Integral as area and accumulation

The accumulation problem

If a constant flow rate is 3 L/min for 4 min, the volume transferred is 12 L. Multiplying rate by duration works because the rate does not change. If the flow rate varies, a single product does not describe the whole interval exactly. We can divide it into short parts, approximate the rate by one value in each part and add the small quantities transferred.

The same structure appears in velocity and displacement, power and energy, linear density and mass. The accumulated output has unit equal to the rate's unit multiplied by the integration variable's unit. A velocity graph in m/s against time in s produces accumulation in metres. Therefore, the word area in the drawing does not necessarily mean square metres: it depends on the quantities on the axes. Identifying these units is part of the interpretation, not just the final presentation.

Construct the integral through sums

Divide [a,b] into n equal parts of width Δx=(b−a)/n. In each subinterval choose an input xᵢ* and calculate f(xᵢ*)Δx. The sum Σf(xᵢ*)Δx approximates the total accumulation. If the sums converge to the same value as the greatest subinterval width tends to zero, independently of the admissible choices, that value is the definite integral ∫_a^bf(x) dx.

A continuous function on a closed interval is integrable. There are also integrable functions with certain discontinuities, but continuity is a simple sufficient condition for this unit. A left- or right-endpoint rectangle sum is a finite approximation, not generally the exact integral. For an increasing function, the left sum underestimates and the right sum overestimates. Refining the partition brings these two estimates closer, provided the model meets the conditions.

The sign changes the question

If f is nonnegative, the integral represents the area between the graph and the horizontal axis on the interval. If f is negative, the sum's terms are negative. The integral then measures signed accumulation: area above the axis minus the magnitude of the area below it. A zero integral can correspond to two large regions that cancel each other.

Total geometric area is obtained by integrating |f|, or by splitting the interval at zeros and making the geometric contributions positive. In kinematics, integrating velocity gives displacement; integrating speed, that is, |v|, gives distance travelled. An object may return to its starting point and have zero displacement after travelling a positive distance. Before calculating, ask whether the quantity sought preserves signs or requires adding magnitudes.

Antiderivatives and the constant of integration

An antiderivative F of f on an interval satisfies F′=f. For example, x³/3 is an antiderivative of x², because its derivative is x². Adding a constant does not change the derivative: all functions x³/3+C are antiderivatives. On an interval, two antiderivatives of the same function differ by a constant. The notation ∫f(x) dx, without bounds, represents this family, called the indefinite integral.

The rule ∫xⁿ dx=xⁿ⁺¹/(n+1)+C applies for n≠−1 on the appropriate real domain. For 1/x, an antiderivative is ln|x|+C on intervals that do not cross zero. Writing the power rule with n=−1 would produce division by zero. The safest check of an antiderivative is to differentiate it. The symbol dx identifies the variable; it should not be treated as dispensable decoration when several quantities are involved.

Why accumulation and differentiation connect

Define A(x)=∫_a^xf(t) dt, with continuous f. When x increases by h, the change A(x+h)−A(x) accumulates only the small strip between x and x+h. Dividing by h gives an average rate of f over that strip. By continuity, this average approaches f(x). Thus A′(x)=f(x): the instantaneous rate of accumulation recovers the function being accumulated.

The second part of the fundamental theorem enables calculation: if f is continuous on [a,b] and F is an antiderivative, then ∫_a^bf(x) dx=F(b)−F(a). The constants cancel. The definite integral is a number, whereas the indefinite integral is a family of functions. You do not need to add +C to the final number of a definite integral. If there are singularities within the interval, this direct application may fail and requires further study.

Properties and limits of use

Accumulations add: integrating f+g means adding the integrals, and a constant factor may be taken outside the integral. Dividing the interval at a point c lets us write ∫ from a to b as the sum from a to c and from c to b. Swapping the bounds changes the sign; integrating between equal bounds gives zero. These properties reflect the direction and composition of the small sums.

The average value of an integrable function on [a,b], with a<b, is the integral divided by b−a. It is not generally the average of just the endpoint values. For experimental data, methods such as trapezoids give approximations; quality depends on variation between measurements, resolution and uncertainties. An exact integral of a model does not make the measured phenomenon exact. Always state which function, interval, units and physical or geometric interpretation are being used.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

A definite integral is always positive area.

It is signed accumulation; geometric area uses magnitudes.

Every integral gets +C in the final result.

The constant belongs to the family of antiderivatives; it cancels in the definite integral.

Finding an antiderivative at each endpoint is enough, even with a singularity between them.

The theorem requires conditions throughout the interval; singularities cannot be ignored.