14 Applications of Integration
In the preceding chapters, we established the Riemann integral as a rigorous mathematical object—a limit of approximating sums that quantifies accumulation. The Fundamental Theorem connected this global concept to local rates of change through differentiation. But integration’s reach extends far beyond computing areas under curves or reversing derivatives. (Our integral calculator can help verify the computations in this chapter.)
The integral is a universal accumulation mechanism. Whenever a quantity accumulates continuously—whether energy from varying force, mass over density variation, or probability over intervals—the mathematical structure is integration. The generality stems from the definition itself: partition, approximate, refine, take limits.
This chapter examines how integration manifests across disciplines. We focus on applications that both illuminate mathematical structure and demonstrate integration’s indispensability in quantitative reasoning. The selection is neither encyclopedic nor arbitrary—each application reveals something essential about the integral’s nature.
14.1 Work and Energy
14.1.1 Physical Motivation
Newton’s second law states that force produces acceleration: F = ma. When a constant force F acts on an object through displacement d, the work done is W = Fd. This is energy transferred to the object.
But what if force varies with position? Consider compressing a spring. Hooke’s law states F(x) = -kx where k is the spring constant and x is displacement from equilibrium. As we compress further, force increases—constant-force intuition fails.
The resolution: divide displacement into small intervals where force is approximately constant, compute work on each piece, sum contributions. This is a Riemann sum.
Definition 14.1 (Work) Let F : [a,b] \to \mathbb{R} be a continuous force function acting along a line. The work done moving an object from position a to position b is W = \int_a^b F(x)\,dx.
Justification. Partition [a,b] with points x_0 < x_1 < \cdots < x_n. On each subinterval [x_{i-1}, x_i], force varies little, so F(x) \approx F(x_i^*) for some sample point x_i^*. Work on this interval is approximately F(x_i^*)\Delta x_i. Total work: W \approx \sum_{i=1}^n F(x_i^*)\Delta x_i.
As mesh \|\mathcal{P}\| \to 0, this Riemann sum converges to the integral.
Example 14.1 (Spring Compression Work) For F(x) = kx (taking the magnitude), compressing from x = 0 to x = d: W = \int_0^d kx\,dx = \left[\frac{kx^2}{2}\right]_0^d = \frac{kd^2}{2}.
The work increases quadratically with displacement—doubling compression quadruples energy stored. This energy is potential energy: U(x) = \frac{1}{2}kx^2. The force is the negative derivative of potential: F = -U'.
Gravitational work. Near Earth’s surface, gravitational force is approximately constant: F = mg. Lifting mass m through height h requires work W = mgh. But at astronomical distances, gravity follows the inverse-square law: F(r) = -\frac{GMm}{r^2} where G is the gravitational constant, M is Earth’s mass, m is the object’s mass, and r is distance from Earth’s center. To lift the object from radius R (Earth’s surface) to radius r: W = -\int_R^r \frac{GMm}{s^2}\,ds = GMm\left[\frac{1}{s}\right]_R^r = GMm\left(\frac{1}{r} - \frac{1}{R}\right).
As r \to \infty, this approaches GMm/R—the escape energy. Setting this equal to kinetic energy \frac{1}{2}mv^2 yields the escape velocity v_{\text{esc}} = \sqrt{2GM/R} \approx 11.2 km/s for Earth.
14.1.2 Conservation of Energy
The work-energy theorem states that work equals change in kinetic energy: W = \int_a^b F(x)\,dx = \frac{1}{2}mv_b^2 - \frac{1}{2}mv_a^2.
For conservative forces (those derivable from a potential U with F = -U'), this becomes: \int_a^b (-U'(x))\,dx = -[U(b) - U(a)] = \frac{1}{2}mv_b^2 - \frac{1}{2}mv_a^2.
Rearranging: \frac{1}{2}mv_a^2 + U(a) = \frac{1}{2}mv_b^2 + U(b).
Total mechanical energy E = \frac{1}{2}mv^2 + U(x) is conserved. Integration encodes this fundamental physical principle.
14.2 Volumes of Revolution
Let f : [a,b] \to \mathbb{R}_{\geq 0} be continuous. Rotating the region under the graph about the x-axis generates a solid. Its volume is V = \pi \int_a^b [f(x)]^2\,dx.
Partition [a,b] with points x_0 < x_1 < \cdots < x_n. On each subinterval [x_{i-1}, x_i], the solid is approximately a disk of radius f(x_i^*) and thickness \Delta x_i, with volume \pi[f(x_i^*)]^2\Delta x_i. The total volume is approximately V \approx \sum_{i=1}^n \pi[f(x_i^*)]^2\Delta x_i. As \|\mathcal{P}\| \to 0, this Riemann sum converges to the integral.
For rotation about the y-axis (assuming f is invertible), use V = 2\pi \int_a^b x f(x)\,dx (shell method), where each shell at radius x has height f(x) and “thickness” dx.
Example 14.2 (Volume of a Sphere) Rotate f(x) = \sqrt{R^2 - x^2} on [-R, R] about the x-axis: V = \pi \int_{-R}^R (R^2 - x^2)\,dx = \pi\left[R^2x - \frac{x^3}{3}\right]_{-R}^R = \frac{4\pi R^3}{3}.
Remark. These are special cases of the general change-of-variables formula in \mathbb{R}^3. The natural setting is multivariable calculus: the volume element dV transforms via the Jacobian determinant. Cylindrical coordinates (r,\theta,z) give dV = r\,dr\,d\theta\,dz, and solids of revolution are simply regions with \theta-independence. The “disk” and “shell” methods are artifacts of reducing three-dimensional integrals to one dimension by exploiting symmetry.
14.3 Center of Mass and Moments
14.3.1 Discrete Masses
For point masses m_1, \ldots, m_n at positions x_1, \ldots, x_n along a line, the center of mass is the weighted average: \bar{x} = \frac{\sum_{i=1}^n m_i x_i}{\sum_{i=1}^n m_i} = \frac{1}{M}\sum_i m_i x_i where M = \sum m_i is total mass. This is the balance point—if the masses were replaced by a single mass M at \bar{x}, the moment about any point would be unchanged.
14.3.2 Continuous Mass Distributions
For an object with continuously varying density, we cannot sum discrete masses. Consider a rod of length L with linear density \rho(x) (mass per unit length). The mass of a small segment [x, x+\Delta x] is approximately \rho(x)\Delta x. Its moment about the origin is x \cdot \rho(x)\Delta x.
Total mass: M = \int_0^L \rho(x)\,dx.
Total moment: M_0 = \int_0^L x\rho(x)\,dx.
Center of mass: \bar{x} = M_0/M.
Definition 14.2 (Center of Mass (Continuous)) For an object occupying [a,b] with density \rho(x), the center of mass is \bar{x} = \frac{\int_a^b x\rho(x)\,dx}{\int_a^b \rho(x)\,dx}.
Example 14.3 (Conical Rod Center of Mass) A rod of length L has density \rho(x) = \rho_0(1 + x/L) increasing linearly. Where is the center of mass?
M = \int_0^L \rho_0\left(1 + \frac{x}{L}\right)dx = \rho_0\left(L + \frac{L}{2}\right) = \frac{3\rho_0 L}{2}.
M_0 = \int_0^L x\rho_0\left(1 + \frac{x}{L}\right)dx = \rho_0\int_0^L \left(x + \frac{x^2}{L}\right)dx= \rho_0\left(\frac{L^2}{2} + \frac{L^2}{3}\right) = \frac{5\rho_0 L^2}{6}.
Thus \bar{x} = \frac{5\rho_0 L^2/6}{3\rho_0 L/2} = \frac{5L}{9}, shifted toward the denser end as expected.
14.3.3 Moments of Inertia
The moment of inertia quantifies rotational inertia—resistance to angular acceleration. For a point mass m at distance r from an axis, I = mr^2. For continuous distributions: I = \int r^2\,dm where dm is an infinitesimal mass element.
For the rod with density \rho(x) rotating about x = 0: I = \int_0^L x^2 \rho(x)\,dx.
This appears in rotational dynamics: \tau = I\alpha (torque equals moment of inertia times angular acceleration), the rotational analogue of F = ma.
14.4 Probability and Expectation
14.4.1 From Weighted Averages to Probability
The center of mass formula \bar{x} = \frac{\sum m_i x_i}{\sum m_i} is a weighted average. If we interpret m_i/M as the “weight” or “probability” of position x_i, this becomes an expected value.
For a discrete random variable X taking values x_1, \ldots, x_n with probabilities p_1, \ldots, p_n (where \sum p_i = 1): \mathbb{E}[X] = \sum_{i=1}^n p_i x_i.
For continuous random variables, we replace sums with integrals.
Definition 14.3 (Probability Density Function) A probability density function (pdf) on [a,b] is a function f(x) \geq 0 satisfying \int_a^b f(x)\,dx = 1.
The probability that X lies in [c,d] \subseteq [a,b] is P(c \leq X \leq d) = \int_c^d f(x)\,dx.
The integral accumulates probability mass over intervals. The condition \int_a^b f = 1 ensures that total probability equals one—certainty that the outcome lies somewhere in [a,b].
Example 14.4 (Uniform Distribution) On [0,1], the uniform pdf is f(x) = 1. Every subinterval of length \ell has probability \ell. The probability that X \in [0.3, 0.7] is \int_{0.3}^{0.7} 1\,dx = 0.4.
Example 14.5 (Exponential Distribution) For a random variable representing time until an event (radioactive decay, arrival times), the exponential pdf is: f(t) = \lambda e^{-\lambda t}, \quad t \geq 0 where \lambda > 0 is the rate parameter. Verify normalization: \int_0^\infty \lambda e^{-\lambda t}\,dt = \lambda\left[-\frac{e^{-\lambda t}}{\lambda}\right]_0^\infty = \lambda \cdot \frac{1}{\lambda} = 1.
The probability of waiting more than time T is:
P(X > T) = \int_T^\infty \lambda e^{-\lambda t}\,dt = e^{-\lambda T}.
This decays exponentially—longer waits are exponentially less likely, reflecting the “memoryless” property of exponential processes.
14.4.2 Expected Value and Variance
Definition 14.4 (Expected Value) For a continuous random variable X with pdf f on [a,b]: \mathbb{E}[X] = \int_a^b x f(x)\,dx.
This is the continuous analogue of the center of mass: the “average” value of X, weighted by probability density. For the uniform distribution on [0,1], \mathbb{E}[X] = \int_0^1 x \cdot 1\,dx = 1/2. For the exponential, \mathbb{E}[X] = \int_0^\infty t \lambda e^{-\lambda t}\,dt = 1/\lambda (computed via integration by parts).
Variance measures spread around the mean: \text{Var}(X) = \mathbb{E}[(X - \mu)^2] = \int_a^b (x-\mu)^2 f(x)\,dx where \mu = \mathbb{E}[X]. Alternatively, \text{Var}(X) = \mathbb{E}[X^2] - \mu^2.
These moments—first moment (mean), second moment (variance)—parallel the physical moments we computed for mass distributions. The mathematical structure is identical: integration of weighted quantities.
14.4.3 The Normal Distribution
The most important pdf in statistics is the normal (Gaussian) distribution: f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/(2\sigma^2)} for x \in \mathbb{R}, where \mu is the mean and \sigma^2 is the variance. The normalization constant ensures \int_{-\infty}^\infty f = 1 (the Gaussian integral \int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}, proven via polar coordinates).
The normal distribution arises from the Central Limit Theorem: sums of independent random variables converge (under mild conditions) to a normal distribution. This universality makes it ubiquitous in science and statistics.
Computing probabilities requires integrating f. But e^{-x^2} has no elementary antiderivative—we need numerical methods or tables. The error function \text{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\,dt is defined specifically for this purpose, relating to the normal cumulative distribution function.
14.4.4 Numerical Integration and Monte Carlo Methods
When antiderivatives are unavailable or integrals exist in high dimensions, numerical methods become essential. The simplest is the midpoint rule: approximate \int_a^b f by f(x_i^*)\Delta x where x_i^* is the midpoint of [x_{i-1}, x_i]. More sophisticated methods (Simpson’s rule, Gaussian quadrature) improve accuracy.
In high dimensions—computing \int_{\mathbb{R}^d} f(x_1, \ldots, x_d)\,dx_1\cdots dx_d—deterministic methods become computationally prohibitive (the “curse of dimensionality”). Monte Carlo integration offers an alternative:
Sample N random points \mathbf{x}_1, \ldots, \mathbf{x}_N uniformly from the integration domain. Approximate: \int_D f \approx \frac{V}{N}\sum_{i=1}^N f(\mathbf{x}_i) where V is the domain’s volume. By the Law of Large Numbers, this converges to the true integral as N \to \infty, with error decreasing as 1/\sqrt{N} regardless of dimension.
This underlies computational physics (quantum mechanics simulations), computer graphics (rendering images via ray tracing), and machine learning (stochastic gradient descent samples the loss landscape).
14.4.5 Neural Networks and Probability
Training a neural network minimizes a loss function L(\theta) over parameters \theta. For classification, this often involves integrating a probability distribution defined by the network’s output (a softmax layer). The cross-entropy loss: \mathcal{L} = -\sum_{i} y_i \log(\hat{y}_i) for discrete distributions generalizes to continuous distributions via integration: \mathcal{L} = -\int p(x)\log(q(x))\,dx where p is the true distribution and q is the model’s approximation. This is the Kullback-Leibler divergence, a measure of distributional distance central to information theory and variational inference.
Generative models like Variational Autoencoders (VAEs) explicitly integrate over latent variables: p(x) = \int p(x|z)p(z)\,dz where z is a latent code and x is observed data. Since this integral is intractable, VAEs use variational inference—approximating the integral via optimization, a technique rooted in calculus of variations.
14.4.6 Feynman Path Integrals
In quantum mechanics, Feynman’s path integral formulation expresses the probability amplitude for a particle to move from point A to point B as an integral over all possible paths: \langle B | A \rangle = \int \mathcal{D}[x(t)] \, e^{iS[x(t)]/\hbar} where S[x(t)] = \int L(x, \dot{x}, t)\,dt is the action (integral of the Lagrangian over time), and \mathcal{D}[x(t)] denotes a “measure” over the space of paths. This is a functional integral—integrating over an infinite-dimensional space of functions, not just \mathbb{R}^n.
The rigorous mathematical foundation involves measure theory and functional analysis (Wiener measure, Feynman-Kac formula). But the conceptual structure is familiar: partition, approximate, sum, take limits. The Riemann integral on \mathbb{R} extends conceptually to infinite dimensions, though the technical machinery changes.
14.4.7 Finance: Options Pricing
The Black-Scholes formula for option pricing involves computing expected values under a probability distribution representing stock price evolution. For a European call option with strike price K and expiry T, the price is: C = \mathbb{E}[\max(S_T - K, 0)] where S_T is the stock price at time T, modeled as a random variable. Under the Black-Scholes assumptions (geometric Brownian motion), this expectation reduces to: C = S_0 \Phi(d_1) - Ke^{-rT}\Phi(d_2) where \Phi is the cumulative distribution function of the standard normal distribution: \Phi(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}}e^{-t^2/2}\,dt.
The integral \Phi cannot be computed in closed form, but numerical methods and tables provide accurate approximations. The entire edifice of quantitative finance rests on integrating probability distributions—managing uncertainty through mathematics.
14.5 Remarks on Structure
These applications share a common pattern:
Continuous accumulation: A quantity varies continuously, and we want its total over an interval (or region).
Partition and approximate: Divide into small pieces where the quantity is approximately constant.
Sum and limit: Form a Riemann sum and pass to the limit as partition refinement increases.
This is the integral as universal accumulator. Whether accumulating mass, probability, arc length, energy, or quantum amplitudes, the mathematical operation is the same. The integral is the fundamental tool for quantifying continuous aggregation.
The Riemann integral, while conceptually elementary (limits of sums), provides the foundation. Extensions—Lebesgue integration for measure theory, functional integrals for quantum field theory, stochastic integrals for random processes—generalize the idea to more abstract settings. But the core principle persists: integration accumulates, differentiation extracts rates, and together they form the calculus that models continuous change.
Area between curves via the general principle.
A circle of radius r is inscribed in a square of side 2r. Compute the area between the square’s boundary and the circle in the first quadrant by setting up the integral \int_0^r [r - \sqrt{r^2 - x^2}]\,dx. Evaluate it to show this area equals \frac{r^2(4-\pi)}{4}.
For general functions f(x) \geq g(x) on [a,b], explain why the area between the curves is \int_a^b [f(x) - g(x)]\,dx by considering the Riemann sum interpretation: on each subinterval [x_{i-1}, x_i], the area is approximately (f(x_i^*) - g(x_i^*))\Delta x_i.
Find the area enclosed between y = x^2 and y = \sqrt{x} on [0,1] by determining which function is larger and applying the formula from part (b).
Work and potential energy in variable force fields.
A spring with constant k is compressed from equilibrium (x = 0) to position x = -d (taking leftward as negative). Compute the work done by the restoring force F(x) = -kx as W = \int_{-d}^0 (-kx)\,dx. Why is this positive?
Define potential energy U(x) such that F = -U'(x). For the spring, show that U(x) = \frac{1}{2}kx^2 + C. By conservation of energy, if a mass m is released from position x = -d, find its velocity v at equilibrium in terms of k, d, and m.
For gravitational force F(r) = -GMm/r^2, find the work required to move a mass m from Earth’s surface at radius R to distance r > R. Show that the escape velocity (velocity at R needed to reach r \to \infty) is v_{\text{esc}} = \sqrt{2GM/R}.
Volumes by revolution and the shell method.
Rotate f(x) = x^2 on [0, h] about the x-axis. Using the disk method V = \pi\int_0^h [f(x)]^2\,dx, show that the volume of the resulting paraboloid is V = \frac{\pi h^5}{5}.
Now rotate the same region about the y-axis using the shell method. Each cylindrical shell at radius x has height f(x) = x^2 and circumference 2\pi x. The volume element is dV = 2\pi x \cdot f(x)\,dx. Evaluate V = 2\pi\int_0^h x \cdot x^2\,dx and verify it equals \frac{\pi h^4}{2}.
Explain why parts (a) and (b) give different volumes: the solids are different because the axis of rotation changed.
Center of mass for variable density distributions.
A rod of length L has density \rho(x) = \rho_0(1 + \alpha x/L) where \alpha > 0 is a dimensionless parameter. Compute the total mass M = \int_0^L \rho(x)\,dx and moment M_0 = \int_0^L x\rho(x)\,dx.
Show that the center of mass is \bar{x} = \frac{L(2 + \alpha)}{3(1 + \alpha/2)}. Verify that when \alpha = 0 (uniform density), \bar{x} = L/2, and as \alpha \to \infty (concentration at x = L), \bar{x} \to 2L/3.
For this rod rotating about x = 0, compute the moment of inertia I = \int_0^L x^2\rho(x)\,dx in terms of \rho_0, L, and \alpha.
Area between curves and general formulas.
Circle inscribed in square. In the first quadrant, the square has boundary at y = r for 0 \leq x \leq r, and the circle has equation x^2 + y^2 = r^2, giving y = \sqrt{r^2 - x^2} for the upper semicircle.
The area between them is A = \int_0^r [r - \sqrt{r^2 - x^2}]\,dx = \int_0^r r\,dx - \int_0^r \sqrt{r^2 - x^2}\,dx.
The first integral is r^2. For the second, use substitution x = r\sin\theta, dx = r\cos\theta\,d\theta: \int_0^r \sqrt{r^2 - x^2}\,dx = \int_0^{\pi/2} r\cos\theta \cdot r\cos\theta\,d\theta = r^2\int_0^{\pi/2}\cos^2\theta\,d\theta = r^2 \cdot \frac{\pi}{4}.
Therefore, A = r^2 - \frac{\pi r^2}{4} = \frac{r^2(4-\pi)}{4}. Multiplying by 4 for all quadrants gives total area r^2(4-\pi) = 4r^2 - \pi r^2, which is square area minus circle area. \square
General area formula. Partition [a,b] with points x_0 < x_1 < \cdots < x_n. On each subinterval [x_{i-1}, x_i], approximate both functions by their values at sample point x_i^* \in [x_{i-1}, x_i].
The area between the curves on this subinterval is approximately a rectangle with height f(x_i^*) - g(x_i^*) and width \Delta x_i = x_i - x_{i-1}, giving area (f(x_i^*) - g(x_i^*))\Delta x_i.
Summing over all subintervals: A \approx \sum_{i=1}^n (f(x_i^*) - g(x_i^*))\Delta x_i.
This is a Riemann sum for the integrand f(x) - g(x). As mesh \|\mathcal{P}\| \to 0, it converges to A = \int_a^b [f(x) - g(x)]\,dx.
The key insight: integration of the difference function accumulates vertical distances between curves. \square
Area between parabola and square root. Compare f(x) = \sqrt{x} and g(x) = x^2 on [0,1].
At x = 0: f(0) = 0 = g(0).
At x = 1: f(1) = 1 = g(1).
At x = 1/4: f(1/4) = 1/2 while g(1/4) = 1/16, so f > g in the interior.Alternatively, \sqrt{x} \geq x^2 on [0,1] iff x \geq x^4 iff x^4 - x \leq 0 iff x(x^3 - 1) \leq 0, which holds for x \in [0,1].
By part (b): A = \int_0^1 (\sqrt{x} - x^2)\,dx = \int_0^1 x^{1/2}\,dx - \int_0^1 x^2\,dx = \left[\frac{2x^{3/2}}{3}\right]_0^1 - \left[\frac{x^3}{3}\right]_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}. \square
Work, energy, and conservative forces.
Spring compression work. The work done by force F(x) = -kx from x = -d to x = 0 is W = \int_{-d}^0 (-kx)\,dx = -k\int_{-d}^0 x\,dx = -k\left[\frac{x^2}{2}\right]_{-d}^0 = -k\left(0 - \frac{d^2}{2}\right) = \frac{kd^2}{2}.
This is positive because the restoring force acts rightward (positive direction) when x < 0, so force and displacement have the same sign. The spring does positive work on the mass as it returns to equilibrium. \square
Potential energy and velocity. If F = -U'(x), then -kx = -U'(x), so U'(x) = kx.
Integrating: U(x) = \frac{1}{2}kx^2 + C. Taking U(0) = 0 gives C = 0.
Conservation of energy: E = \frac{1}{2}mv^2 + U(x) is constant. At x = -d, the mass starts from rest (v = 0), so E = 0 + U(-d) = \frac{1}{2}kd^2.
At equilibrium x = 0: E = \frac{1}{2}mv^2 + U(0) = \frac{1}{2}mv^2.
Equating: \frac{1}{2}mv^2 = \frac{1}{2}kd^2, so v = d\sqrt{k/m}. \square
Gravitational escape velocity. Work to move from R to r against gravity: W = -\int_R^r \frac{GMm}{s^2}\,ds = GMm\int_R^r s^{-2}\,ds = GMm\left[-\frac{1}{s}\right]_R^r = GMm\left(\frac{1}{R} - \frac{1}{r}\right).
For escape (r \to \infty), W_{\text{esc}} = GMm/R.
Initial kinetic energy must equal escape work: \frac{1}{2}mv_{\text{esc}}^2 = GMm/R. Solving: v_{\text{esc}} = \sqrt{\frac{2GM}{R}}. \square
Volumes of revolution by different methods.
Disk method about x-axis. For f(x) = x^2 rotated about the x-axis on [0,h]: V = \pi\int_0^h [f(x)]^2\,dx = \pi\int_0^h x^4\,dx = \pi\left[\frac{x^5}{5}\right]_0^h = \frac{\pi h^5}{5}. \square
Shell method about y-axis. Rotating about the y-axis, a shell at radius x has height f(x) = x^2 and circumference 2\pi x. Volume element: dV = 2\pi x \cdot x^2\,dx = 2\pi x^3\,dx.
Total volume: V = 2\pi\int_0^h x^3\,dx = 2\pi\left[\frac{x^4}{4}\right]_0^h = \frac{\pi h^4}{2}. \square
Why volumes differ. In part (a), we rotate about the x-axis, creating a paraboloid of revolution with vertical axis. In part (b), we rotate the same region about the y-axis, creating a different solid with horizontal axis.
These are distinct geometric objects. The disk method in (a) stacks circular disks perpendicular to the x-axis. The shell method in (b) uses nested cylinders around the y-axis. The formulas \pi h^5/5 vs \pi h^4/2 reflect fundamentally different solids. \square
Center of mass with variable density.
Computing mass and moment. For \rho(x) = \rho_0(1 + \alpha x/L): M = \int_0^L \rho_0\left(1 + \frac{\alpha x}{L}\right)dx = \rho_0\left[x + \frac{\alpha x^2}{2L}\right]_0^L = \rho_0\left(L + \frac{\alpha L}{2}\right) = \rho_0 L\left(1 + \frac{\alpha}{2}\right).
M_0 = \int_0^L x\rho_0\left(1 + \frac{\alpha x}{L}\right)dx = \rho_0\int_0^L \left(x + \frac{\alpha x^2}{L}\right)dx = \rho_0\left[\frac{x^2}{2} + \frac{\alpha x^3}{3L}\right]_0^L = \rho_0\left(\frac{L^2}{2} + \frac{\alpha L^2}{3}\right). \square
Center of mass and limiting cases. From part (a): \bar{x} = \frac{M_0}{M} = \frac{\rho_0(\frac{L^2}{2} + \frac{\alpha L^2}{3})}{\rho_0 L(1 + \frac{\alpha}{2})} = \frac{L^2(\frac{3 + 2\alpha}{6})}{L(\frac{2+\alpha}{2})} = \frac{L(3+2\alpha)}{3(2+\alpha)} = \frac{L(2+\alpha) + L}{3(2+\alpha)} = \frac{L}{3} \cdot \frac{3+2\alpha}{2+\alpha}.
Simplify: \bar{x} = L \cdot \frac{3+2\alpha}{3(2+\alpha)}. Let me recalculate more carefully: \bar{x} = \frac{L^2/2 + \alpha L^2/3}{L(1 + \alpha/2)} = \frac{L(1/2 + \alpha/3)}{1 + \alpha/2} = L \cdot \frac{3 + 2\alpha}{6} \cdot \frac{2}{2+\alpha} = \frac{L(3+2\alpha)}{3(2+\alpha)}.
Wait, let me be more careful: numerator is L^2(3/6 + 2\alpha/6) = L^2(3+2\alpha)/6, denominator is L(2+\alpha)/2. So: \bar{x} = \frac{L^2(3+2\alpha)/6}{L(2+\alpha)/2} = \frac{L(3+2\alpha)}{3(2+\alpha)}.
When \alpha = 0: \bar{x} = \frac{3L}{3 \cdot 2} = \frac{L}{2}.
As \alpha \to \infty: \bar{x} \to \frac{2\alpha L}{3\alpha} = \frac{2L}{3}.
The center shifts toward the denser end as expected. \square
Moment of inertia. For rotation about x = 0: I = \int_0^L x^2\rho(x)\,dx = \rho_0\int_0^L x^2\left(1 + \frac{\alpha x}{L}\right)dx = \rho_0\int_0^L \left(x^2 + \frac{\alpha x^3}{L}\right)dx.
I = \rho_0\left[\frac{x^3}{3} + \frac{\alpha x^4}{4L}\right]_0^L = \rho_0\left(\frac{L^3}{3} + \frac{\alpha L^3}{4}\right) = \rho_0 L^3\left(\frac{1}{3} + \frac{\alpha}{4}\right) = \frac{\rho_0 L^3(4+3\alpha)}{12}. \square