2.4 Integral Calculus & Applications

PRC Electronics Engineering Licensure Examination Study Guide Exam

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1 Chapter 1: Introduction & Board Exam Strategy

1.1 PRC ECE Licensure Exam Overview & Regulation 1.2 Table of Specifications & Exam Strategy

2 Chapter 2: Mathematics Foundations

2.1 College & Advanced Algebra 2.2 Trigonometry & Solid Mensuration 2.3 Analytic Geometry & Differential Calculus 2.4 Integral Calculus & Applications

3 Chapter 3: Advanced Mathematics for ECE

3.1 Differential Equations & Circuit Transients 3.2 Complex Numbers & Laplace Transforms 3.3 Fourier Series, Fourier Transforms & Z-Transforms 3.4 Vector Analysis, Matrices & Probability

4 Chapter 4: GEAS - Engineering Sciences & Economics

4.1 Engineering Physics & Materials Chemistry 4.2 Engineering Mechanics & Strength of Materials 4.3 Engineering Economics & Financial Analysis 4.4 Computer Fundamentals & C/C++ Programming

5 Chapter 5: GEAS - Management, Materials, Environment & CAD

5.1 Engineering Management & Technopreneurship 101 5.2 Materials Science & Engineering 5.3 Environment Science & Engineering 5.4 Computer-Aided Drawing (CAD) & Engineering Graphics

6 Chapter 6: GEAS - ECE Laws, Ethics & Contracts

6.1 Republic Act No. 9292 (Electronics Engineering Law of 2004) 6.2 ECE Code of Ethics & Professional Conduct 6.3 Contracts, Specifications & National Building Code (PD 1096)

7 Chapter 7: Electronics Engineering - Circuits & Semiconductor Diodes

7.1 DC & AC Circuit Analysis & Network Theorems 7.2 Electromagnetics & Passive Components 7.3 Semiconductor Physics & Diode Circuits 7.4 Power Supply Design & Voltage Regulators

8 Chapter 8: Electronics Engineering - Transistors & Analog Circuits

8.1 BJT Operation, Biasing & Small-Signal Amplifiers 8.2 FET Characteristics, Biasing & Amplifier Circuits 8.3 Multistage Amplifiers, Frequency Response & Power Amplifiers 8.4 Operational Amplifiers & Active Filter Design

9 Chapter 9: Electronics Engineering - Digital Electronics & Control Systems

9.1 Feedback Topologies & Oscillator Circuits 9.2 Boolean Algebra, Logic Gates & Combinational Circuits 9.3 Sequential Logic, Flip-Flops & Microprocessor Architecture 9.4 Industrial Power Electronics & Control System Theory

10 Chapter 10: EST - Communications Fundamentals & Transmission Lines

10.1 Signal Fundamentals, Decibel Calculations & Noise Analysis 10.2 Amplitude Modulation (AM) Principles & Receivers 10.3 Frequency Modulation (FM), Phase Modulation & Discriminators 10.4 Transmission Line Parameters, SWR & Smith Chart

11 Chapter 11: EST - Digital Communications, Wireless & Optical Systems

11.1 Pulse & Digital Modulation Systems 11.2 Data Communications, OSI Model & Protocols 11.3 Antenna Principles & Electromagnetic Wave Propagation 11.4 Optical Fiber, Microwave, Satellite & Philippine NTC Regulations

2.4 Integral Calculus & Applications

Key Takeaways Explain This Topic AI

Execute fundamental integration techniques including u-substitution, integration by parts, and trigonometric substitution.

Calculate RMS values, average power, and total stored charge for periodic electrical waveforms using definite integrals.

Determine areas between curves and volumes of revolution using disk, washer, and cylindrical shell methods.

Compute centroids, moments of inertia, and arc lengths for physical electronic components and antenna profiles.

Last updated: July 2026

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 : ∫ x n d x = x n + 1 n + 1 + C ( n ≠ − 1 ) \int x^n dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1) ∫ x n d x = n + 1 x n + 1 ​ + C ( n  = − 1 )

Logarithmic Integration : ∫ 1 x d x = ln ⁡ ∣ x ∣ + C \int \frac{1}{x} dx = \ln|x| + C ∫ x 1 ​ d x = ln ∣ x ∣ + C

Exponential Integration : ∫ e a x d x = 1 a e a x + C \int e^{ax} dx = \frac{1}{a} e^{ax} + C ∫ e a x d x = a 1 ​ e a x + C

Trigonometric Integration : ∫ sin ⁡ ( a x ) d x = − 1 a cos ⁡ ( a x ) + C ∣ ∫ cos ⁡ ( a x ) d x = 1 a sin ⁡ ( a x ) + C \int \sin(ax) dx = -\frac{1}{a}\cos(ax) + C \quad | \quad \int \cos(ax) dx = \frac{1}{a}\sin(ax) + C ∫ sin ( a x ) d x = − a 1 ​ cos ( a x ) + C ∣ ∫ cos ( a x ) d x = a 1 ​ sin ( a x ) + C

Key Techniques

Integration by Substitution (u-sub) : Transform ∫ f ( g ( x ) ) g ′ ( x ) d x \int f(g(x))g'(x)dx ∫ f ( g ( x )) g ′ ( x ) d x into ∫ f ( u ) d u \int f(u)du ∫ f ( u ) d u by letting u = g ( x ) u = g(x) u = g ( x ) .

Integration by Parts (IBP) : Derived from the product rule: ∫ u d v = u v − ∫ v d u \int u \, dv = uv - \int v \, du ∫ u d v = uv − ∫ v d u

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 ∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x) dx = F(b) - F(a) ∫ a b ​ f ( x ) d x = F ( b ) − F ( a ) , where F ′ ( x ) = f ( x ) F'(x) = f(x) F ′ ( x ) = f ( x ) .

Electrical Quantities Defined by Definite Integrals

Accumulated Electric Charge : Q = ∫ t 1 t 2 i ( t ) d t Q = \int_{t_1}^{t_2} i(t) dt Q = ∫ t 1 ​ t 2 ​ ​ i ( t ) d t

Average Value of Periodic Signal (Period T T T ): V avg = 1 T ∫ 0 T v ( t ) d t V_{\text{avg}} = \frac{1}{T} \int_0^T v(t) dt V avg ​ = T 1 ​ ∫ 0 T ​ v ( t ) d t

Root-Mean-Square (RMS) Value of Periodic Signal : V rms = 1 T ∫ 0 T [ v ( t ) ] 2 d t V_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T [v(t)]^2 dt} V rms ​ = T 1 ​ ∫ 0 T ​ [ v ( t ) ] 2 d t ​

Energy Stored in Capacitor / Inductor : E C = ∫ 0 V c ⋅ v d v = 1 2 C V 2 ∣ E L = ∫ 0 I l ⋅ i d i = 1 2 L I 2 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 E C ​ = ∫ 0 V ​ c ⋅ v d v = 2 1 ​ C V 2 ∣ E L ​ = ∫ 0 I ​ l ⋅ i d i = 2 1 ​ L I 2

Worked ECE Board Example 1: RMS Calculation of Triangular Waveform

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

Solution:

Step 1: Due to symmetry over the full period, evaluate the mean-square value over half-period t 0 t_0 t 0 ​ : V rms 2 = 1 t 0 ∫ 0 t 0 ( V m t 0 t ) 2 d t = V m 2 t 0 3 ∫ 0 t 0 t 2 d t 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 V rms 2 ​ = t 0 ​ 1 ​ ∫ 0 t 0 ​ ​ ( t 0 ​ V m ​ ​ t ) 2 d t = t 0 3 ​ V m 2 ​ ​ ∫ 0 t 0 ​ ​ t 2 d t

Step 2: Perform definite integration: ∫ 0 t 0 t 2 d t = [ t 3 3 ] 0 t 0 = t 0 3 3 \int_0^{t_0} t^2 dt = \left[ \frac{t^3}{3} \right]_0^{t_0} = \frac{t_0^3}{3} ∫ 0 t 0 ​ ​ t 2 d t = [ 3 t 3 ​ ] 0 t 0 ​ ​ = 3 t 0 3 ​ ​

Step 3: Substitute back: V rms 2 = V m 2 t 0 3 ( t 0 3 3 ) = V m 2 3 ⟹ V rms = V m 3 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}} V rms 2 ​ = t 0 3 ​ V m 2 ​ ​ ( 3 t 0 3 ​ ​ ) = 3 V m 2 ​ ​ ⟹ V rms ​ = 3 ​ V m ​ ​

Step 4: For V m = 12 V V_m = 12\text{ V} V m ​ = 12 V : V rms = 12 3 = 4 3 ≈ 6.93 V V_{\text{rms}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \approx 6.93\text{ V} V rms ​ = 3 ​ 12 ​ = 4 3 ​ ≈ 6.93 V

• 3. Geometric Applications: Area & Volume of Revolution
• Area Between Curves

The area A A A bounded between top curve y 1 = f ( x ) y_1 = f(x) y 1 ​ = f ( x ) and bottom curve y 2 = g ( x ) y_2 = g(x) y 2 ​ = g ( x ) from x = a x = a x = a to x = b x = b x = b is: A = ∫ a b [ f ( x ) − g ( x ) ] d x A = \int_a^b \left[ f(x) - g(x) \right] dx A = ∫ a b ​ [ f ( x ) − g ( x ) ] d x

Volumes of Solids of Revolution

Disk Method (Rotation about x-axis): V = π ∫ a b [ f ( x ) ] 2 d x V = \pi \int_a^b [f(x)]^2 dx V = π ∫ a b ​ [ f ( x ) ] 2 d x

Washer Method (Rotation about x-axis between two curves): V = π ∫ a b ( [ f ( x ) ] 2 − [ g ( x ) ] 2 ) d x V = \pi \int_a^b \left( [f(x)]^2 - [g(x)]^2 \right) dx V = π ∫ a b ​ ( [ f ( x ) ] 2 − [ g ( x ) ] 2 ) d x

Cylindrical Shell Method (Rotation about y-axis): V = 2 π ∫ a b x f ( x ) d x V = 2\pi \int_a^b x f(x) dx V = 2 π ∫ a b ​ x f ( x ) d x

Worked ECE Board Example 2: Area Bounded by Signal Asymptotes

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

Solution:

Step 1: Find points of intersection by setting x 2 = 2 x ⟹ x ( x − 2 ) = 0 ⟹ x = 0 x^2 = 2x \implies x(x - 2) = 0 \implies x = 0 x 2 = 2 x ⟹ x ( x − 2 ) = 0 ⟹ x = 0 and x = 2 x = 2 x = 2 .

Step 2: Over 0 ≤ x ≤ 2 0 \le x \le 2 0 ≤ x ≤ 2 , 2 x ≥ x 2 2x \ge x^2 2 x ≥ x 2 . Set up definite integral for area: A = ∫ 0 2 ( 2 x − x 2 ) d x = [ x 2 − x 3 3 ] 0 2 = ( 2 2 − 2 3 3 ) − 0 = 4 − 8 3 = 4 3 square units 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} A = ∫ 0 2 ​ ( 2 x − x 2 ) d x = [ x 2 − 3 x 3 ​ ] 0 2 ​ = ( 2 2 − 3 2 3 ​ ) − 0 = 4 − 3 8 ​ = 3 4 ​ square units

• 4. Physical Applications: Centroids & Arc Length
• Centroid of a Planar Region ( x ˉ , y ˉ ) (\bar{x}, \bar{y}) ( x ˉ , y ˉ ​ )

For a region bounded by y = f ( x ) y = f(x) y = f ( x ) above the x-axis from x = a x = a x = a to x = b x = b x = b with total area A A A : x ˉ = 1 A ∫ a b x f ( x ) d x ∣ y ˉ = 1 2 A ∫ a b [ f ( x ) ] 2 d x \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 x ˉ = A 1 ​ ∫ a b ​ x f ( x ) d x ∣ y ˉ ​ = 2 A 1 ​ ∫ a b ​ [ f ( x ) ] 2 d x

Arc Length of a Curve

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

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

Problem: Evaluate the energy integral I = ∫ 0 π / 2 t cos ⁡ ( t ) d t I = \int_0^{\pi/2} t \cos(t) dt I = ∫ 0 π /2 ​ t cos ( t ) d t .

Solution:

Step 1: Use Integration by Parts: Let u = t ⟹ d u = d t u = t \implies du = dt u = t ⟹ d u = d t , and d v = cos ⁡ ( t ) d t ⟹ v = sin ⁡ ( t ) dv = \cos(t)dt \implies v = \sin(t) d v = cos ( t ) d t ⟹ v = sin ( t ) .

Step 2: Apply formula ∫ u d v = u v − ∫ v d u \int u dv = uv - \int v du ∫ u d v = uv − ∫ v d u : ∫ t cos ⁡ ( t ) d t = t sin ⁡ ( t ) − ∫ sin ⁡ ( t ) d t = t sin ⁡ ( t ) + cos ⁡ ( t ) \int t \cos(t) dt = t \sin(t) - \int \sin(t) dt = t \sin(t) + \cos(t) ∫ t cos ( t ) d t = t sin ( t ) − ∫ sin ( t ) d t = t sin ( t ) + cos ( t )

Step 3: Evaluate limits from 0 0 0 to π / 2 \pi/2 π /2 : I = [ t sin ⁡ ( t ) + cos ⁡ ( t ) ] 0 π / 2 = ( π 2 sin ⁡ ( π 2 ) + cos ⁡ ( π 2 ) ) − ( 0 ⋅ sin ⁡ ( 0 ) + cos ⁡ ( 0 ) ) 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 = [ t sin ( t ) + cos ( t ) ] 0 π /2 ​ = ( 2 π ​ sin ( 2 π ​ ) + cos ( 2 π ​ ) ) − ( 0 ⋅ sin ( 0 ) + cos ( 0 )) I = ( π 2 ( 1 ) + 0 ) − ( 0 + 1 ) = π 2 − 1 ≈ 0.5708 I = \left( \frac{\pi}{2}(1) + 0 \right) - (0 + 1) = \frac{\pi}{2} - 1 \approx 0.5708 I = ( 2 π ​ ( 1 ) + 0 ) − ( 0 + 1 ) = 2 π ​ − 1 ≈ 0.5708

5. Summary Table of Integration Formulas & Applications

Engineering Quantity Integral Formula ECE Application Context

RMS Value V rms = 1 T ∫ 0 T v 2 ( t ) d t V_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T v^2(t) dt} V rms ​ = T 1 ​ ∫ 0 T ​ v 2 ( t ) d t ​ AC effective power calculation for non-sinusoidal waves

Integration by Parts ∫ u d v = u v − ∫ v d u \int u dv = uv - \int v du ∫ u d v = uv − ∫ v d u Evaluating Fourier & Laplace transforms of t e − a t t e^{-at} t e − a t or t cos ⁡ ( ω t ) t \cos(\omega t) t cos ( ω t )

Area Between Curves A = ∫ a b [ f ( x ) − g ( x ) ] d x A = \int_a^b [f(x) - g(x)] dx A = ∫ a b ​ [ f ( x ) − g ( x )] d x Determining probability density areas, signal power envelope

Disk Volume V = π ∫ a b [ f ( x ) ] 2 d x V = \pi \int_a^b [f(x)]^2 dx V = π ∫ a b ​ [ f ( x ) ] 2 d x Component packaging volume, coaxial shielding envelope

Arc Length L = ∫ a b 1 + ( y ′ ) 2 d x L = \int_a^b \sqrt{1 + (y')^2} dx L = ∫ a b ​ 1 + ( y ′ ) 2 ​ d x Determining exact wire length of helical or parabolic antenna elements