Math for ME · Chapter 5 of 19 · Beginner
Single-Variable Calculus: Integrals
Work, impulse, mass, energy, and resultants of distributed loads: integration is how engineers add up infinitely many small contributions.
The thread: A derivative breaks change into instants. Integration runs the other way, adding the instants back into a total: distance from velocity, work from force, a resultant from a distributed load.
Readiness check
From Derivatives. Tick only what you can do closed-notes.
- Differentiate polynomials, exponentials, and sin/cos fluently.
- Apply the chain rule without prompting.
- Compute areas of rectangles and triangles instantly.
- Read a definite quantity off a graph as "area under the curve".
- Keep units through multi-step calculations.
The core idea
An integral is a sum of slices: area, accumulation, total effect.
W = ∫F(x)dx∫xⁿdx = xⁿ⁺¹/(n+1) + CThe fundamental theorem links the two halves of calculus: integration undoes differentiation. For engineers the definite integral is the star: it turns a varying quantity (force, load, flow) into a total (work, resultant, volume).
The skills, taught in order
5.1 Two readings: area and accumulation
A definite integral has two equivalent meanings. Geometrically it is the signed area between a curve and the axis. Physically it is the total accumulated from a rate: distance from velocity, charge from current, work from force. Both come from one idea, summing slices F(x)·dx and letting the slice width shrink to zero.
5.2 Antiderivatives and the fundamental theorem
Integration is differentiation run backwards. The fundamental theorem says a definite integral is found by evaluating an antiderivative at the two limits and subtracting. The basic antiderivatives mirror the derivative table:
| ∫ f dx | xⁿ (n ≠ −1) | 1/x | ex | sin x | cos x |
|---|---|---|---|---|---|
| result | xn+1/(n+1) | ln|x| | ex | −cos x | sin x |
Every indefinite integral carries a +C; a definite integral does not, because the constant cancels on subtraction.
5.3 Substitution: the chain rule in reverse
When the integrand contains a function together with its own derivative, substitution untangles it. Set u = g(x), so that du = g′(x) dx, and rewrite the whole integral, limits included, in terms of u. The most common error is changing x to u but leaving dx unchanged.
5.4 Definite integrals build engineering totals
Most engineering use follows one pattern: name the physical meaning of a single slice, then integrate it. A few standard slices:
| Total | Integral |
|---|---|
| Work by a varying force | W = ∫ F dx |
| Impulse of a force over time | J = ∫ F dt |
| Resultant of a distributed load | R = ∫ w dx |
| Distance from velocity | s = ∫ v dt |
5.5 Signed area and numerical integration
The integral counts area below the axis as negative, so net accumulation and total magnitude differ when the curve crosses zero; split at the crossings when you need magnitude. When a function has no clean antiderivative, the same area is estimated numerically with trapezoids or Simpson's rule, the subject of Numerical Methods.
Engineering connection: Statics (distributed loads, centroids), Dynamics (impulse), Mechanics of Materials, Thermodynamics (boundary work).
Worked example: resultant of a triangular load, by integration
A beam carries a distributed load that grows linearly: w(x) = 200x N/m from x = 0 to x = 3 m. Find the total force and where it acts. (Statics Module 4 quoted this result; now prove it.)
- ProblemFind the resultant force of the load in Figure 1 and its line of action.
- Given / findw(x) = 200x N/m, 0 ≤ x ≤ 3 m. Find R and x̄.
- AssumptionsLoad intensity varies only with x; the beam is rigid.
- ModelOne slice at position x carries force w(x)·dx and moment x·w(x)·dx about the left end. Integrate both.
- EquationsR = ∫₀³ w dx x̄ = ∫₀³ x·w dx / R
- SolveR = ∫₀³ 200x dx = 200·x²/2 |₀³ = 100(9) = 900 N. ∫₀³ 200x² dx = 200·x³/3 |₀³ = 1800 N·m. So x̄ = 1800/900 = 2.0 m.
- CheckGeometry shortcut: triangle area = ½ × 3 × 600 = 900 N, centroid at 2/3 of 3 m = 2.0 m. Integration and geometry agree exactly.
- ConclusionThe "place the resultant at the centroid" rule used throughout Statics is not a recipe: it is this integral. When the load curve is not a clean triangle, the integral still works and the recipe does not.
Worked example 2: impulse changes momentum
A time-varying thrust F(t) = 12t N pushes a 2 kg cart along a track from rest, acting from t = 0 to t = 3 s. Find the impulse delivered and the cart's final speed.
- Given / findF(t) = 12t N, m = 2 kg, starts from rest, 0 ≤ t ≤ 3 s. Find the impulse J and the final speed v.
- ModelOne time slice delivers impulse F(t)·dt. Summing the slices is the integral, and impulse equals the change in momentum.
- EquationsJ = ∫₀³ F dt J = m Δv
- IntegrateJ = ∫₀³ 12t dt = 6t² |₀³ = 6(9) = 54 N·s.
- Final speedΔv = J/m = 54/2 = 27 m/s, and since it began at rest, v = 27 m/s.
- CheckUnits: N·s = kg·m/s, so dividing by kg gives m/s. The force grows to 36 N, averaging 18 N over 3 s, which is 54 N·s; the integral matches the average-force shortcut for this linear case.
- ConclusionImpulse is the time-integral of force, just as work is the distance-integral of force. The same slicing idea, integrated over a different variable, gives a different physical total.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Forgetting the limits | A definite question answered with "+C" | "Was I asked for a number or a function?" | Totals need limits. Evaluate at both ends and subtract. |
| Power rule applied to 1/x | ∫x⁻¹dx reported as x⁰/0 | "What is the one exception to the power rule?" | ∫dx/x = ln|x| + C. The n = −1 case is special. |
| Substitution without changing dx | Answers off by constant factors | "Did I replace dx along with x?" | If u = g(x), then du = g'(x)dx. Substitute both or substitute nothing. |
| Area below the axis ignored | Net totals wrong when the curve dips negative | "Do I want net accumulation or total magnitude?" | The integral counts signed area. Split at the zero crossings if you need magnitudes. |
Practice ladder
A spring force is F(x) = 300x² N. Find the work done stretching it from x = 0 to x = 2 m.
Show answer
W = ∫₀² 300x² dx = 100x³ |₀² = 800 J.
Then evaluate ∫₁⁴ (6x − 2) dx.
Show answer
= [3x² − 2x]₁⁴ = (48 − 8) − (3 − 2) = 40 − 1 = 39.
A vehicle's velocity is v(t) = 3t² m/s. How far does it travel between t = 1 s and t = 3 s?
Show answer
s = ∫₁³ 3t² dt = t³ |₁³ = 27 − 1 = 26 m. Accumulating a rate gives the total: the inverse of Derivatives.
A current i(t) = 4t A flows into a capacitor for 5 s. How much charge accumulates?
Show answer
Q = ∫₀⁵ 4t dt = 2t² |₀⁵ = 50 C. Charge is the time-integral of current, the same accumulation pattern.
Use substitution to evaluate ∫₀¹ 2t·et² dt, and state which derivative rule this reverses.
Show answer
u = t², du = 2t dt: ∫₀¹ eu du = e − 1 ≈ 1.718. Substitution is the chain rule run backwards.
Estimate the energy in your phone battery from a charging session: record charge percentage every 10 minutes, convert to power assumptions, and integrate numerically (trapezoids). Compare with the battery's rated watt-hours.
What good work looks like
A data table, the trapezoid sums shown, a final energy in Wh with stated assumptions, and a percent comparison to the rating with one error source named.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Write a half-page derivation, in your own words and notation, of why a distributed load's resultant acts at the centroid of the load diagram (the worked example, generalized to any w(x)). Add one worked non-triangular case, w(x) = 50x² over 0 to 2 m.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. State the fundamental theorem of calculus in engineering words.
The total accumulated from a rate equals the antiderivative evaluated between the limits: integration undoes differentiation.
2. What does ∫w(x)dx mean physically for a beam load?
The sum of force slivers w·dx: the total force of the distributed load.
3. Power rule for integrals, and its exception?
∫xⁿdx = xⁿ⁺¹/(n+1) + C for n ≠ −1; the exception is ∫dx/x = ln|x| + C.
4. Which derivative rules do substitution and integration by parts reverse?
Substitution reverses the chain rule; integration by parts reverses the product rule.
5. Velocity is the derivative of position. State the integral counterpart for impulse and momentum.
Impulse = ∫F dt = change in momentum. Accumulating force over time changes motion: Dynamics in one line.
Textbook mapping
| Item | Mapping |
|---|---|
| Main sources | Stewart, Calculus: Early Transcendentals (integration and applications chapters) |
| Core topics | 5.1 Area · 5.2 Accumulation · 5.3 Antiderivatives · 5.4 Definite and indefinite · 5.5 Substitution · 5.6 Parts (basic) · 5.7 Numerical intuition · 5.8 Work, impulse, mass, energy, area moments |
| Engineering connection | Statics Modules 4 and 9 (distributed loads, centroids), Dynamics (impulse-momentum), Thermodynamics (boundary work), Mechanics of Materials. |
| Skip on first pass | Trig-substitution zoo, partial fractions beyond the simplest case, improper-integral theory. |
| Read next | Sequences, Series, and Taylor Approximation. |