Integral Calculus - Detailed Explanation
Integral Calculus is one of the most heavily tested topics in the Engineering Mathematics component of the PRC Civil Engineer Licensure Examination. It is the mathematical inverse of differentiation and serves as the backbone for computing areas, volumes, centroids, moments of inertia, and arc lengths — all of which appear directly in structural analysis, fluid mechanics, and geotechnical engineering. Board exam questions typically test (1) evaluation of definite and indefinite integrals using standard formulas and techniques, (2) area between curves, (3) volumes of revolution by the disk/washer and shell methods, and (4) centroids by integration. Mastery of this chapter is non-negotiable: expect 8–15 items per board examination day covering these topics. This review is structured to build from fundamental integration rules through applied engineering problems, with board-style worked examples in every section.
Basic Integration Formulas and Techniques
Examples
Evaluate ∫(3x² − 4x + 5) dx.
Solution
∫(3x² − 4x + 5) dx
= 3·x³/3 − 4·x²/2 + 5x + C
= x³ − 2x² + 5x + C
Apply the Power Rule term by term. Each term's exponent increases by 1, and you divide by the new exponent. The constant 5 integrates to 5x (since 5 = 5x⁰ and ∫x⁰ dx = x).
Evaluate ∫x·cos x dx using integration by parts.
Solution
Let u = x → du = dx
Let dv = cos x dx → v = sin x
∫x cos x dx = uv − ∫v du
= x·sin x − ∫sin x dx
= x sin x − (−cos x) + C
= x sin x + cos x + C
By LIATE: x is Algebraic, cos x is Trigonometric — choose u = x (Algebraic comes first). After one application of integration by parts, the remaining integral ∫sin x dx is a standard formula.
Evaluate ∫(6x+3)/(x²+x−2) dx by partial fractions.
Solution
Step 1 — Factor denominator: x²+x−2 = (x+2)(x−1).
Step 2 — Decompose: (6x+3)/[(x+2)(x−1)] = A/(x+2) + B/(x−1)
→ 6x+3 = A(x−1) + B(x+2)
Step 3 — Solve: Set x=1: 9 = 3B → B=3. Set x=−2: −9 = −3A → A=3.
Step 4 — Integrate:
∫[3/(x+2) + 3/(x−1)] dx
= 3 ln|x+2| + 3 ln|x−1| + C
= 3 ln|(x+2)(x−1)| + C
Partial fractions break a complex rational function into simpler ones that can be integrated using the ln|x| formula. The cover-up method (substituting the root of each factor) quickly yields the constants A and B.
Integration (antidifferentiation) reverses differentiation. The fundamental rule for power functions is the Power Rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C, valid for n ≠ −1. For n = −1, ∫(1/x)dx = ln|x| + C. Standard transcendental integrals include ∫eˣ dx = eˣ + C, ∫sin x dx = −cos x + C, and ∫cos x dx = sin x + C. The constant of integration C must always be written for indefinite integrals — omitting it is a common board exam deduction.
Three major techniques extend the basic rules:
- U-SUBSTITUTION: When the integrand contains a composite function, let u = inner function, compute du, and rewrite entirely in terms of u. Example: ∫2x(x²+1)⁴ dx — let u = x²+1, du = 2x dx → ∫u⁴ du = u⁵/5 + C = (x²+1)⁵/5 + C.
- INTEGRATION BY PARTS: Used when the integrand is a product of two different function types (polynomial × exponential, polynomial × trigonometric, etc.). Formula: ∫u dv = uv − ∫v du. Choose u using LIATE order: Logarithmic, Inverse trig, Algebraic (polynomial), Trigonometric, Exponential. Example: ∫x eˣ dx — let u = x, dv = eˣ dx → du = dx, v = eˣ → xeˣ − ∫eˣ dx = xeˣ − eˣ + C = eˣ(x−1) + C.
- PARTIAL FRACTIONS: Used for rational functions P(x)/Q(x) where degree of P < degree of Q. Decompose into simpler fractions before integrating. Example: ∫dx/(x²−1) = ∫[1/2·1/(x−1) − 1/2·1/(x+1)]dx = (1/2)ln|x−1| − (1/2)ln|x+1| + C.
Additional trigonometric integrals frequently tested:
| ∫tan x dx = ln | sec x | + C |
|---|---|---|
| ∫sec x dx = ln | sec x + tan x | + C |
∫sec²x dx = tan x + C
∫1/(a²+x²) dx = (1/a)arctan(x/a) + C
∫1/√(a²−x²) dx = arcsin(x/a) + C
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.
Common 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
Example
∫(4x³ − 2/x + e²ˣ) dx = x⁴ − 2ln|x| + (1/2)e²ˣ + C
Approach
Apply Power Rule, trig formulas, or exponential formulas directly. Always add + C.
Example
∫cos(3x+1) dx — let u = 3x+1, du = 3 dx → (1/3)∫cos u du = (1/3)sin(3x+1) + C
Approach
Identify u = inner function, compute du/dx, substitute, integrate in terms of u, back-substitute.
Example
∫ln x dx — u = ln x, dv = dx → v = x → x ln x − ∫x·(1/x) dx = x ln x − x + C
Approach
Apply LIATE to choose u, write dv, compute v, apply ∫u dv = uv − ∫v du.
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.
Definite Integrals and Area Between Curves
Examples
Evaluate ∫₁³ (2x + 1) dx.
Solution
∫₁³ (2x + 1) dx = [x² + x]₁³
= (3² + 3) − (1² + 1)
= (9 + 3) − (1 + 1)
= 12 − 2
= 10
Apply Power Rule to get the antiderivative F(x) = x² + x, then evaluate at the upper limit (x=3) minus the lower limit (x=1). This is a direct application of the Fundamental Theorem of Calculus.
Find the area bounded by y = 4 − x² and the x-axis.
Solution
Step 1 — Find intersections with x-axis (y = 0):
4 − x² = 0 → x = ±2
So limits: a = −2, b = 2.
Step 2 — Confirm the parabola is above the x-axis:
At x = 0: y = 4 > 0. ✓ (upper curve is y = 4−x², lower is y = 0)
Step 3 — Integrate:
A = ∫₋₂² (4 − x²) dx
= [4x − x³/3]₋₂²
= [4(2) − 8/3] − [4(−2) − (−8)/3]
= [8 − 8/3] − [−8 + 8/3]
= 8 − 8/3 + 8 − 8/3
= 16 − 16/3
= 48/3 − 16/3
= 32/3 ≈ 10.67 sq. units
The parabola y = 4−x² opens downward with vertex at (0,4). It crosses the x-axis at x = ±2, which become the integration limits. Since the parabola lies above the x-axis throughout, subtract y=0 from y=4−x².
Find the area enclosed between y = x and y = x² (intersecting at x = 0 and x = 1).
Solution
Step 1 — Identify upper curve:
At x = 0.5: y = x gives 0.5; y = x² gives 0.25. So y = x is above y = x².
Step 2 — Integrate:
A = ∫₀¹ (x − x²) dx
= [x²/2 − x³/3]₀¹
= (1/2 − 1/3) − 0
= 3/6 − 2/6
= 1/6 ≈ 0.1667 sq. units
This is a classic board exam problem. The two curves y=x (line) and y=x² (parabola) intersect at x=0 and x=1. Between these limits, the line is always above the parabola, so f(x)=x is the upper curve and g(x)=x² is the lower curve.
The Fundamental Theorem of Calculus links antiderivatives to definite integrals:
∫ₐᵇ f(x) dx = F(b) − F(a)
where F is any antiderivative of f. The result is a number (no + C needed) representing the net signed area under f(x) from x = a to x = b.
KEY PROPERTIES:
- ∫ₐᵃ f(x) dx = 0
- ∫ₐᵇ f(x) dx = −∫ᵦᵃ f(x) dx
- ∫ₐᵇ [f(x) ± g(x)] dx = ∫ₐᵇ f(x) dx ± ∫ₐᵇ g(x) dx
- ∫ₐᵇ c·f(x) dx = c·∫ₐᵇ f(x) dx
AREA BETWEEN TWO CURVES:
If f(x) ≥ g(x) on [a, b], the area enclosed is:
A = ∫ₐᵇ [f(x) − g(x)] dx
This is always positive. The critical step is identifying which curve is on top (upper) and which is on bottom (lower). To find intersection points (the limits a and b when not given), set f(x) = g(x) and solve.
For curves expressed as x = f(y) (horizontal strips):
A = ∫꜀ᵈ [f(y) − g(y)] dy
where f(y) is the right curve and g(y) is the left curve, integrated with respect to y.
When the curves switch positions within the interval, split the integral at the crossing point:
A = ∫ₐᶜ [f(x)−g(x)] dx + ∫꜀ᵇ [g(x)−f(x)] dx
(always subtract smaller from larger in each sub-interval).
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).
Common 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
Example
∫₀π sin x dx = [−cos x]₀π = −cos π − (−cos 0) = −(−1)+1 = 2
Approach
Find antiderivative F(x), apply F(b) − F(a). Watch signs carefully.
Example
Area under y = x² from x = 0 to x = 3: ∫₀³ x² dx = [x³/3]₀³ = 9 sq. units
Approach
Set y = 0 to find limits, then integrate |f(x)| over the interval.
Example
Area between y = x+2 and y = x²: set x+2 = x² → x = −1, 2; A = ∫₋₁² (x+2−x²) dx = 9/2
Approach
Find intersections, determine upper/lower curves, integrate (upper − lower).
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.
Volumes of Revolution — Disk/Washer and Shell Methods
Examples
The region under y = x² from x = 0 to x = 1 is revolved about the x-axis. Find the volume.
Solution
Using the Disk Method:
R(x) = x² (radius of each disk)
V = π ∫₀¹ (x²)² dx
= π ∫₀¹ x⁴ dx
= π [x⁵/5]₀¹
= π (1/5 − 0)
= π/5
≈ 0.6283 cubic units
The function y = x² becomes the radius of each disk when revolved about the x-axis. The radius must be squared inside the integral — (x²)² = x⁴, NOT x². This is the most tested board exam setup for disk volumes.
The region between y = √x and y = x (from x = 0 to x = 1) is revolved about the x-axis. Find the volume using the Washer Method.
Solution
Check which is outer: at x = 0.25, √0.25 = 0.5 > 0.25 = x. So R(x) = √x and r(x) = x.
V = π ∫₀¹ {(√x)² − x²} dx
= π ∫₀¹ (x − x²) dx
= π [x²/2 − x³/3]₀¹
= π (1/2 − 1/3)
= π (1/6)
= π/6
≈ 0.5236 cubic units
The washer method subtracts the hole (inner radius squared) from the full disk (outer radius squared). Always verify which curve has a larger y-value to determine the outer radius. Note: (√x)² = x simplifies the calculation elegantly.
Using the Shell Method, find the volume when y = x² (from x = 0 to x = 2) is revolved about the y-axis.
Solution
Shell radius = x, shell height = f(x) = x²
V = 2π ∫₀² x · x² dx
= 2π ∫₀² x³ dx
= 2π [x⁴/4]₀²
= 2π (16/4)
= 2π (4)
= 8π
≈ 25.13 cubic units
The shell method is ideal when revolving about the y-axis and the function is given as y = f(x). The shell radius is x (distance from y-axis), and the shell height is f(x) = x². The product x·f(x) = x·x² = x³ is integrated. This avoids solving for x in terms of y.
When a planar region is revolved about an axis, it generates a three-dimensional solid. Two main methods compute the volume:
METHOD 1 — DISK METHOD (revolution about x-axis):
Stack thin circular disks perpendicular to the axis. Each disk has radius R(x) = f(x) and thickness dx.
V = π ∫ₐᵇ [R(x)]² dx
METHOD 2 — WASHER METHOD (hollow solid, revolution about x-axis):
When there is a hole (the region between two curves is revolved), each cross-section is a washer with outer radius R(x) = f(x) and inner radius r(x) = g(x).
V = π ∫ₐᵇ {[R(x)]² − [r(x)]²} dx
METHOD 3 — SHELL METHOD (revolution about y-axis):
Stack thin cylindrical shells with radius x, height f(x), and thickness dx.
V = 2π ∫ₐᵇ x·f(x) dx
For revolution about y-axis using the disk/washer method, express x as a function of y and integrate with respect to y:
V = π ∫꜀ᵈ [x(y)]² dy
CHOOSING THE METHOD:
- Revolving about x-axis, region bounded by a single curve and x-axis → Disk.
- Revolving about x-axis, region between two curves → Washer.
- Revolving about y-axis, function given as y = f(x) → Shell (easier, stays in x).
- Revolving about a line y = k or x = h → Shift the radius: R(x) = f(x) − k or R = h − x accordingly.
CRITICAL MEMORY AID:
- Disk/Washer: ∫ of radius² (perpendicular to axis, multiply by π)
- Shell: ∫ of radius × height (parallel to axis, multiply by 2π)
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.
Common 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
Example
y = √x from 0 to 4 revolved about x-axis: V = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π[x²/2]₀⁴ = 8π
Approach
V = π∫[f(x)]² dx. Square the function, multiply by π, integrate over the given limits.
Example
Region between y = x and y = x² (0 to 1) about x-axis: V = π∫₀¹(x²−x⁴) dx = π[x³/3−x⁵/5]₀¹ = 2π/15
Approach
Identify outer (R) and inner (r) radii, V = π∫(R²−r²) dx.
Example
y = x from 0 to 3 revolved about y-axis: V = 2π∫₀³ x·x dx = 2π∫₀³ x² dx = 2π[x³/3]₀³ = 18π
Approach
V = 2π∫x·f(x) dx. No need to square — multiply radius × height × 2π.
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.
Centroids, Moments, and Moment of Inertia by Integration
Examples
Find the centroid (x̄, ȳ) of the area under y = x² from x = 0 to x = 2.
Solution
Step 1 — Total Area:
A = ∫₀² x² dx = [x³/3]₀² = 8/3
Step 2 — Moment about y-axis (Mᵧ):
Mᵧ = ∫₀² x·x² dx = ∫₀² x³ dx = [x⁴/4]₀² = 4
x̄ = Mᵧ/A = 4 ÷ (8/3) = 4 × (3/8) = 3/2 = 1.5
Step 3 — Moment about x-axis (Mₓ):
Mₓ = ∫₀² (y/2)·y dx = ∫₀² (x²/2)·x² dx = ∫₀² x⁴/2 dx
= [x⁵/10]₀² = 32/10 = 16/5
ȳ = Mₓ/A = (16/5) ÷ (8/3) = (16/5) × (3/8) = 6/5 = 1.2
Centroid: (1.5, 1.2)
Note carefully: for Mₓ, the y-coordinate of the strip centroid is y/2 = x²/2, NOT x². The element moment is (y/2)·dA = (y/2)·y dx = y²/2 dx. Using y instead of y/2 would give ȳ = 2.4, which is wrong — this is the #1 error on board exams for centroid problems.
Find the arc length of y = x^(3/2) from x = 0 to x = 4.
Solution
Step 1 — Compute dy/dx:
y = x^(3/2) → dy/dx = (3/2)x^(1/2)
Step 2 — Set up arc length integral:
L = ∫₀⁴ √(1 + [(3/2)x^(1/2)]²) dx
= ∫₀⁴ √(1 + (9/4)x) dx
Step 3 — U-substitution:
Let u = 1 + (9/4)x → du = (9/4) dx → dx = (4/9) du
When x=0: u=1; when x=4: u=1+9=10
L = ∫₁¹⁰ √u · (4/9) du
= (4/9) · [u^(3/2) / (3/2)]₁¹⁰
= (4/9) · (2/3) [u^(3/2)]₁¹⁰
= (8/27) [10^(3/2) − 1]
= (8/27) [31.623 − 1]
= (8/27)(30.623)
≈ 9.073 units
Arc length requires computing the derivative first. The expression 1 + (dy/dx)² under the square root typically requires u-substitution for integration. The result (≈9.07 units) is greater than the straight-line distance from (0,0) to (4,8) = √(16+64) ≈ 8.94, which is a reasonable sanity check.
The centroid (geometric center) of a plane area A is the point (x̄, ȳ) where:
x̄ = (∫x dA) / (∫dA) = Mᵧ / A
ȳ = (∫yₑₗ dA) / (∫dA) = Mₓ / A
where Mᵧ is the first moment about the y-axis and Mₓ is the first moment about the x-axis.
For a VERTICAL STRIP of width dx and height y = f(x):
- dA = y dx
- x-coordinate of strip centroid = x (use x as the lever arm for Mᵧ)
- y-coordinate of strip centroid = y/2 (the midpoint of the strip's height, NOT y)
So the formulas become:
A = ∫ₐᵇ y dx
Mᵧ = ∫ₐᵇ x·y dx → x̄ = Mᵧ/A
Mₓ = ∫ₐᵇ (y/2)·y dx = ∫ₐᵇ (y²/2) dx → ȳ = Mₓ/A
For HORIZONTAL STRIPS (function expressed as x = g(y)):
dA = x dy, x̄ = (∫x/2·x dy)/(∫x dy), ȳ = (∫y·x dy)/(∫x dy)
MOMENT OF INERTIA (Second Moment of Area):
For vertical strips about the x-axis:
Iₓ = ∫y² dA = ∫ₐᵇ y³/3 dx
For vertical strips about the y-axis:
Iᵧ = ∫x² dA = ∫ₐᵇ x²·y dx
ARC LENGTH:
L = ∫ₐᵇ √(1 + [f'(x)]²) dx
This formula computes the actual length of a curve from x = a to x = b.
PAPPUS' THEOREM (powerful shortcut):
- Volume of solid of revolution = 2π × ȳ × A (revolving about x-axis)
- Volume of solid of revolution = 2π × x̄ × A (revolving about y-axis)
This connects centroids directly to volumes of revolution.
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.
Common 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
Example
Centroid x̄ under y = 3x from 0 to 2: A = ∫₀² 3x dx = 6; Mᵧ = ∫₀² x·3x dx = ∫₀² 3x² dx = 8; x̄ = 8/6 = 4/3
Approach
Compute A = ∫y dx, then Mᵧ = ∫x·y dx, then x̄ = Mᵧ/A.
Example
ȳ under y = 2x from 0 to 1: A = 1; Mₓ = ∫₀¹ (2x)²/2 dx = ∫₀¹ 2x² dx = 2/3; ȳ = 2/3
Approach
Compute A = ∫y dx, then Mₓ = ∫(y²/2) dx, then ȳ = Mₓ/A. Use y/2 for strip centroid y-coordinate.
Example
A triangle with centroid 3 cm from y-axis, area = 6 cm² revolved about y-axis: V = 2π(3)(6) = 36π cm³
Approach
Find centroid of the region first, then apply V = 2π × (distance of centroid from axis) × A.
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.
Practice problems
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
Integrate term by term using the Power Rule: ∫2x dx = x², ∫1 dx = x. Evaluate at upper limit (x=3) minus lower limit (x=1). No constant C needed for definite integrals.
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
Identify that the derivative of the inner function (x³+5) is 3x², which appears as a factor in the integrand. This perfect match indicates u-substitution. After substitution, the integral becomes ∫u⁴ du, a simple Power Rule application.
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
This is a classic board exam problem. Find intersections by setting the equations equal, then integrate (upper − lower) between the intersection points. A positive result (4.5) confirms the setup is correct.
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
The disk method is appropriate since a single curve defines the region and we revolve about the x-axis. The key simplification: (√x)² = x, making the integral straightforward. The result 8π is exact.
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
The washer method requires both an outer and inner radius. R² = x² and r² = x⁴. These are subtracted (not the radii, but the SQUARES of the radii) before integrating. Note: do not subtract R from r — subtract R² from R² inside the integral.
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
The Shell Method for revolution about the y-axis uses V = 2π∫x·f(x) dx. There is no squaring of f(x) — this is the key difference from the Disk Method. The shell radius is simply x (horizontal distance from y-axis).
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 ✓
Always use y/2 (not y) as the centroid y-coordinate of a vertical strip. The verification using the known triangle centroid formula confirms the integration result. By symmetry of this particular triangle, x̄ = ȳ.
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 ✓
By LIATE, x (Algebraic) is chosen as u because its derivative is simpler (du = dx). After one integration by parts, the remaining integral ∫eˣ dx is elementary. Always verify by differentiating the answer.
Exam preparation tips
- MASTER THE STANDARD FORMULAS FIRST: Memorize all basic integral formulas (Power Rule, trig, exponential, logarithmic) — at least 3–5 board exam items are direct applications of these with no technique required.
- SKETCH THE REGION ALWAYS: For area and volume problems, draw a quick sketch of the curves. This takes 30 seconds but prevents the most common error of inverting upper/lower or outer/inner curves.
- REMEMBER THE 'y/2 RULE': In centroid ȳ calculations with vertical strips, the element centroid is at y/2, making Mₓ = ∫(y/2)(y dx) = ∫y²/2 dx. Writing ∫y·y dx = ∫y² dx (without the 1/2) is one of the most common centroid errors in board exams.
- DISK vs SHELL CHECKLIST: If revolving about x-axis → Disk/Washer (integrate f(x)² in terms of x). If revolving about y-axis and function is given as y=f(x) → Shell is almost always easier (integrate x·f(x)).
- FIND INTERSECTION POINTS FIRST: For area-between-curves and volume problems, always find where the curves intersect to establish the limits of integration. Set f(x) = g(x) and solve — this is non-negotiable.
- VERIFY WITH DIFFERENTIATION: After evaluating an indefinite integral, differentiate your answer to confirm it matches the original integrand. This 10-second check can save you from choosing a wrong answer choice.
- WATCH SIGNS IN DEFINITE INTEGRALS: F(b) − F(a) — upper limit evaluates first, then subtract lower limit evaluation. Writing F(a) − F(b) flips the sign.
- USE PAPPUS' THEOREM WHEN CENTROIDS ARE GIVEN: If a problem gives you the centroid of a region and asks for the volume of revolution, use V = 2π·d̄·A immediately — no integration needed.
- PRACTICE MIXED-TYPE PROBLEMS: Board exams often combine techniques: e.g., u-substitution within a definite integral, or computing area then using it for centroid. Practice problems that require two integration techniques.
- TIME MANAGEMENT — ATTEMPT ORDER: Start with direct definite integral evaluations (fastest), then area problems (moderate), then volume problems (moderate), then centroid problems (most time-consuming). Flag difficult items and return to them.
- ANTI-DERIVATIVE CHECK LIST: For each problem, identify: (1) Is it indefinite or definite? (2) What technique is needed — direct, substitution, by parts, partial fractions? (3) After integrating, have I correctly applied limits or included +C?
- KNOW YOUR SPECIAL INTEGRALS: ∫sec²x dx = tan x + C; ∫tan x dx = ln|sec x| + C; ∫1/(a²+x²) dx = (1/a)arctan(x/a) + C; ∫1/√(a²−x²) dx = arcsin(x/a) + C — these appear in board exams without warning.
Diagrams
Integral Calculus — Complete Topic Overview
A comprehensive mind map showing all major topics in Integral Calculus: basic integration formulas and techniques, definite integrals, area between curves, three volume of revolution methods, and centroid/moment applications. Use this to see how all sub-topics connect before diving into individual concepts.
Volume of Revolution — Method Selection Decision Tree
A decision tree flowchart guiding students to select the correct volume of revolution method based on the axis of revolution and the geometry of the region. Includes the key formula reminder for each method and critical checkpoints to avoid the most common board exam errors.
Step-by-Step Procedure for Area Between Curves
A step-by-step procedural flowchart for solving area between curves problems. Includes the critical decision point of identifying which curve is the upper curve, plus a built-in verification check (the result must be positive). This mirrors the exact process a board exam solver should follow.
Centroid Computation — Step-by-Step Guide with Common Error Warning
A detailed flowchart for computing centroids of plane areas by integration. The diagram explicitly highlights the most critical board exam error: using y instead of y/2 for the y-coordinate of the vertical strip centroid when computing ȳ. The warning branch visually reinforces the correct procedure.
Integration Technique Selection Guide
A decision tree for selecting the correct integration technique based on the form of the integrand. Covers direct formula application, u-substitution, integration by parts (with LIATE reminder), and partial fractions with the polynomial long division prerequisite check. Essential for efficient board exam problem solving.
Distribution of Integral Calculus Topics in PRC Board Exams
An estimated pie chart showing the approximate distribution of Integral Calculus questions by topic type in PRC Civil Engineer Licensure Examinations, based on historical board exam analysis. Basic integration and techniques form the largest single category (30%), followed jointly by area and volume problems (25% each). This guides study prioritization.
In summary
Integral Calculus is a cornerstone of the Engineering Mathematics component of the PRC Civil Engineer Licensure Examination. The chapter builds logically from foundational integration rules → definite integrals and area → volumes of revolution → centroids and moments. Each level builds on the previous: you must be fluent in evaluating integrals before computing areas, and you must compute areas correctly before finding centroids.
The five most critical skills to develop for board exam success are:
- Flawless execution of the Power Rule, including the special case ∫(1/x)dx = ln|x| + C.
- Correct identification of upper vs. lower curves for area problems — always sketch first.
- The Disk/Washer distinction and the habit of squaring R inside the integral: V = π∫R² dx, NOT π∫R dx.
- The y/2 rule for centroid ȳ computations — the #1 most missed centroid fact on board exams.
- Verification by differentiation — always check your antiderivative by differentiating it back.
Approach board exam problems systematically: identify the problem type, recall the correct formula or technique, set up the integral carefully (especially the limits and the integrand), evaluate step by step, and verify the sign and magnitude of the answer. With consistent practice on the 8 board-style problems in this chapter and mastery of the decision trees provided, you will solve Integral Calculus items with speed and confidence on examination day. Good luck, Engineer!