition pins down the arbitrary constant that appears after integration. Think of it as: the equation tells you the shape of the curve, and the initial condition tells you exactly which curve.
Volume of Revolution (Disk Method)
The volume of the solid formed by rotating y = f(x) about the x-axis from x = a to x = b is pi times the integral from a to b of [f(x)]^2 dx. Each thin slice is a disk of radius f(x), and you sum them up.
Improper Integral (Type 1, infinite interval)
An integral where one or both limits of integration are infinite. You replace the infinite limit with a variable t, evaluate the integral, then take the limit as t approaches infinity. If the limit exists and is finite, the integral converges.
Improper Integral (Type 2, discontinuous integrand)
An integral where the integrand has a vertical asymptote within the interval of integration. You split the integral at the asymptote, replace the problematic endpoint with a variable, and take a one-sided limit. If both pieces converge, the whole integral converges.
Convergent vs Divergent
An improper integral is convergent if its limiting value is a finite number. It is divergent if the limit does not exist or is infinite. In simple terms, convergent means the area under the curve is finite despite the infinite extent or the blow-up.
Given dE/dt = 18000(2t + 4)^(-3/2) with E(0) = 2000.
Integrate both sides with respect to t.
Use substitution u = 2t + 4, du = 2 dt.
The antiderivative of (2t + 4)^(-3/2) is -1/(2t + 4)^(1/2), scaled appropriately.
Result: E(t) = C - 18000/(2t + 4)^(1/2).
Apply the initial condition E(0) = 2000 to find C.
E(0) = C - 18000/2 = 2000, so C = 11000.
E(t) = 18000(1/2 - 1/(2t + 4)^(1/2)) + 2000.
Long-term behaviour: as t approaches infinity, 1/(2t + 4)^(1/2) approaches 0, so lim E(t) = 18000(1/2) + 2000 = 11000.
To find the volume of the solid obtained by rotating y = f(x) about the x-axis from x = a to x = b:
Volume = pi times integral from a to b of [f(x)]^2 dx.
Example: y = 6 sqrt(x), from x = 0 to x = 1.
[f(x)]^2 = 36x.
V = pi times integral from 0 to 1 of 36x dx = pi times [18x^2] from 0 to 1 = 18 pi.
Replace the infinite limit with a finite variable t, evaluate, then take the limit.
Example: integral from 0 to infinity of 4/(2x + 1)^2 dx.
Evaluate integral from 0 to t: use u = 2x + 1, result is 2 - 2/(2t + 1).
As t approaches infinity: limit = 2.
The integral converges and equals 2.
When the integrand has a vertical asymptote inside the interval, split the integral at that point.
Example: integral from 0 to 2 of 6/(x - 1)^(2/3) dx.
The integrand blows up at x = 1.
Split into I1 (from 0 to 1) and I2 (from 1 to 2).
For I1: evaluate from 0 to t, take the limit as t approaches 1 from the left. Antiderivative is 18(x - 1)^(1/3). Result: I1 = 18.
For I2: evaluate from t to 2, take the limit as t approaches 1 from the right. Result: I2 = 18.
Total: I = 36. Both pieces converge, so the whole integral converges.
V = \pi \int_a^b [f(x)]^2\,dx \quad \text{(disk method, rotation about x-axis)}\int_a^{\infty} f(x)\,dx = \lim_{t \to \infty} \int_a^t f(x)\,dx\int_a^b f(x)\,dx = \lim_{t \to c^-} \int_a^t f(x)\,dx + \lim_{t \to c^+} \int_t^b f(x)\,dx \quad \text{(asymptote at } x = c \text{)}The enrollment growth problem in this exam is a direct model of population or adoption growth with diminishing returns: the rate of growth decays over time, and the long-run limit is the carrying capacity. Engineers use volumes of revolution to design axially symmetric components such as nozzles, bottles and turbine housings. Improper integrals underpin probability theory: the total area under a probability density function over the entire real line must equal 1, which is an improper integral.
Students often forget to split a Type 2 improper integral at the asymptote. If the integrand blows up at a point inside the interval, you cannot just integrate straight through. Both pieces must converge independently.
When computing volumes of revolution, students sometimes forget to square f(x). The disk method uses [f(x)]^2 inside the integral, not f(x).
With initial value problems, students sometimes solve for the general antiderivative but forget to apply the initial condition to pin down the constant C. Without this step, the answer is a family of solutions, not a unique one.
For Type 1 improper integrals, students occasionally evaluate the antiderivative at infinity directly instead of writing the limit. The limit process is essential and must be shown.
Expect at least one improper integral on the exam. Know both types: infinite interval and discontinuous integrand.
Volume of revolution is a standard exam question. The disk method formula is non-negotiable: memorise it.
The enrollment/population growth problem tests whether you can integrate a given rate function and apply an initial condition. It is a classic exam format.
When splitting a Type 2 improper integral, both halves must independently converge. If either diverges, the whole integral diverges.
True or False: If dE/dt = 18000(2t + 4)^(-3/2) and E(0) = 2000, the long-term enrollment approaches 11000. _____ (True.)
Fill in the blank: The disk method formula for volume when rotating y = f(x) about the x-axis is V = ____ times integral from a to b of ____ dx. (pi; [f(x)]^2)
True or False: The integral from 0 to 2 of 1/(x - 1)^2 dx converges. _____ (False: 1/(x-1)^2 has a non-integrable singularity at x = 1 because the exponent 2 is at least 1.)
True or False: An improper integral with an infinite upper limit always diverges. _____ (False: it converges if the integrand decreases fast enough, e.g. 1/x^2.)
Q: The rate of increase in enrollment is E'(t) = 18000(2t + 4)^(-3/2) with E(0) = 2000. What is the long-term enrollment?
A: Integrate E'(t) to get E(t) = C - 18000/(2t+4)^(1/2). Apply E(0) = 2000: C = 11000. As t approaches infinity, 1/(2t+4)^(1/2) approaches 0, so lim E(t) = 11000.
Q: Find the volume of the solid generated by rotating y = 6 sqrt(x) about the x-axis from x = 0 to x = 1.
A: V = pi times integral from 0 to 1 of 36x dx = pi times 18x^2 evaluated from 0 to 1 = 18 pi cubic units.
Q: Determine whether the integral from 0 to infinity of 4/(2x + 1)^2 dx converges, and if so, find its value.
A: Substitute u = 2x + 1. The integral from 0 to t becomes (1/2) integral of 4/u^2 du = 2 - 2/(2t+1). As t approaches infinity, the limit is 2. Convergent, I = 2.
Q: Evaluate the integral from 0 to 2 of 6/(x - 1)^(2/3) dx.
A: Split at x = 1. Both I1 (0 to 1) and I2 (1 to 2) yield 18 each (antiderivative is 18(x-1)^(1/3)). Total: I = 36.
Improper integrals connect directly to infinite series: the Integral Test uses the convergence or divergence of an improper integral to determine whether a related series converges. Volumes of revolution extend to the shell method (rotation about the y-axis), which is typically covered alongside the disk method. The DE/IVP material here is a stepping stone toward separable and linear first-order differential equations in later courses.
Improper integral, convergent, divergent, Type 1 improper integral, Type 2 improper integral, vertical asymptote, volume of revolution, disk method, solid of revolution, initial value problem, IVP, differential equation, antidifferentiation, long-term behaviour, carrying capacity, MATH 101, integral calculus, UT Austin