2.4 Integral Calculus & Applications

2.4 Integral Calculus & Applications

Integral calculus is the mathematical inverse of differentiation. In electronics engineering, integration is used to evaluate total accumulated quantities, such as electric charge, stored energy, average power, Root-Mean-Square (RMS) voltage, and geometric centroids.

---

1. Fundamental Integration Techniques

Standard Integration Rules

• Power Rule: $\int x^n dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$
• Logarithmic Integration: $\int \frac{1}{x} dx = \ln|x| + C$
• Exponential Integration: $\int e^{ax} dx = \frac{1}{a} e^{ax} + C$
• Trigonometric Integration: $\int \sin(ax) dx = -\frac{1}{a}\cos(ax) + C \quad | \quad \int \cos(ax) dx = \frac{1}{a}\sin(ax) + C$

Key Techniques

1. Integration by Substitution (u-sub): Transform $\int f(g(x))g'(x)dx$ into $\int f(u)du$ by letting $u = g(x)$.
2. Integration by Parts (IBP): Derived from the product rule: $\int u \, dv = uv - \int v \, du$
3. Partial Fraction Integration: Resolves rational functions into simpler integrable logarithmic/exponential fractions.

---

2. Definite Integrals & ECE Electrical Signal Applications

The Fundamental Theorem of Calculus states that $\int_a^b f(x) dx = F(b) - F(a)$, where $F'(x) = f(x)$.

Electrical Quantities Defined by Definite Integrals

• Accumulated Electric Charge: $Q = \int_{t_1}^{t_2} i(t) dt$
• Average Value of Periodic Signal (Period $T$): $V_{\text{avg}} = \frac{1}{T} \int_0^T v(t) dt$
• Root-Mean-Square (RMS) Value of Periodic Signal: $V_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T [v(t)]^2 dt}$
• Energy Stored in Capacitor / Inductor: $E_C = \int_0^V c \cdot v , dv = \frac{1}{2} C V^2 \quad | \quad E_L = \int_0^I l \cdot i , di = \frac{1}{2} L I^2$

Worked ECE Board Example 1: RMS Calculation of Triangular Waveform

Problem: A symmetric triangular voltage waveform rises linearly from $0\text{ V}$ to peak voltage $V_m = 12\text{ V}$ over half-period $t_0$ (so $v(t) = \frac{V_m}{t_0}t$ for $0 \le t \le t_0$). Compute its RMS voltage $V_{\text{rms}}$.

Solution:

• Step 1: Due to symmetry over the full period, evaluate the mean-square value over half-period $t_0$: $V_{\text{rms}}^2 = \frac{1}{t_0} \int_0^{t_0} \left(\frac{V_m}{t_0} t\right)^2 dt = \frac{V_m^2}{t_0^3} \int_0^{t_0} t^2 dt$
• Step 2: Perform definite integration: $\int_0^{t_0} t^2 dt = \left[ \frac{t^3}{3} \right]_0^{t_0} = \frac{t_0^3}{3}$
• Step 3: Substitute back: $V_{\text{rms}}^2 = \frac{V_m^2}{t_0^3} \left( \frac{t_0^3}{3} \right) = \frac{V_m^2}{3} \implies V_{\text{rms}} = \frac{V_m}{\sqrt{3}}$
• Step 4: For $V_m = 12\text{ V}$: $V_{\text{rms}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \approx 6.93\text{ V}$

---

3. Geometric Applications: Area & Volume of Revolution

Area Between Curves

The area $A$ bounded between top curve $y_1 = f(x)$ and bottom curve $y_2 = g(x)$ from $x = a$ to $x = b$ is: $A = \int_a^b \left[ f(x) - g(x) \right] dx$

Volumes of Solids of Revolution

1. Disk Method (Rotation about x-axis): $V = \pi \int_a^b [f(x)]^2 dx$
2. Washer Method (Rotation about x-axis between two curves): $V = \pi \int_a^b \left( [f(x)]^2 - [g(x)]^2 \right) dx$
3. Cylindrical Shell Method (Rotation about y-axis): $V = 2\pi \int_a^b x f(x) dx$

Worked ECE Board Example 2: Area Bounded by Signal Asymptotes

Problem: Calculate the area of the region bounded by the curves $y = 2x$ and $y = x^2$.

Solution:

• Step 1: Find points of intersection by setting $x^2 = 2x \implies x(x - 2) = 0 \implies x = 0$ and $x = 2$.
• Step 2: Over $0 \le x \le 2$, $2x \ge x^2$. Set up definite integral for area: $A = \int_0^2 (2x - x^2) dx = \left[ x^2 - \frac{x^3}{3} \right]_0^2 = \left( 2^2 - \frac{2^3}{3} \right) - 0 = 4 - \frac{8}{3} = \frac{4}{3}\text{ square units}$

---

4. Physical Applications: Centroids & Arc Length

Centroid of a Planar Region $(\bar{x}, \bar{y})$

For a region bounded by $y = f(x)$ above the x-axis from $x = a$ to $x = b$ with total area $A$: $\bar{x} = \frac{1}{A} \int_a^b x f(x) dx \quad | \quad \bar{y} = \frac{1}{2A} \int_a^b [f(x)]^2 dx$

Arc Length of a Curve

The physical length $L$ of a curve $y = f(x)$ from $x = a$ to $x = b$ (used to find physical wire lengths in antenna coils or transmission lines) is: $L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$

Worked ECE Board Example 3: Integration by Parts for Signal Energy

Problem: Evaluate the energy integral $I = \int_0^{\pi/2} t \cos(t) dt$.

Solution:

• Step 1: Use Integration by Parts: Let $u = t \implies du = dt$, and $dv = \cos(t)dt \implies v = \sin(t)$.
• Step 2: Apply formula $\int u dv = uv - \int v du$: $\int t \cos(t) dt = t \sin(t) - \int \sin(t) dt = t \sin(t) + \cos(t)$
• Step 3: Evaluate limits from $0$ to $\pi/2$: $I = \left[ t \sin(t) + \cos(t) \right]_0^{\pi/2} = \left( \frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) \right) - (0 \cdot \sin(0) + \cos(0))$ $I = \left( \frac{\pi}{2}(1) + 0 \right) - (0 + 1) = \frac{\pi}{2} - 1 \approx 0.5708$

---

5. Summary Table of Integration Formulas & Applications