CELE Engineering Mathematics — Integral CalculusRevision Notes
Condensed revision notes for Integral Calculus, built for the final weeks before the CELE 2026. These are the distilled key points you need when there is no time left for full study notes — just the concepts, formulas, and traps Professional Regulation Commission (PRC) — Board of Civil Engineering tests.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Engineering Mathematics subtest is marked as "Core" in the official pattern, and Integral Calculus appears in position 6th of 10 in the CELE Engineering Mathematics review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Integral Calculus - Revision Notes
Integral Calculus is one of the most heavily tested topics in the Engineering Mathematics component of the PRC Civil Engineer Licensure Examination. Integration is the inverse operation of differentiation and serves as the mathematical engine behind computing areas, volumes, centroids, moments of inertia, arc lengths, and work — all of which appear directly in structural, hydraulic, and geotechnical engineering applications. These revision notes cover the four major exam clusters: (1) Basic Integration Rules and Techniques, (2) Definite Integrals and Areas Between Curves, (3) Volumes of Revolution, and (4) Centroids and Moments. Each section is packed with board-style worked examples, critical formulas, common pitfalls, and exam strategy tips to maximize your score on examination day.
Sections
Formulas
Example
∫3x² dx = 3·(x³/3) + C = x³ + C. Verify by differentiating: d/dx(x³) = 3x². ✓
Formula
∫xⁿ dx = xⁿ⁺¹/(n+1) + C, n ≠ −1
Variables
x = variable of integration; n = any real exponent except −1; C = constant of integration
Application
Evaluating polynomial, radical, and reciprocal-power integrals — the single most frequently used formula in the board exam.
Example
∫(2/x)dx = 2 ln|x| + C. At x = 1: 2 ln 1 = 0; at x = e: 2 ln e = 2. ✓
Formula
∫dx/x = ln|x| + C
Variables
x = variable; absolute value bars are required because ln is defined only for positive arguments
Application
Arises whenever n = −1; also appears after u-substitution when du/u is produced.
Example
∫x eˣ dx: let u = x, dv = eˣdx → du = dx, v = eˣ. Then = x·eˣ − ∫eˣdx = xeˣ − eˣ + C = eˣ(x−1) + C.
Formula
∫u dv = uv − ∫v du
Variables
u = chosen function (differentiated); dv = remaining differential (integrated); v = antiderivative of dv
Application
Products of unlike function families: x·eˣ, x·sin x, x²·ln x, etc.
Example
∫₀^(π/2) cos x dx = [sin x]₀^(π/2) = sin(π/2) − sin(0) = 1 − 0 = 1.
Formula
∫sin x dx = −cos x + C; ∫cos x dx = sin x + C
Variables
x = angle in radians
Application
Direct integration of sinusoidal functions; foundation for trig-substitution problems.
Exam Tips
- Always verify your antiderivative by differentiating your answer — this takes 10 seconds and confirms correctness.
- When you see a composite function (e.g., (2x+3)⁵), immediately attempt u-substitution with u = inner function.
- For integrals of the form ∫sin²x dx or ∫cos²x dx, apply the half-angle identity first before integrating.
- Memorize the five standard integrals (power, ln, eˣ, sin, cos) perfectly — they appear as sub-steps in every application problem.
- In multiple-choice format, substitute x = 1 (or another simple value) into both your answer and the answer choices to eliminate wrong options quickly.
Key Points
- Integration reverses differentiation: if d/dx[F(x)] = f(x), then ∫f(x)dx = F(x) + C, where C is the constant of integration — ALWAYS include C for indefinite integrals.
- The Power Rule ∫xⁿdx = xⁿ⁺¹/(n+1) + C is valid for all n ≠ −1; the exception ∫dx/x = ln|x| + C must be memorized separately.
- Standard trigonometric integrals: ∫sin x dx = −cos x + C and ∫cos x dx = sin x + C; note the sign change for sine.
- Exponential integral: ∫eˣdx = eˣ + C; for ∫eᵃˣdx = (1/a)eᵃˣ + C.
- Substitution (u-substitution) is the most tested technique: let u = inner function, compute du, convert everything to u, integrate, then back-substitute.
- Integration by Parts formula: ∫u dv = uv − ∫v du. Use the LIATE priority rule to select u: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential.
- Partial Fractions: decompose a rational function P(x)/Q(x) (degree of P < degree of Q) into simpler fractions before integrating.
- Trigonometric identities frequently simplify integrals: sin²x = (1−cos2x)/2 and cos²x = (1+cos2x)/2 are essential.
Definitions
Term
Indefinite Integral
Definition
The family of all antiderivatives of f(x), written ∫f(x)dx = F(x) + C, representing a general solution with an arbitrary constant C.
Importance
Baseline concept — every definite integral and application formula is built on antidifferentiation.
Term
Constant of Integration (C)
Definition
An arbitrary real constant appended to every indefinite integral result because differentiation of any constant gives zero.
Importance
Omitting C in an indefinite integral is a direct deduction in board-exam graded problems.
Term
U-Substitution
Definition
A technique where a new variable u replaces a composite inner function to simplify the integrand. The differential du must replace dx completely before integrating.
Importance
Solves the majority of composite-function integrals in the licensure exam.
Term
LIATE Rule
Definition
A mnemonic for selecting u in integration by parts: Logarithmic > Inverse trigonometric > Algebraic > Trigonometric > Exponential — the first applicable type in the product becomes u.
Importance
Prevents choosing the wrong u, which leads to a more complex remaining integral.
Section Title
Basic Integration Rules and Techniques
Common Mistakes
- Omitting the constant of integration C in indefinite integrals — automatic point deduction.
- Applying the power rule to n = −1: ∫x⁻¹dx ≠ x⁰/0 — this is undefined; the correct answer is ln|x| + C.
- Forgetting to back-substitute after u-substitution: leaving the answer in terms of u instead of the original variable x.
- Sign error in ∫sin x dx: the result is −cos x + C, not +cos x + C.
- In integration by parts, choosing u as the exponential function when an algebraic or logarithmic term is present — violates LIATE and complicates the integral.
- Not fully converting the integral to u after substitution: mixing x and u terms in the integrand.
Formulas
Example
∫₁³ (2x+1) dx = [x²+x]₁³ = (9+3) − (1+1) = 12 − 2 = 10.
Formula
∫ₐᵇ f(x) dx = F(b) − F(a)
Variables
a, b = lower and upper limits of integration; F(x) = antiderivative of f(x)
Application
Evaluating any definite integral; computing exact area under a curve.
Example
Area between y = x and y = x² from 0 to 1: A = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6 ≈ 0.167 sq. units.
Formula
A = ∫ₐᵇ [f(x) − g(x)] dx
Variables
f(x) = upper curve; g(x) = lower curve; a, b = x-coordinates of intersection or given limits
Application
Computing the area of a region enclosed between two curves — the most common board-exam area problem type.
Example
For x = y² (left) and x = y+2 (right): intersect at y²=y+2 → y = −1 and y = 2. A = ∫₋₁² (y+2−y²)dy = [y²/2+2y−y³/3]₋₁² = (2+4−8/3)−(1/2−2+1/3) = 4.5 sq. units.
Formula
A = ∫_c^d [h(y) − k(y)] dy
Variables
h(y) = right boundary expressed in y; k(y) = left boundary expressed in y; c, d = y-limits
Application
Horizontal strip method — used when curves are more naturally expressed as x = f(y).
Exam Tips
- Always sketch a rough graph of both curves and mark intersection points before setting up the integral — 30 seconds of sketching prevents major errors.
- Check: if the computed area is negative, you reversed the upper and lower curves — swap and re-evaluate.
- For parabola-vs-line problems (extremely common in board exams), the intersection points are the limits; area = ∫(line − parabola)dx in most cases.
- When limits are given (e.g., 0 to 2), verify which curve is above the other by testing a midpoint x-value.
- Memorize the result: area under y = 4 − x² above x-axis = ∫₋₂² (4−x²)dx = 32/3 ≈ 10.67 — this specific problem type recurs in Philippine board exams.
Key Points
- The Fundamental Theorem of Calculus: ∫ₐᵇ f(x)dx = F(b) − F(a), where F is any antiderivative of f. No constant C needed.
- Geometrically, the definite integral gives the net signed area between f(x) and the x-axis from x = a to x = b.
- Area is always positive: if f(x) < 0 on [a,b], take the absolute value or integrate |f(x)|.
- Area between two curves (upper minus lower): A = ∫ₐᵇ [f(x) − g(x)] dx, where f(x) ≥ g(x) on [a,b].
- Find intersection points first by solving f(x) = g(x) — these become the limits of integration when not given.
- If the curves switch positions (one becomes upper, then lower), split the integral at the crossing point and add absolute values.
- Horizontal strip method: sometimes integrating with respect to y (with rightmost minus leftmost functions expressed in x = h(y)) is simpler for sideways parabolas or regions bounded by y-axis.
- Properties: ∫ₐᵇ f dx = −∫ᵦᵃ f dx; ∫ₐᵃ f dx = 0; ∫ₐᶜ f dx = ∫ₐᵇ f dx + ∫ᵦᶜ f dx.
Definitions
Term
Definite Integral
Definition
The limit of a Riemann sum ∑f(xᵢ*)Δx as Δx→0, evaluated exactly as F(b)−F(a). It represents the net signed area.
Importance
Foundation of all area, volume, centroid, and moment calculations in engineering.
Term
Limits of Integration
Definition
The values a (lower) and b (upper) that define the interval over which integration is performed.
Importance
Incorrect limits — especially when not finding intersection points first — is the top source of wrong answers in area problems.
Term
Net Signed Area vs. Total Area
Definition
Net signed area counts regions below the x-axis as negative; total area uses |f(x)| to make all contributions positive.
Importance
Board exams almost always ask for total (positive) area — integrate |upper − lower| to ensure positivity.
Section Title
Definite Integrals and Area Between Curves
Common Mistakes
- Not finding intersection points first — using arbitrary or wrong limits of integration produces a completely incorrect area.
- Subtracting in the wrong order (lower minus upper instead of upper minus upper) — gives a negative area.
- Failing to split the integral when curves cross within the interval — underestimates the true enclosed area.
- Applying the Fundamental Theorem without computing F(b) − F(a) correctly — arithmetic errors in substituting limits.
- Forgetting to set g(x) = 0 when the lower curve is the x-axis — treat it explicitly as y = 0.
Formulas
Example
Revolve y = x² from 0 to 1 about the x-axis: V = π∫₀¹ (x²)² dx = π∫₀¹ x⁴ dx = π[x⁵/5]₀¹ = π/5 ≈ 0.628 m³.
Formula
V_disk = π ∫ₐᵇ [R(x)]² dx
Variables
R(x) = radius of disk = distance from curve to axis of revolution (x-axis); a, b = limits
Application
Volume of a solid formed by rotating a curve about the x-axis where no hole exists (region touches the axis).
Example
Region between y = √x (outer) and y = x² (inner), 0 to 1, revolved about x-axis: V = π∫₀¹ [(√x)² − (x²)²] dx = π∫₀¹ [x − x⁴] dx = π[x²/2 − x⁵/5]₀¹ = π(1/2 − 1/5) = 3π/10 ≈ 0.942 m³.
Formula
V_washer = π ∫ₐᵇ {[R(x)]² − [r(x)]²} dx
Variables
R(x) = outer radius (farther curve from axis); r(x) = inner radius (closer curve to axis); a, b = limits
Application
Volume of a hollow solid of revolution — region is between two curves, creating a washer cross-section.
Example
Revolve y = x², 0 to 2, about y-axis: V = 2π∫₀² x·x² dx = 2π∫₀² x³ dx = 2π[x⁴/4]₀² = 2π(4) = 8π ≈ 25.13 m³.
Formula
V_shell = 2π ∫ₐᵇ x · f(x) dx
Variables
x = shell radius (distance from y-axis); f(x) = shell height = y = f(x); a, b = x-limits of region
Application
Revolution about the y-axis using vertical strips — avoids solving for x in terms of y.
Exam Tips
- Always write out the radius R(x) explicitly before squaring — this one step prevents the [R−r]² mistake.
- Quick check: disk volume about x-axis for y = f(x) from a to b must be π times the integral of f(x)² — if your result lacks π, something is wrong.
- For revolution about the y-axis, compare effort: shell method integrates directly in x; disk method requires solving for x = g(y) and re-expressing limits in y. Shell is usually faster.
- When the axis is NOT the x-axis or y-axis (e.g., y = 2), the radius becomes R(x) = 2 − f(x) (if below y = 2) or f(x) − 2 (if above). Redraw the axis explicitly.
- Memorize: volume of sphere = (4/3)πr³ — you can derive this by revolving a semicircle y = √(r²−x²) from −r to r using the disk method. If your formula gives this, your setup is correct.
Key Points
- A solid of revolution is formed by rotating a plane region about a line (axis of revolution).
- Disk Method (solid of revolution, about x-axis): V = π∫ₐᵇ [R(x)]² dx, where R(x) is the radius (distance from curve to axis).
- Washer Method (hollow solid, about x-axis): V = π∫ₐᵇ {[R(x)]² − [r(x)]²} dx, where R = outer radius, r = inner radius.
- Shell Method (about y-axis): V = 2π∫ₐᵇ x·f(x) dx. The integrand is (shell radius)×(shell height) = x·f(x).
- For revolution about y-axis using disk/washer, express x as a function of y and integrate with respect to y.
- For revolution about x-axis using shell, express y as a function of y (shell height is in y), integrate w.r.t. y.
- Choosing disk vs. shell: disk is natural when revolving about the x-axis with vertical strips; shell is natural when revolving about the y-axis with vertical strips — no need to re-express the function in terms of y.
- Always confirm the axis of revolution and whether the region generates a solid (disk) or hollow (washer) cross-section.
Definitions
Term
Disk Method
Definition
A method of computing the volume of a solid of revolution by summing infinitely thin circular disk slices, each with area πR², perpendicular to the axis of revolution.
Importance
Standard method for revolving a single curve about the x-axis — tested in nearly every board exam set.
Term
Washer Method
Definition
An extension of the disk method for hollow solids, subtracting the inner disk area πr² from the outer disk area πR² to get the washer area π(R²−r²).
Importance
Applied when the region to be revolved does not touch the axis, creating a hole in the solid.
Term
Shell Method
Definition
A method computing volume by summing infinitely thin cylindrical shells of radius x, height f(x), and thickness dx, giving dV = 2πx·f(x)dx.
Importance
Preferred over disk/washer when revolving about the y-axis with functions expressed in x — no need to invert the function.
Term
Axis of Revolution
Definition
The line about which the plane region is rotated to generate the solid — could be the x-axis, y-axis, or any horizontal/vertical line.
Importance
Determines which method (disk, washer, shell) is most convenient and how the radius is expressed.
Section Title
Volumes of Revolution
Common Mistakes
- Forgetting to square the radius in the disk/washer method — integrating R(x) instead of [R(x)]².
- In the washer method, squaring the difference instead of the difference of squares: [R − r]² ≠ R² − r².
- Using the shell method formula 2π∫x·f(x)dx but integrating with respect to y instead of x.
- Not identifying which curve is outer vs. inner in the washer method — setting up R and r incorrectly.
- Revolving about a shifted axis (e.g., x = 3 or y = −1) without adjusting the radius expression: R(x) = (3 − x) or R(y) = (y + 1), not simply x or y.
- Applying the disk formula when a washer is needed — ignoring the hollow region produces an overestimate.
Formulas
Example
Centroid of y = x² from 0 to 2: A = ∫₀² x²dx = 8/3. Mᵧ = ∫₀² x·x²dx = ∫₀² x³dx = 4. x̄ = 4/(8/3) = 3/2 = 1.5 m.
Formula
x̄ = (∫ₐᵇ x·f(x) dx) / (∫ₐᵇ f(x) dx)
Variables
x̄ = x-coordinate of centroid; f(x) = height of area element; limits a to b
Application
Finding the horizontal position of the centroid of an area under a curve — required in beam loading and structural analysis.
Example
Continuing above (y = x², 0 to 2): Mₓ = ∫₀² (1/2)(x²)²dx = (1/2)∫₀² x⁴dx = (1/2)(32/5) = 16/5. ȳ = (16/5)/(8/3) = (16/5)·(3/8) = 6/5 = 1.2 m.
Formula
ȳ = (∫ₐᵇ (1/2)[f(x)]² dx) / (∫ₐᵇ f(x) dx)
Variables
ȳ = y-coordinate of centroid; (1/2)[f(x)]² comes from integrating y_el·dA where y_el = f(x)/2 and dA = f(x)dx
Application
Finding the vertical position of the centroid — note the (1/2)[f(x)]² in the numerator, not just f(x).
Example
Length of y = (2/3)x^(3/2) from 0 to 3: f'(x) = x^(1/2). L = ∫₀³ √(1+x)dx = [(2/3)(1+x)^(3/2)]₀³ = (2/3)(8 − 1) = 14/3 ≈ 4.67 m.
Formula
L = ∫ₐᵇ √(1 + [f'(x)]²) dx
Variables
L = arc length; f'(x) = dy/dx = derivative of the curve; a, b = x-limits
Application
Length of a curved path — used in road/cable geometry and surveying computations.
Example
If centroid of a region is at x̄ = 3 and A = 4 m², volume revolved about y-axis = 2π(3)(4) = 24π ≈ 75.4 m³.
Formula
V = 2π·x̄·A (Pappus, about y-axis); V = 2π·ȳ·A (Pappus, about x-axis)
Variables
x̄, ȳ = centroid coordinates; A = area of the region
Application
Quick computation of volume of revolution once the centroid is known — excellent cross-check tool.
Exam Tips
- Always compute A first, then Mᵧ and Mₓ separately — never skip the area computation even if it seems obvious.
- The formula ȳ = [∫(1/2)(f(x))²dx] / [∫f(x)dx] is a direct shortcut — memorize it exactly in this form to avoid the y vs. y/2 error.
- Pappus's Theorem is your best cross-check: after computing a volume by disk/shell, verify using V = 2π·x̄·A if you know the centroid.
- For composite areas (T-beams, L-shapes), use the weighted average formula x̄ = ΣAᵢx̄ᵢ / ΣAᵢ — this is the discrete analogue of the integral centroid.
- If the problem gives a region between two curves, the total area A = ∫(f−g)dx, and Mᵧ = ∫x(f−g)dx, Mₓ = ∫(1/2)(f²−g²)dx.
Key Points
- The centroid (x̄, ȳ) is the geometric center of a plane area, equivalent to the center of gravity for a uniform density lamina.
- Total area: A = ∫ dA = ∫ₐᵇ f(x) dx (for area under a single curve above the x-axis).
- Moment about y-axis (first moment): Mᵧ = ∫ x dA = ∫ₐᵇ x·f(x) dx → centroid x̄ = Mᵧ/A.
- Moment about x-axis: Mₓ = ∫ y_el dA where y_el = f(x)/2 (centroid of the vertical strip) → ȳ = Mₓ/A.
- Critical insight: the centroid of a vertical strip of height y = f(x) is at y/2, NOT y. Using y instead of y/2 for Mₓ is the most common centroid error.
- Moment of Inertia (second moment of area): Iₓ = ∫ y² dA = ∫ₐᵇ [f(x)]³/3 dx (for vertical strip about x-axis).
- Arc length: L = ∫ₐᵇ √(1 + [f'(x)]²) dx — integration of the differential arc length element ds.
- Pappus's Theorem (shortcut): Volume = 2π·ȳ·A (revolution about x-axis) or V = 2π·x̄·A (about y-axis). Useful for verification.
Definitions
Term
Centroid
Definition
The geometric center of a plane region — the point (x̄, ȳ) where the area could be concentrated as a single point with equivalent first moments about any axis.
Importance
Used in computing resultant forces on surfaces, beam neutral axes, and moment arm calculations in structural engineering.
Term
First Moment of Area
Definition
The integral ∫x dA or ∫y dA — the product of area element and its distance from a reference axis. Dividing by total area gives the centroid coordinate.
Importance
Direct application in determining centroidal locations of composite sections in structural design (NSCP 2015 beam calculations).
Term
Second Moment of Area (Moment of Inertia)
Definition
I = ∫y² dA — measures resistance to bending. For a vertical strip of width dx and height y: dI = (1/3)y³dx.
Importance
Fundamental to beam bending stress calculation σ = Mc/I and deflection formulas — directly in NSCP 2015 structural analysis provisions.
Term
Arc Length
Definition
The actual length along a curve between two points, computed as L = ∫√(1+(dy/dx)²)dx.
Importance
Applied in cable geometry (catenary), road alignment design, and pipe-routing length calculations.
Section Title
Centroids and Moments by Integration
Common Mistakes
- Using y instead of y/2 as the y-centroid of a vertical strip: the strip's own centroid is at its midpoint y/2, so Mₓ = ∫(y/2)(y dx) = ∫(1/2)y² dx, NOT ∫y·y dx = ∫y² dx.
- Computing ȳ = ∫y dA / A using the wrong expression for y — students sometimes plug in f(x) directly without the 1/2 factor.
- In arc length, forgetting to differentiate f(x) before squaring — using f(x)² instead of [f'(x)]² inside the square root.
- Applying Pappus's Theorem for revolution about the x-axis using x̄ instead of ȳ — must use the centroid coordinate perpendicular to the axis.
- Setting up moment integrals without a common denominator of A — computing numerator alone and reporting it as the centroid.
Connections
- Differential Calculus ↔ Integral Calculus: Integration is the inverse of differentiation. The power rule in differentiation (d/dx[xⁿ] = nxⁿ⁻¹) directly gives the power rule for integration (∫xⁿdx = xⁿ⁺¹/(n+1)).
- Integral Calculus ↔ Structural Engineering: Moment of inertia I = ∫y²dA (second moment of area) is central to beam bending stress σ = Mc/I, referenced in NSCP 2015 Sections 502–506 for flexural member design.
- Integral Calculus ↔ Fluid Mechanics / Hydraulics: Hydrostatic force on a submerged surface F = γ∫y·dA where γ is unit weight of fluid — a direct application of first moment of area.
- Integral Calculus ↔ Geotechnical Engineering: Stress distribution under foundations involves integration of pressure distributions over contact areas; settlement computations involve ∫(σ/E)dz.
- Volumes of Revolution ↔ Geometry of Engineering Objects: Water tanks, pipes, domes, and retaining wall footings often have geometries that are solids of revolution — volume calculations use disk/washer/shell methods.
- Centroids ↔ Composite Sections in NSCP 2015: The discrete centroid formula x̄ = ΣAᵢx̄ᵢ/ΣAᵢ is the practical engineering version of the integral definition — used to locate neutral axes of T-beams and built-up steel sections.
- Arc Length ↔ Transportation/Route Surveying: Curve lengths in horizontal alignment design (circular and parabolic curves) are special cases of the arc length integral L = ∫√(1+(y')²)dx.
- Integration Techniques ↔ Differential Equations: U-substitution and integration by parts are the primary tools for solving separable and linear first-order ODEs — which appear in transient structural and hydraulic analysis.
Exam Strategy
For the PRC Civil Engineer Licensure Examination, Integral Calculus questions typically comprise 8–12 items per Mathematics set. Prioritize mastery in this order: (1) Basic definite integrals — fastest points, almost always solvable in under 2 minutes. (2) Area between curves — solve 4–5 practice problems per day using the upper-minus-lower template until the setup is automatic. (3) Volumes of revolution — the disk and washer methods are tested more frequently than the shell method; however, know the shell method for y-axis revolution to avoid inverting functions. (4) Centroids — the single most error-prone topic due to the y/2 factor; write the formula ȳ = [∫(1/2)f²dx]/[∫fdx] on your scratch paper at the start of every exam. Time management: spend no more than 3 minutes per item. If a volume or centroid problem exceeds 3 minutes, mark it and move on — return with remaining time. For multiple-choice items, use elimination by dimension: a volume must be in cubic units; an area in square units — choices with wrong dimensions are automatically wrong. Always verify antiderivatives by mental differentiation before substituting limits. Practice board-exam problems from the last 10 years (2014–2024) — the same problem archetypes recur with different numerical values, so pattern recognition is a powerful scoring strategy.
Quick Review Questions
Evaluate ∫₁³ (2x + 1) dx.
Antiderivative: F(x) = x² + x. Evaluate: F(3) − F(1) = (9+3) − (1+1) = 12 − 2 = 10. This is a straight application of the Fundamental Theorem.
Find the area enclosed between y = 4 − x² and the x-axis.
Set y = 0: 4 − x² = 0 → x = ±2. A = ∫₋₂² (4−x²)dx = [4x − x³/3]₋₂² = (8−8/3) − (−8+8/3) = 16 − 16/3 = 32/3.
The region under y = √x from x = 0 to x = 4 is revolved about the x-axis. Find the volume.
Disk method: V = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π[x²/2]₀⁴ = π(8) = 8π. Note: R(x) = √x, so R² = x.
What is the y-coordinate of the centroid of the area under y = x² from x = 0 to x = 2?
A = ∫₀² x² dx = 8/3. Mₓ = ∫₀² (1/2)(x²)² dx = (1/2)(32/5) = 16/5. ȳ = Mₓ/A = (16/5)/(8/3) = (16×3)/(5×8) = 48/40 = 6/5.
Evaluate ∫ x·eˣ dx using integration by parts.
LIATE: u = x (algebraic), dv = eˣdx. Then du = dx, v = eˣ. Apply formula: ∫x eˣ dx = x·eˣ − ∫eˣ dx = xeˣ − eˣ + C. Factor: eˣ(x−1) + C.
The region between y = x (upper) and y = x² (lower) from x = 0 to x = 1 is revolved about the y-axis using the shell method. Find the volume.
V = 2π∫₀¹ x(x − x²)dx = 2π∫₀¹ (x² − x³)dx = 2π[x³/3 − x⁴/4]₀¹ = 2π(1/3 − 1/4) = 2π(1/12) = π/6.
State the LIATE rule and identify u and dv for ∫ x² ln x dx.
LIATE: Logarithmic functions take priority over Algebraic. So u = ln x (Logarithmic) and dv = x² dx (Algebraic). This gives du = dx/x and v = x³/3, leading to a manageable ∫v du = ∫(x³/3)(dx/x) = ∫(x²/3)dx.
Find the volume generated by revolving the region between y = x² (inner) and y = x (outer) from 0 to 1, about the x-axis using the washer method.
R = x (outer), r = x² (inner). V = π∫₀¹ [x² − x⁴] dx = π[x³/3 − x⁵/5]₀¹ = π(1/3 − 1/5) = π(2/15) = 2π/15.
A region has centroid at x̄ = 2 and area A = 5 m². Using Pappus's Theorem, find the volume generated by revolving this region about the y-axis.
Pappus's Theorem (y-axis): V = 2π·x̄·A = 2π(2)(5) = 20π. This theorem avoids re-computing the integral if the centroid is already known.
What integral gives the arc length of y = (1/3)x³ from x = 0 to x = 3?
Arc length formula: L = ∫ₐᵇ √(1 + [f'(x)]²)dx. Compute f'(x) = x². Then [f'(x)]² = x⁴. So L = ∫₀³ √(1+x⁴)dx. (This integral has no closed form — board exams that ask this usually provide numerical answer choices.)
Ready to practise for the CELE 2026?
Super Tutor's AI review plan adapts to your weak areas and builds a weekly practice schedule around your target CELE exam date.