CELE Engineering Mathematics — Integral CalculusDetailed Explanation

Examples

Applications

• Computing work done by a variable force F(x): W = ∫F(x) dx — essential in structural mechanics.
• Finding displacement from a velocity function: s = ∫v(t) dt.
• Hydraulic pressure on a submerged surface: P = ∫γ·h·w(h) dh, where γ is the unit weight of fluid.
• Integration by parts appears in deriving beam deflection formulas from moment equations.
• Partial fractions arise in Laplace transform problems used in structural dynamics.

Misconceptions

• Writing ∫xⁿ dx = xⁿ⁺¹ (forgetting to divide by n+1).
• Using the Power Rule for n = −1: ∫x⁻¹ dx ≠ x⁰/0; it equals ln|x| + C.
• Forgetting the constant of integration C in indefinite integrals.
• In u-substitution, forgetting to replace dx with du/[du/dx] — the entire integrand must be in terms of u.
• In integration by parts, choosing u = eˣ when a polynomial is present — this creates a more complex, not simpler, remaining integral.
• ∫eˣ² dx cannot be integrated using elementary functions — do not attempt to apply the Power Rule or u-substitution directly to this form.

Related Concepts

• Differentiation (inverse operation — used to verify integration results)
• Definite Integrals and Area
• Fundamental Theorem of Calculus
• Trigonometric Identities (needed for trig integration)
• Algebraic Factoring (needed for partial fractions)

Common Exam Questions

Key Points To Remember

• Power Rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C — the exponent increases by 1 and you divide by the new exponent.
• ∫(1/x)dx = ln|x| + C — the absolute value bars are required; never write ln(x) without them.
• Always include + C for indefinite integrals; board exams deduct for missing constants.
• U-substitution: identify the 'inside' function whose derivative also appears in the integrand.
• LIATE mnemonic guides which factor to call 'u' in integration by parts.
• For partial fractions: always perform polynomial long division first if degree of numerator ≥ degree of denominator.
• ∫eᵃˣ dx = (1/a)eᵃˣ + C — the 1/a factor is frequently missed.
• Check your answer by differentiating — if d/dx[F(x)] = f(x), the integration is correct.

Examples

Applications

• Computing cross-sectional areas of irregular structural members.
• Computing flow volumes in hydraulics: Q = ∫A v(y) dy across a channel cross-section.
• Determining net displacement in kinematics from a velocity-time graph.
• Foundation area calculations for irregularly shaped footings.
• Computing the area of a pressure diagram in soil mechanics (e.g., triangular or trapezoidal earth pressure distributions).

Misconceptions

• Forgetting to find intersection points first when limits are not given — randomly choosing limits gives wrong answers.
• Not checking which curve is on top — subtracting upper from lower gives a negative area.
• Evaluating F(b) + F(a) instead of F(b) − F(a).
• Thinking ∫₋₂² (−x²+4) dx gives the area when the function is negative somewhere — must check the sign of f(x) throughout.
• Confusing the area (always positive) with the net signed area (can be negative or zero for symmetric functions about the x-axis).

Related Concepts

• Fundamental Theorem of Calculus
• Finding Roots/Zeros of Functions (for intersection points)
• Volumes of Revolution (next logical step after area)
• Centroids by Integration (uses area as denominator)
• Riemann Sums (conceptual basis of definite integrals)

Common Exam Questions

Key Points To Remember

• Definite integral = F(b) − F(a); evaluate the antiderivative at the upper limit minus the lower limit.
• No constant of integration C in definite integrals — it cancels out.
• Area is always positive; if you get a negative value, you have the upper and lower curves switched.
• Find intersection points by setting the two functions equal — these become your limits of integration.
• For area problems, always sketch the region (even a rough sketch) to identify upper and lower curves.
• When curves intersect within the interval, split the integral at each crossing point.
• For integration with respect to y, subtract left curve from right curve.
• Check: ∫ₐᵇ f(x) dx gives net signed area — if the function dips below the x-axis, that portion is counted negative in the integral but positive in the physical area.

Examples

Applications

• Design of tank and vessel volumes in chemical/water treatment plant design.
• Computation of volumes of concrete piles or footings with circular cross-sections.
• Structural design: determining the volume (and hence mass/weight) of parabolic arch ribs or domes.
• Piping system design: computing volumes of tapered pipe sections.
• Computing section moduli of non-standard structural cross-sections through integration.

Misconceptions

• Forgetting to square the radius in the Disk/Washer method: writing V = π∫f(x) dx instead of π∫[f(x)]² dx.
• Using the Disk formula when there is a hole — must use the Washer formula (subtract inner radius squared).
• Confusing which method to use: Shell (2π∫x·f dx) versus Disk (π∫f² dx) — Shell does NOT square the function.
• Not switching outer and inner radii correctly in the Washer method.
• Forgetting the factor of 2π in the Shell method (writing π instead of 2π).
• Attempting to use Disk method about the y-axis without expressing x as a function of y — the Shell method is usually far simpler in these cases.

Related Concepts

• Area Between Curves (forms the basis of both methods)
• Cross-sectional area formulas (circle area = πr²)
• Integration techniques (Power Rule, substitution)
• Centroids of Solids of Revolution (Pappus' Theorem)
• Washer vs. Disk — special case of washer where inner radius = 0

Common Exam Questions

Key Points To Remember

• Disk: V = π∫[R(x)]² dx — the radius is squared, then multiplied by π (cross-section is a circle).
• Washer: V = π∫{[R(x)]² − [r(x)]²} dx — outer radius squared minus inner radius squared.
• Shell: V = 2π∫x·f(x) dx — circumference × height × thickness, no squaring.
• Disk/Washer method integrates perpendicular to the axis of revolution.
• Shell method integrates parallel to the axis of revolution.
• For revolution about the y-axis using disks, rewrite x = g(y) and integrate ∫[g(y)]² dy with respect to y.
• When revolving about a line y = k (not the x-axis), the radius is R(x) = f(x) − k (adjust accordingly).
• Always square the radius — the most common board exam error is forgetting to square R.

Examples

Applications

• Structural engineering: locating the neutral axis of composite beams, which requires the centroid of the cross-section.
• Determining shear center for thin-walled open sections (critical for torsion design per NSCP 2015).
• Computing section modulus S = I/c, where I (moment of inertia) is found by integration.
• Soil mechanics: centroid of pressure diagram determines the point of action of resultant lateral earth pressure on retaining walls.
• Hydraulics: centroid of submerged plane surface determines the location of the force resultant for gate design.
• Arc length computations for curved bridge alignments, vertical curves in highway design.

Misconceptions

• Using y instead of y/2 as the y-coordinate of the vertical strip's centroid when computing ȳ — gives double the correct answer.
• Confusing Mₓ (first moment about x-axis) with Iₓ (second moment / moment of inertia about x-axis).
• Forgetting that the centroid formulas divide by total area A, not by the integral of x or y alone.
• In Pappus' Theorem, using the wrong distance — it must be the distance from the CENTROID of the area to the axis of revolution.
• In arc length, forgetting to add 1 under the square root: writing ∫√(dy/dx)² dx instead of ∫√(1+(dy/dx)²) dx.

Related Concepts

• Area Between Curves (provides the denominator A in centroid formulas)
• Volumes of Revolution (Pappus' Theorem links centroids to volumes)
• Moment of Inertia of Sections (structural design — NSCP 2015 steel and concrete design use I values)
• Parallel Axis Theorem (extends integration-derived I to translated axes)
• Statics — Center of Gravity and Center of Mass

Common Exam Questions

Key Points To Remember

• For centroid ȳ, the y-coordinate of the strip centroid is y/2 (midpoint of vertical strip), NOT y — this is the most common board exam error.
• x̄ = Mᵧ/A and ȳ = Mₓ/A — first moments divided by total area.
• Always compute the total area A first, then compute the moments.
• Moment of inertia Iₓ = ∫y³/3 dx for vertical strips (derived from integrating y² over the strip height).
• For symmetric figures about an axis, the centroid lies on that axis — exploit symmetry to save computation.
• Pappus' Theorem: V = 2π·d̄·A where d̄ is the distance from the centroid to the axis of revolution.
• Arc length formula: L = ∫√(1+(dy/dx)²) dx — compute dy/dx first, then substitute.
• Centroids appear in structural engineering when computing shear center, neutral axis location, and section properties.

Problem

PROBLEM 1 (Basic Integration): Evaluate ∫₁³ (2x + 1) dx.

Solution

Antiderivative: F(x) = x² + x Apply Fundamental Theorem: F(3) − F(1) = (9 + 3) − (1 + 1) = 12 − 2 = 10 Answer: 10

Problem

PROBLEM 2 (U-Substitution): Evaluate ∫(3x²)(x³+5)⁴ dx.

Solution

Let u = x³ + 5 du = 3x² dx ← this matches exactly! ∫u⁴ du = u⁵/5 + C Back-substitute: = (x³+5)⁵ / 5 + C

Problem

PROBLEM 3 (Area Between Curves): Find the area enclosed by y = x+2 and y = x².

Solution

Step 1 — Intersections: x+2 = x² → x²−x−2 = 0 → (x−2)(x+1) = 0 → x = −1, x = 2 Step 2 — Determine upper curve (check x=0): y=x+2 gives 2; y=x² gives 0. Line is above parabola. Step 3 — Integrate: A = ∫₋₁² [(x+2) − x²] dx = [x²/2 + 2x − x³/3]₋₁² At x = 2: 4/2 + 4 − 8/3 = 2 + 4 − 8/3 = 6 − 8/3 = 10/3 At x = −1: 1/2 − 2 + 1/3 = 3/6 − 12/6 + 2/6 = −7/6 A = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2 = 4.5 sq. units

Problem

PROBLEM 4 (Volume — Disk): The region bounded by y = √x, x = 4, and the x-axis is revolved about the x-axis. Find the volume.

Solution

R(x) = √x (radius of disk) V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π [x²/2]₀⁴ = π (16/2) = 8π ≈ 25.13 cubic units

Problem

PROBLEM 5 (Volume — Washer): Find the volume when the region between y = x and y = x² (0 to 1) is revolved about the x-axis.

Solution

Outer radius R(x) = x (line is above parabola between 0 and 1) Inner radius r(x) = x² V = π ∫₀¹ [(x)² − (x²)²] dx = π ∫₀¹ [x² − x⁴] dx = π [x³/3 − x⁵/5]₀¹ = π (1/3 − 1/5) = π (5/15 − 3/15) = π (2/15) = 2π/15 ≈ 0.419 cubic units

Problem

PROBLEM 6 (Volume — Shell): Using the Shell Method, find the volume generated by revolving y = x², from x=0 to x=1, about the y-axis.

Solution

Shell radius = x Shell height = f(x) = x² V = 2π ∫₀¹ x · x² dx = 2π ∫₀¹ x³ dx = 2π [x⁴/4]₀¹ = 2π (1/4) = π/2 ≈ 1.571 cubic units

Problem

PROBLEM 7 (Centroid): Find the centroid (x̄, ȳ) of the triangular area bounded by y = 2x, x = 2, and y = 0.

Solution

Step 1 — Total Area (using integration): A = ∫₀² 2x dx = [x²]₀² = 4 sq. units (Check: triangle with base=2, height=4: A=(1/2)(2)(4)=4 ✓) Step 2 — x̄: Mᵧ = ∫₀² x · 2x dx = ∫₀² 2x² dx = [2x³/3]₀² = 16/3 x̄ = Mᵧ/A = (16/3)/4 = 4/3 ≈ 1.333 Step 3 — ȳ: Mₓ = ∫₀² (y/2)·y dx = ∫₀² (2x)²/2 dx = ∫₀² 2x² dx = [2x³/3]₀² = 16/3 ȳ = Mₓ/A = (16/3)/4 = 4/3 ≈ 1.333 Centroid: (4/3, 4/3) Note: For a triangle, the centroid is at (2/3 × base, 1/3 × height) from vertex at origin: x̄ = (2/3)(2) = 4/3 ✓, ȳ = (1/3)(4) = 4/3 ✓

Problem

PROBLEM 8 (Integration by Parts): Evaluate ∫x·eˣ dx.

Solution

Choose by LIATE: Let u = x → du = dx Let dv = eˣ dx → v = eˣ ∫x·eˣ dx = uv − ∫v du = x·eˣ − ∫eˣ dx = xeˣ − eˣ + C = eˣ(x − 1) + C Verification: d/dx[eˣ(x−1)] = eˣ(x−1) + eˣ(1) = eˣ(x−1+1) = xeˣ ✓