Integrals and multivariable calculus

Accumulate a rate in small intervals

A derivative describes a local rate. An integral can accumulate that rate over an interval. If signed velocity is exactly constant at +2 m/s for 0.5 s, its displacement is (+2 m/s)(0.5 s) = +1 m. For a changing velocity, multiplying one chosen value by the whole elapsed time need not give the correct displacement.

Divide a time interval into small pieces of positive duration Δt. Choose a sample time in each piece, multiply the velocity there by that piece's duration, then add the signed products. The symbol Σ means add the indicated terms, so this weighted sum can be written Σ v(sample time) Δt. Each term has units (m/s)s = m. Such sums are called Riemann sums.

A finite sampled sum is an approximation in general. For a continuous rate on a closed finite interval, these sums approach a unique value as the largest piece width tends to zero. Continuous means nearby outputs approach the output at each point, using the available side at an interval endpoint. The mathematical refinement does not claim that an instrument resolves arbitrarily short times.

Read the definite integral

Write the accumulated value as ∫_a^b f(u) du, the definite integral of f from a to b. The function f is the integrand; u is the variable of integration; a and b are the bounds. The notation du marks the variable whose interval widths enter the sum. A fixed pair of bounds gives one value, carrying the units of f multiplied by the units of u.

For a ≤ b, positive function values give positive contributions and negative values give negative contributions. On a graph these correspond to the region above the horizontal axis minus the region below it. But the result is not automatically a physical area: a velocity–time graph accumulates metres because its axes have m/s and s units, rather than two length units.

Adjacent intervals can be added: ∫_a^b f(u) du = ∫_a^c f(u) du + ∫_c^b f(u) du when a ≤ c ≤ b and the function is integrable. Reversing the bounds reverses the sign, and equal bounds give zero. These are mathematical bound conventions. Reversing time bounds alone does not describe a different body's motion backward through time.

Separate displacement and distance

Supply an original exact model for one chosen point on a body along a straight coordinate line. Fix the origin, take right positive, and measure t from the declared start. The signed velocity is v(t) = 3 m/s − (1 m/s²)t, only for 0 ≤ t ≤ 5 s, including both endpoints. The rule applies at every time in this interval; the table and graph illustrate that stipulation. They are not measurements.

For one-dimensional motion, speed is the nonnegative magnitude |v(t)|. Net displacement is ∫_0s^5s v(t) dt; path distance is ∫_0s^5s |v(t)| dt. Here v = 0 at t = 3 s, with positive velocity before and negative velocity after. The triangular positive contribution is (1/2)(3 s)(3 m/s) = +4.5 m. The negative contribution is −(1/2)(2 s)(2 m/s) = −2 m.

The net displacement is +4.5 m − 2 m = +2.5 m. The path distance adds magnitudes, 4.5 m + 2 m = 6.5 m. In the supplied model the point first advances right, then returns partway left; its final position need not equal the distance traveled. Path distance is nonnegative, and could be zero for a point that remains at rest throughout an interval.

Find an antiderivative and its constant

An antiderivative F of f satisfies F′(u) = f(u) on the interval of interest. The prime marks the derivative. On a connected interval, adding a constant C changes no derivative, and the antiderivatives form the family F(u) + C. This is written ∫ f(u) du = F(u) + C, an indefinite integral without fixed bounds. It gives a family of functions, unlike a definite integral's single accumulated value.

For our velocity model, choose F(t) = (3 m/s)t − (1/2)(1 m/s²)t². Its derivative is 3 m/s − (1 m/s²)t. The elementary check uses the derivative of t as 1 and of t² as 2t: for h ≠ 0, [(t + h)² − t²]/h = 2t + h, which tends to 2t. A constant's derivative is zero. These checks concern interior times within the supplied domain.

Thus x(t) = F(t) + C has the supplied velocity, where C has metre units. To select one position history, also supply x(0) = −1 m. Since F(0) = 0 m, C = −1 m. Velocity alone would leave this reference-position constant undetermined. A change in the constant shifts all positions but leaves their differences unchanged.

Apply the fundamental theorem carefully

One form of the fundamental theorem connects accumulation with the local rate. If f is continuous on [a, b], define A(u) = ∫_a^u f(s) ds; then A′(u) = f(u) at interior a < u < b. The letter s is a temporary integration variable, distinct from the upper-bound input u. For each chosen u, A(u) is one definite-integral value; varying u gives an accumulation function.

For evaluation, use the following sufficient conditions: f is continuous on [a, b]; F is continuous on [a, b], differentiable in its interior, and F′ = f there. Then ∫_a^b f(u) du = F(b) − F(a). Differentiable means that the finite two-sided difference-quotient limit exists. These conditions suffice for our examples; they are not a claim that every integrable function must be continuous.

Our linear v is continuous over the stated interval and its polynomial F meets these conditions. F(5 s) = 15 m − 12.5 m = 2.5 m, and F(0) = 0 m. Their difference confirms the signed integral. The same C cancels from (F(b) + C) − (F(a) + C). With the supplied x(0) = −1 m, the final position is x(5 s) = −1 m + 2.5 m = +1.5 m.

Change one input of a field

Now switch to a separate original static sheet model. Use an ideal flat plane with perpendicular metre coordinates: origin at the sheet's lower-left corner, positive x right and positive y up. The supplied rectangular region R is 0 ≤ x ≤ 3 m and 0 ≤ y ≤ 2 m. Its surface mass density, meaning mass per unit area, is σ(x, y) = 2 kg/m² + (1 kg/m³)x + (0.5 kg/m³)y. The Greek letter σ names this field; each location supplies two inputs and one density output.

A partial derivative changes one input while keeping the other fixed. For an interior point, ∂σ/∂x = lim as h → 0 of [σ(x + h, y) − σ(x, y)]/h, when this finite two-sided limit exists; ∂ is the partial-derivative symbol. Here h is a length increment. The y term stays constant, so ∂σ/∂x = 1 kg/m³. Similarly, holding x fixed gives ∂σ/∂y = 0.5 kg/m³.

These units are (kg/m²)/m: change in surface mass density per coordinate length. They do not turn σ into a volume mass density merely because the written derivative unit is kg/m³. At the interior point (1 m, 1 m), σ = 3.5 kg/m². Both partial derivatives describe changes across this static field, not temporal change or motion of sheet material. The two-sided statements here use 0 < x < 3 m and 0 < y < 2 m; edge derivatives would need a stated one-sided convention or a domain extension.

Accumulate over a declared region

To accumulate mass across the sheet, divide R into small rectangles. Each has positive area ΔA = Δx Δy. A sample density times ΔA approximates that rectangle's mass when the density varies. Add these products and refine both coordinate widths toward zero. The limit is the double integral M = ∬_R σ(x, y) dA. The double integral marks accumulation over two inputs; M names the model's total mass.

The measure dA has square-metre units in these perpendicular Cartesian coordinates. Thus (kg/m²)m² = kg. It is a positive area measure for the stated region; choosing one face of the sheet does not assign its mass a positive or negative sign. Both coordinate integrations below run from their smaller to larger bounds. Other coordinate systems or curved regions require appropriate measures and bounds; the rectangular formula is not a universal replacement for those checks.

For a continuous function on a closed rectangle, the double integral can be evaluated as two successive, or iterated, integrals in either order. Our linear density is continuous and positive on R. Choosing x first, then y, gives M = ∫_0m^2m [∫_0m^3m σ(x, y) dx] dy. Read the inner dx first: integrate across x while holding y fixed. Then integrate the resulting function of y.

Check the accumulated quantity

Write the density as σ = B + D x + E y, with B = 2 kg/m², D = 1 kg/m³ and E = 0.5 kg/m³. For fixed y, an antiderivative with respect to x is (B + E y)x + (D/2)x². Evaluating from 0 to 3 m gives H(y) = 10.5 kg/m + (1.5 kg/m²)y. The inner result has units kg/m, since one length integration remains.

Integrating H from y = 0 to 2 m gives (10.5 kg/m)(2 m) + (1/2)(1.5 kg/m²)(2 m)² = 21 kg + 3 kg = 24 kg. This is exact for the supplied density model. The region's area is instead (3 m)(2 m) = 6 m². The same region can support different integral questions because the integrand determines what is accumulated.

Before interpreting any result, identify the quantity, its units, the interval or region, the sign convention and the theorem conditions used. Our supplied continuous models make these calculations possible. Exact arithmetic does not establish an actual body's velocity, a real sheet's density or measurement precision; those physical questions require evidence.

Worked examples

Try it yourself

1. A separate exact model supplies signed velocity −2 m/s throughout a 3 s interval on a line with right positive. Find its net displacement and path distance.

   Show the answer and reasoning

   Net displacement is −6 m; path distance is 6 m.

   A constant signed velocity integrates to (−2 m/s)(3 s) = −6 m. Speed is |-2 m/s| = 2 m/s, so its integral is 6 m. The negative sign belongs to displacement, not a negative distance.

2. For the supplied v(t) on 0–5 s, take five 1 s pieces and use left sample times 0, 1, 2, 3 and 4 s. Calculate the Riemann sum. Is it the exact signed integral?

   Show the answer and reasoning

   The sum is (3 + 2 + 1 + 0 − 1)(m/s)(1 s) = 5 m. It differs from the exact signed integral +2.5 m.

   Each sample represents its entire finite piece, while v changes across that piece. This left-sampled sum is an approximation. The stipulated continuous rule, together with the limiting definition or fundamental theorem, supplies the exact integral; the five samples alone do not.

3. For the supplied velocity, ∫_0s^5s v(t) dt = +2.5 m. What is ∫_5s^0s v(t) dt? If the same 7 m constant is added to F at both endpoints, does the forward definite integral change?

   Show the answer and reasoning

   The reversed-bounds integral is −2.5 m. Adding the same constant to F at both endpoints leaves the forward integral +2.5 m.

   Reversing bounds negates a definite integral. In [F(5 s) + 7 m] − [F(0) + 7 m], the added constants cancel. This bound reversal is a mathematical convention, not an extra physical trajectory.

4. For continuous f on [a, b], let A(u) = ∫_a^u f(s) ds. What does the stated fundamental theorem give at an interior a < u < b? Does it claim an ordinary two-sided derivative at an endpoint using only this domain?

   Show the answer and reasoning

   A′(u) = f(u) in the interior. The stated result does not claim an ordinary two-sided endpoint derivative from this domain alone.

   Continuity supplies the local accumulation-rate result. Interior inputs can approach from both sides. An endpoint has only one available side in the supplied domain, so any endpoint derivative must use an explicitly stated convention or a justified extension.

5. For the supplied σ(x, y), evaluate σ at the interior point (1 m, 1 m), and give ∂σ/∂x and ∂σ/∂y there. Does ∂σ/∂x hold y fixed?

   Show the answer and reasoning

   σ = 3.5 kg/m², ∂σ/∂x = 1 kg/m³ and ∂σ/∂y = 0.5 kg/m³. The x partial holds y fixed.

   The density is 2 kg/m² + 1 kg/m² + 0.5 kg/m². For the x partial, the y contribution is constant; for the y partial, the x contribution is constant. The derivative units describe density change per coordinate length, and do not relabel σ as volume mass density.

6. On the same 3 m by 2 m rectangle, a separate exact sheet model has uniform surface mass density 4 kg/m². Evaluate ∬_R 1 dA and ∬_R (4 kg/m²) dA. Which gives area and which gives mass?

   Show the answer and reasoning

   ∬_R 1 dA = 6 m² gives area; ∬_R (4 kg/m²) dA = 24 kg gives mass.

   The constant 1 is dimensionless, so its product with dA retains square-metre units. Density times the same area gives (4 kg/m²)(6 m²) = 24 kg. A region alone does not determine the quantity being accumulated; the integrand and units matter.

What to carry forward

• A definite integral accumulates weighted contributions over stated bounds; finite sampled sums are approximations in general.
• Retain signs and units: integrating a rate multiplies its units by the integration variable's units.
• In the stated line model, integrating velocity gives net displacement, while integrating its nonnegative magnitude gives path distance.
• An indefinite integral includes a constant; the same constant cancels in an endpoint difference under the stated fundamental-theorem conditions.
• A partial derivative changes one input while holding the others fixed, where the corresponding limit exists.
• A continuous surface mass density integrated over the stated rectangle gives mass; the measure, region, bounds and integrand determine the interpretation.