1.2 Differential and Integral Calculus

1.2 Differential and Integral Calculus

Calculus is named in the NCEES Mathematics specification as single-variable calculus, and it underpins nearly every other area of the exam: centroids and moments of inertia are integrals, shear and moment diagrams are derivatives of the load, and every rate process is a differential equation. The Reference Handbook supplies the derivative and integral tables, so exam speed comes from recognizing which rule applies rather than from recall.

Calculus and differential equations represent core tools for modeling continuous physical processes in engineering, including fluid flow, dynamic vibrations, heat transfer, rate reactions, and stress gradients.

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1. Differential Calculus

Core Derivative Rules

Let $u(x)$ and $v(x)$ be differentiable functions of $x$:

• Product Rule: $\frac{d}{dx}(uv) = u'v + uv'$
• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$
• Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$
• Exponential & Logarithmic Derivatives: $$\frac{d}{dx}(e^{ax}) = a e^{ax}, \quad \frac{d}{dx}(\ln x) = \frac{1}{x}$$
• Trigonometric Derivatives: $$\frac{d}{dx}(\sin x) = \cos x, \quad \frac{d}{dx}(\cos x) = -\sin x, \quad \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1 + x^2}$$

Critical Points & Optimization

• Critical Points: Occur where $f'(x) = 0$ or $f'(x)$ does not exist.
• Second Derivative Test:
  • If $f'(x_0) = 0$ and $f''(x_0) > 0$, $x_0$ is a local minimum.
  • If $f'(x_0) = 0$ and $f''(x_0) < 0$, $x_0$ is a local maximum.
  • If $f''(x_0) = 0$, the test is inconclusive (inflection point candidate).

Multivariable Calculus: Gradient & Directional Derivatives

For a scalar field $f(x, y, z)$:

• Gradient Vector: $$\nabla f = \frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j} + \frac{\partial f}{\partial z}\mathbf{k}$$
• Directional Derivative: The rate of change of $f$ in the direction of unit vector $\mathbf{u}$: $$D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u}$$ The gradient $\nabla f$ points in the direction of maximum rate of increase, and $|\nabla f|$ is the maximum directional derivative.

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2. Integral Calculus & Applications

Integration Techniques

• Integration by Substitution (u-substitution): $\int f(g(x))g'(x)dx = \int f(u)du$
• Integration by Parts: $\int u \, dv = uv - \int v \, du$

Engineering Applications of Definite Integrals

• Area Between Curves: $A = \int_a^b [f(x) - g(x)] dx$
• Centroid Coordinate $(\bar{x}, \bar{y})$ of Planar Area: $$\bar{x} = \frac{1}{A} \int x [f(x) - g(x)] dx, \quad \bar{y} = \frac{1}{2A} \int \left([f(x)]^2 - [g(x)]^2\right) dx$$
• Area Moment of Inertia about x-axis ($I_x$): $$I_x = \iint y^2 \, dA$$
• Volume of Revolution (Disk Method about x-axis): $$V = \pi \int_a^b [f(x)]^2 dx$$

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Limits, Continuity, and L'Hopital's Rule

A limit describes the value a function approaches, whether or not it is defined there. The FE exam tests limits almost exclusively through indeterminate forms.

$$\text{Indeterminate forms: } \frac{0}{0}, \; \frac{\infty}{\infty}, \; 0\cdot\infty, \; \infty-\infty, \; 1^\infty, \; 0^0, \; \infty^0$$

L'Hopital's rule applies only to $\tfrac{0}{0}$ and $\tfrac{\infty}{\infty}$: differentiate the numerator and denominator separately — not as a quotient — and re-evaluate.

$$\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}$$

Repeat if the result is still indeterminate.

Worked example.

$$\lim_{x\to 0}\frac{1-\cos x}{x^2} \;\to\; \frac{0}{0}$$

First application: $\displaystyle\lim_{x\to0}\frac{\sin x}{2x} \to \frac{0}{0}$ — still indeterminate.

Second application: $\displaystyle\lim_{x\to0}\frac{\cos x}{2} = \boxed{\frac{1}{2}}$

Trap: applying L'Hopital's rule when the form is not indeterminate gives a wrong answer. $\displaystyle\lim_{x\to0}\frac{\sin x}{1+x} = \frac{0}{1} = 0$ directly; differentiating both parts would give $\cos(0)/1 = 1$, which is wrong. Always confirm the indeterminate form before differentiating.

Taylor and Maclaurin Series

$$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$$

A Maclaurin series is the special case $a = 0$. The four expansions worth recognizing on sight:

$$\ne^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}+\cdots \qquad \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}-\cdots$$ $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}-\cdots \qquad \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \;\;(|x|<1)$$

Why this matters beyond mathematics. The small-angle approximations $\sin\theta \approx \theta$ and $\cos\theta \approx 1 - \theta^2/2$ are the first terms of these series, and they are what make the simple-pendulum equation linear in the Dynamics chapter. Truncating a series is also the basis of the numerical methods in the next section.

Applications That Recur Across the Exam

Worked Example: Average Value and Work

A spring has a nonlinear force law $F(x) = 400x + 1{,}200x^2$ N, with $x$ in metres. Find the work to compress it from 0 to 0.30 m, and the average force over that interval.

Work:

$$W = \int_0^{0.30}(400x + 1{,}200x^2)\,dx = \left[200x^2 + 400x^3\right]_0^{0.30}$$ $$W = 200(0.09) + 400(0.027) = 18.0 + 10.8 = \boxed{28.8\ \text{J}}$$

Average force:

$$\bar{F} = \frac{W}{\Delta x} = \frac{28.8}{0.30} = 96.0\ \text{N}$$

Check against the endpoint force: $F(0.30) = 400(0.30) + 1{,}200(0.09) = 120 + 108 = 228$ N. The average of 96.0 N is well below half the peak force of 228 N, because the quadratic term concentrates the force near the end of travel. A linear spring of the same final force would store $\tfrac{1}{2}(228)(0.30) = 34.2$ J — more than the actual 28.8 J.

Trap: for a linear spring, $W = \tfrac{1}{2}kx^2$ and the average force is exactly half the peak. That shortcut fails the moment the force law is nonlinear, and the exam supplies nonlinear force laws precisely to catch its misuse. Integrate whenever the force is not proportional to displacement.