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2.4 Integral Calculus & Applications

Key Takeaways

  • 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: ∫xndx=xn+1n+1+C(n≠−1)\int x^n dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)
  • Logarithmic Integration: ∫1xdx=ln⁡∣x∣+C\int \frac{1}{x} dx = \ln|x| + C
  • Exponential Integration: ∫eaxdx=1aeax+C\int e^{ax} dx = \frac{1}{a} e^{ax} + C
  • Trigonometric Integration: ∫sin⁡(ax)dx=−1acos⁡(ax)+C∣∫cos⁡(ax)dx=1asin⁡(ax)+C\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 ∫f(g(x))g′(x)dx\int f(g(x))g'(x)dx into ∫f(u)du\int f(u)du by letting u=g(x)u = g(x).
  2. Integration by Parts (IBP): Derived from the product rule: ∫u dv=uv−∫v du\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 ∫abf(x)dx=F(b)−F(a)\int_a^b f(x) dx = F(b) - F(a), where F′(x)=f(x)F'(x) = f(x).

Electrical Quantities Defined by Definite Integrals

  • Accumulated Electric Charge: Q=∫t1t2i(t)dtQ = \int_{t_1}^{t_2} i(t) dt
  • Average Value of Periodic Signal (Period TT): Vavg=1T∫0Tv(t)dtV_{\text{avg}} = \frac{1}{T} \int_0^T v(t) dt
  • Root-Mean-Square (RMS) Value of Periodic Signal: Vrms=1T∫0T[v(t)]2dtV_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T [v(t)]^2 dt}
  • Energy Stored in Capacitor / Inductor: EC=∫0Vc⋅v dv=12CV2∣EL=∫0Il⋅i di=12LI2E_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 V0\text{ V} to peak voltage Vm=12 VV_m = 12\text{ V} over half-period t0t_0 (so v(t)=Vmt0tv(t) = \frac{V_m}{t_0}t for 0≤t≤t00 \le t \le t_0). Compute its RMS voltage VrmsV_{\text{rms}}.

Solution:

  • Step 1: Due to symmetry over the full period, evaluate the mean-square value over half-period t0t_0: Vrms2=1t0∫0t0(Vmt0t)2dt=Vm2t03∫0t0t2dtV_{\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: ∫0t0t2dt=[t33]0t0=t033\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: Vrms2=Vm2t03(t033)=Vm23  ⟹  Vrms=Vm3V_{\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 Vm=12 VV_m = 12\text{ V}: Vrms=123=43≈6.93 VV_{\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 AA bounded between top curve y1=f(x)y_1 = f(x) and bottom curve y2=g(x)y_2 = g(x) from x=ax = a to x=bx = b is:

A=∫ab[f(x)−g(x)]dxA = \int_a^b \left[ f(x) - g(x) \right] dx

Volumes of Solids of Revolution

  1. Disk Method (Rotation about x-axis): V=π∫ab[f(x)]2dxV = \pi \int_a^b [f(x)]^2 dx
  2. Washer Method (Rotation about x-axis between two curves): V=π∫ab([f(x)]2−[g(x)]2)dxV = \pi \int_a^b \left( [f(x)]^2 - [g(x)]^2 \right) dx
  3. Cylindrical Shell Method (Rotation about y-axis): V=2π∫abxf(x)dxV = 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=2xy = 2x and y=x2y = x^2.

Solution:

  • Step 1: Find points of intersection by setting x2=2x  ⟹  x(x−2)=0  ⟹  x=0x^2 = 2x \implies x(x - 2) = 0 \implies x = 0 and x=2x = 2.
  • Step 2: Over 0≤x≤20 \le x \le 2, 2x≥x22x \ge x^2. Set up definite integral for area: A=∫02(2x−x2)dx=[x2−x33]02=(22−233)−0=4−83=43 square unitsA = \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 (xˉ,yˉ)(\bar{x}, \bar{y})

For a region bounded by y=f(x)y = f(x) above the x-axis from x=ax = a to x=bx = b with total area AA:

xˉ=1A∫abxf(x)dx∣yˉ=12A∫ab[f(x)]2dx\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 LL of a curve y=f(x)y = f(x) from x=ax = a to x=bx = b (used to find physical wire lengths in antenna coils or transmission lines) is:

L=∫ab1+(dydx)2dxL = \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=∫0π/2tcos⁡(t)dtI = \int_0^{\pi/2} t \cos(t) dt.

Solution:

  • Step 1: Use Integration by Parts: Let u=t  ⟹  du=dtu = t \implies du = dt, and dv=cos⁡(t)dt  ⟹  v=sin⁡(t)dv = \cos(t)dt \implies v = \sin(t).
  • Step 2: Apply formula ∫udv=uv−∫vdu\int u dv = uv - \int v du: ∫tcos⁡(t)dt=tsin⁡(t)−∫sin⁡(t)dt=tsin⁡(t)+cos⁡(t)\int t \cos(t) dt = t \sin(t) - \int \sin(t) dt = t \sin(t) + \cos(t)
  • Step 3: Evaluate limits from 00 to π/2\pi/2: I=[tsin⁡(t)+cos⁡(t)]0π/2=(π2sin⁡(π2)+cos⁡(π2))−(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=(π2(1)+0)−(0+1)=π2−1≈0.5708I = \left( \frac{\pi}{2}(1) + 0 \right) - (0 + 1) = \frac{\pi}{2} - 1 \approx 0.5708

5. Summary Table of Integration Formulas & Applications

Engineering QuantityIntegral FormulaECE Application Context
RMS ValueVrms=1T∫0Tv2(t)dtV_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T v^2(t) dt}AC effective power calculation for non-sinusoidal waves
Integration by Parts∫udv=uv−∫vdu\int u dv = uv - \int v duEvaluating Fourier & Laplace transforms of te−att e^{-at} or tcos⁡(ωt)t \cos(\omega t)
Area Between CurvesA=∫ab[f(x)−g(x)]dxA = \int_a^b [f(x) - g(x)] dxDetermining probability density areas, signal power envelope
Disk VolumeV=π∫ab[f(x)]2dxV = \pi \int_a^b [f(x)]^2 dxComponent packaging volume, coaxial shielding envelope
Arc LengthL=∫ab1+(y′)2dxL = \int_a^b \sqrt{1 + (y')^2} dxDetermining exact wire length of helical or parabolic antenna elements
Test Your Knowledge

What is the value of the definite integral ∫₀^{π/2} t cos(t) dt evaluated using integration by parts?

A

π - 1 (~2.142)

B

1.0

C

π/2 (~1.571)

D

π/2 - 1 (~0.571)

Test Your Knowledge

A symmetrical triangular voltage waveform rises linearly from 0 V to peak voltage V_m = 12 V over half-period t_0. What is its Root-Mean-Square (V_rms) voltage?

A

12.0 V

B

6.0 V

C

6√2 V (~8.49 V)

D

4√3 V (~6.93 V)

Test Your Knowledge

What is the area of the region bounded between the curves y = 2x and y = x²?

A

2 square units

B

2/3 square units

C

1/3 square units

D

4/3 square units

Sections you finish are checked off in the contents.