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Integrals & the Fundamental Theorem

Useful preparation: Derivative Tests & Optimization

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Meaning & notation

Integration adds small signed contributions. Rate times interval approximates change; refining partitions produces an integral. The Fundamental Theorem connects accumulation to differentiation and explains antiderivative evaluation.

Definite integral: The limit of weighted sums over an interval when that limit exists independently of sample choices. Its units are integrand units multiplied by input units.
Antiderivative: A function whose derivative equals the integrand on an interval. Any two antiderivatives on the same interval differ by a constant.
Accumulation function: An integral whose upper or lower endpoint varies. It records how the accumulated signed quantity depends on that endpoint.

Conditions before calculation

Continuity on a closed interval is a sufficient condition for integrability and for the standard Fundamental Theorem statements used here.
The derivative of an accumulation function equals the integrand at interior points where the integrand is continuous; moving endpoints also require the chain rule.
A signed integral measures net change. Total geometric area or total distance requires integrating an appropriate nonnegative magnitude.

Why the method works

1. Build accumulation from small pieces

Each rectangle approximates a contribution using a subinterval sample. The largest width tends to zero; negative heights contribute negative quantities.

\int_a^b f(t)\,dt=\lim_{\max\Delta t_i\to0}\sum_i f(t_i^*)\Delta t_i

2. Recover the local rate

Moving the accumulation endpoint adds a thin strip. Continuity makes its average height approach the endpoint height.

A(x)=\int_a^x f(t)\,dt,\qquad A'(x)=f(x)

3. Evaluate through an antiderivative

An antiderivative and accumulation have the same derivative, so they differ by a constant. Endpoint subtraction eliminates it; a definite value needs no arbitrary constant.

F'=f\quad\Longrightarrow\quad\int_a^b f(t)\,dt=F(b)-F(a)

Worked examples

EXAMPLE 1

A particle has the displayed velocity in meters per second. Find its displacement and total distance over the first four seconds.

Locate the change of direction. Velocity is negative before the zero and positive afterward.
$v(t)=3t-6,\quad 0\le t\le4;\qquad v(t)=0\iff t=2$
Integrate velocity for signed displacement. The two directional contributions cancel.
$\int_0^4(3t-6)\,dt=\left[\frac32t^2-6t\right]_0^4=0$
Distance adds magnitudes, so reverse the sign on the interval where velocity is negative and split at the zero.
$\int_0^4|v(t)|\,dt=-\int_0^2(3t-6)\,dt+\int_2^4(3t-6)\,dt=6+6=12$

Conclusion: Displacement is zero meters, while total distance is 12 meters; returning to the start does not mean no motion occurred.

EXAMPLE 2

Differentiate the integral with two moving endpoints without finding an antiderivative.

The integrand is continuous for every real input, so accumulation is defined even when the endpoints reverse order.
$B(x)=\int_x^{x^2}e^{-t^2}\,dt$
Write the interval as a difference of two accumulations from the same fixed base. This exposes both endpoint dependencies.
$B(x)=\int_0^{x^2}e^{-t^2}\,dt-\int_0^x e^{-t^2}\,dt$
Apply the Fundamental Theorem and multiply by each endpoint's derivative. The lower endpoint contributes with a minus sign.
$B'(x)=2x e^{-x^4}-e^{-x^2},\qquad B(1)=0,\quad B'(1)=e^{-1}$

Conclusion: An integral can be differentiated exactly even when an elementary antiderivative is unavailable.

Pitfalls & counterexamples

Tempting claim: A zero definite integral means the integrand is zero everywhere.

Correction: Positive and negative contributions may cancel. The integral only reports their signed total.

Test the claim: The particle's velocity integral is zero even though its velocity is nonzero except at the turning time.

Check your understanding

If a continuous inflow rate is positive throughout an interval, what happens to its accumulation function?

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Compare your reasoning

It is strictly increasing.

Every extension of the endpoint adds a positive contribution, and the accumulation derivative equals the positive rate.

Connect this idea

The theorem powers substitution and integration by parts. The distinction between net change and total magnitude reappears in line integrals, flux, and probability densities.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.