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Integration and the fundamental theorem of calculus | Chapter 8, Essence of calculus thumbnail

Integration and the fundamental theorem of calculus | Chapter 8, Essence of calculus

3Blue1Brown·
5 min read

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TL;DR

Distance from a time-varying velocity can be modeled as the area under the velocity curve.

Briefing

Integration is the inverse of differentiation in a precise sense: the accumulated area under a velocity curve produces a distance function whose derivative at any time equals the original velocity. That link matters because it turns “add up infinitely many tiny contributions” into a computation that only needs two evaluations—at the upper and lower bounds—once an antiderivative is found.

The discussion begins with a car problem. With no view outside the window, the only information available is the speedometer reading over time. If velocity were constant, distance would be velocity times elapsed time—an area rectangle on a graph where time is horizontal and velocity is vertical. Real motion is harder because velocity changes continuously, so the approach is to approximate the changing velocity by treating it as constant over many short intervals. On each small interval of length dt, the car’s distance contribution is approximated by “(velocity at the left endpoint) × dt,” which corresponds to the area of a thin rectangle. Summing these rectangle areas across the time range gives an approximation to total distance.

As dt shrinks toward zero, the rectangle sum converges to the exact area under the velocity curve. This limiting area is written as an integral of v(t) over time. The key intuition is that the integral is built from every input between the bounds—yet the final computation can be expressed in a surprisingly economical way.

To connect area to derivatives, the upper bound is treated as a variable capital T. Define a distance function s(T) as the area under the velocity graph from 0 to T. A small increase in T by dt adds a thin sliver of area ds. For sufficiently small dt, that sliver is essentially a rectangle whose height is v(T) and width is dt, so ds/dt equals v(T). In other words, differentiating the “area-so-far” function recovers the original function that generated the area.

Once this relationship is accepted, the fundamental theorem of calculus follows. If v(t) is the integrand, then finding a function F(t) whose derivative is v(t) (an antiderivative) allows the integral to be computed as F(upper bound) minus F(lower bound). There are infinitely many antiderivatives because adding a constant doesn’t change derivatives, but the subtraction of the lower-bound value cancels that ambiguity automatically.

A concrete example uses v(t)=t(8−t)=8t−t^2. An antiderivative is 4t^2−(1/3)t^3, so the distance from time 1 to 7 comes from evaluating that expression at 7 and subtracting its value at 1.

Finally, the treatment emphasizes signed area. If velocity becomes negative (the car moves backward), the corresponding rectangles lie below the horizontal axis, contributing negative distance change. Integrals therefore measure signed area between the graph and the axis, not just geometric area in the everyday sense. The result is a unified method for turning accumulation problems—like distance from velocity—into derivative/antiderivative relationships that can be computed efficiently.

Cornell Notes

The integral of a velocity function gives the distance traveled: it is the limit of sums of thin rectangles under the velocity graph. If s(T) denotes the area under v(t) from 0 to T, then differentiating s(T) returns v(T). This “area-to-derivative” relationship is the engine behind the fundamental theorem of calculus. It implies that to compute an integral, one can find an antiderivative F with F′(t)=v(t) and then evaluate F at the upper and lower bounds. Any constant added to F cancels out when subtracting the lower-bound value.

Why does distance become an area problem when velocity changes continuously?

Distance over a time interval can be approximated by splitting time into small pieces of width dt. On each piece, velocity is treated as constant at the left endpoint, so the distance contribution is approximately v(t)·dt. On a graph with time on the horizontal axis and velocity on the vertical axis, v(t)·dt is exactly the area of a thin rectangle. Summing these rectangle areas approximates total distance, and refining dt makes the approximation converge to the exact area under the curve.

What does the integral mean as dt approaches 0?

The integral is the limiting value of the rectangle sum as the time step dt shrinks toward zero. Each rectangle’s area gets smaller, but the number of rectangles increases, so the total approaches a definite value. That limit is the precise distance traveled for the true (non-constant) velocity function.

How does differentiating an “area-so-far” function recover the original velocity?

Let s(T) be the area under v(t) from 0 to T. Increasing T by a tiny amount dt adds a thin sliver of area ds. For small dt, that sliver is approximately a rectangle with height v(T) and width dt, so ds ≈ v(T)·dt. Dividing by dt and taking the limit gives ds/dt = v(T). This generalizes: the derivative of the accumulated area function equals the integrand at the boundary point.

Why does the fundamental theorem of calculus let integrals be computed using only two evaluations?

If F′(t)=v(t), then the integral of v(t) from a to b equals F(b)−F(a). Although the integral conceptually aggregates contributions from every point between a and b, the antiderivative encodes that accumulation. The subtraction at the bounds also removes the ambiguity from adding constants to F, since constants vanish under differentiation and cancel in F(b)−F(a).

What changes when velocity is negative?

Negative velocity corresponds to graph portions below the horizontal axis. The rectangle areas below the axis are treated as negative contributions, so the integral measures signed area. That means the computed result reflects net distance change: forward motion adds positive area, backward motion subtracts it.

Review Questions

  1. If s(T) is defined as the area under v(t) from 0 to T, what relationship must hold between s′(T) and v(T)?
  2. Given v(t)=8t−t^2, what antiderivative F(t) satisfies F′(t)=v(t), and how would you compute ∫_1^7 v(t) dt using F(b)−F(a)?
  3. Why does adding a constant to an antiderivative not affect the value of a definite integral?

Key Points

  1. 1

    Distance from a time-varying velocity can be modeled as the area under the velocity curve.

  2. 2

    Approximating velocity as constant on small intervals turns distance into a sum of rectangle areas v(t)·dt.

  3. 3

    As dt→0, the rectangle sum converges to the exact integral, giving precise total distance.

  4. 4

    Defining s(T) as accumulated area from 0 to T implies s′(T)=v(T).

  5. 5

    To compute a definite integral, find an antiderivative F with F′=integrand and use F(upper)−F(lower).

  6. 6

    Antiderivatives differ by constants, but those constants cancel automatically in definite integrals.

  7. 7

    Integrals measure signed area: regions where the graph is below the axis contribute negative values.

Highlights

Treating velocity as constant on tiny intervals turns distance into the sum of thin rectangle areas under the graph.
The derivative of “area accumulated up to T” equals the height of the graph at T, linking integration and differentiation directly.
The fundamental theorem of calculus converts an infinite accumulation problem into two bound evaluations of an antiderivative.
Negative velocity produces negative signed area, so integrals capture net distance change rather than raw geometric area.

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