Integrals: adding small contributions

Water flows faster each minute: hypothetical flow q(t)=2+t litres per minute, 0≤t≤4, with t counting minutes. How much water arrives? Multiplying total time by initial flow will not work. An integral adds small “rate × time” contributions and takes their limit. Use limits and antiderivatives.

From four rectangles to a limit

Split time into n equal pieces, choose a time tᵢ* within each and estimate its contribution:

$Q_n=\sum_{i=1}^nq(t_i^*)\Delta t,\qquad\Delta t=4/n.$

With n=4, left endpoints give 2+3+4+5=14 L and right endpoints 3+4+5+6=18 L. Increasing flow makes the first an underestimate and the second an overestimate. For general n:

$L_n=16-8/n,\qquad R_n=16+8/n;\qquad L_n,R_n\to16.$

The definite integral is the common limit of these sums as the largest piece width tends to zero, independent of partitions and sample points, when that limit exists. Continuous functions on closed intervals have such a limit. Definition and signs.

Four left rectangles

q=2+t; n=4; L₄=14

Contribution signs

f=x−1; ∫ f =0; ∫ |f| =1

Uniform scales. Top: hypothetical flow; shaded left sum 14 L is below exact 16 L; right sum would be 18 L. Bottom: dimensionless x,f, two areas 1/2 with opposite signs.

How an antiderivative simplifies the work

If f is continuous on [a,b] and F is its antiderivative, the fundamental theorem gives:

$\int_a^b f(x)\,dx=F(b)-F(a),\qquad F'=f.$

Adding changes in F across neighboring pieces cancels interior values and leaves the endpoints. The link between differentiation and accumulation justifies the formula, not merely similar integral symbols. OpenStax

$Q=\int_0^4(2+t)\,dt=\left[2t+\frac{t^2}2\right]_0^4=16.$

Thus 16 L. Units agree: L/min × min = L. The constant C cancels from the difference. a,b are integration bounds; dx or dt specifies the variable. A definite integral is a number; an indefinite one is a family of functions.

Signed area versus geometric area

Parts below the axis contribute negatively. For the dimensionless example f(x)=x−1:

$\int_0^2(x-1)\,dx=0,\qquad\int_0^2|x-1|\,dx=1.$

Two triangles each have area 1/2: they cancel in a signed sum and add in geometric area. Similarly, integrating velocity gives displacement, whereas integrating its absolute value gives distance traveled.

Between continuous graphs, subtract the lower function from the upper. On [0,1], x≥x²:

$S=\int_0^1(x-x^2)\,dx=\frac16.$

If the graphs exchange order, split at their intersections or integrate the absolute difference. Areas between curves.

A bridge to volume

Cross-sectional area A(x) times a small thickness gives a small volume. For continuous areas of slices perpendicular to the x-axis:

$V=\int_a^b A(x)\,dx;\qquad\int_0^3 2\,dx=6.$

A constant area 2 m² along 3 m gives 6 m³. This is the same accumulation principle, not another isolated formula. Slicing method.

Choose the integral according to what accumulates: water, signed change, area or volume. Bounds, signs and units determine what the answer means.