CLASSICAL-MECHANICS

Reversing the Derivative: How Integration Recovers Motion

IKey Formulas

Formula	Name	Description
$v(t) + C = \displaystyle\int a(t)\,dt$	Indefinite integral	Recovers velocity as a function (constant $C$ fixed by initial conditions)
$\displaystyle\int_{t_i}^{t_f} a(t)\,dt = v(t_f) - v(t_i)$	Definite integral (FTC)	A single number: the net change in velocity = signed area under $a(t)$
$x(t) = x_i + \displaystyle\int_{t_i}^{t} v(t')\,dt'$	Position from velocity	Integrate velocity to recover position
$\displaystyle\lim_{n\to\infty} S_n = \sum_j a(t_{c_j})\,\Delta t = v(t_f) - v(t_i)$	Riemann sum limit	Rectangle areas converge to the exact velocity change

IIInteractive Visualizations

Visualization 1 — The Constant of Integration

One acceleration integrates to a whole family of velocity curves ( $b = 1$ ); the initial value $v(0)$ singles out one.

Initial velocity v₀ = v(0) 1

Visualization 2 — The Definite Integral as Signed Area

For $a(t) = t - 2$ , the net change $\displaystyle\int_{0}^{t_f} a\,dt$ is signed area: below the axis subtracts, above adds.

Upper limit t_f 3

Visualization 3 — Riemann Sums Approaching the Exact Area

As the rectangles multiply, their total area closes in on the exact integral $\int_0^2 t^2\,dt = \tfrac{8}{3} \approx 2.667$ .

Rectangles n 4

💡 The limit is the same for any sample point in each strip — left end, right end, or midpoint — which is exactly what the Fundamental Theorem of Calculus guarantees.

IIIQuiz Questions

Question 1

An object has acceleration $a(t) = b\,t^2$ with $b$ constant, and initial velocity $v(0) = v_0$ . Integrating $a(t)$ gives $v(t) + C = \frac{b\,t^3}{3}$ .

After applying the initial condition, what is the velocity function $v(t)$ ?

✅ Correct! The initial condition gives $C = -v_0$ , so $v(t) = v_0 + bt^3/3$ .

❌ Not quite. At $t = 0$ the cubic term vanishes, so $v_0 + C = 0$ gives $C = -v_0$ . The constant adds to the cubic term.

Show solution

Solution:

Integrate: $v(t) + C = \dfrac{b\,t^3}{3}$ , so $v(t) = \dfrac{b\,t^3}{3} - C$ .

Apply the initial condition at $t = 0$ : $v(0) + C = \frac{b\,(0)^3}{3} = 0 \quad\Rightarrow\quad v_0 + C = 0 \quad\Rightarrow\quad C = -v_0.$

Substitute back: $v(t) = \frac{b\,t^3}{3} - (-v_0) = v_0 + \frac{b\,t^3}{3}.$

Question 2

A definite integral of acceleration, $\displaystyle\int_{t_i}^{t_f} a(t)\,dt$ , evaluates to a single number. Which statement is true?

✅ Correct! The definite integral is the net change $v(t_f) - v(t_i)$ , not the velocity itself.

❌ Not quite. A common confusion: the definite integral gives the change in velocity, not $v(t_f)$ itself and not a function. You still need the initial velocity to recover $v(t_f)$ .

Show solution

Solution:

The definite integral returns the net change in velocity over the interval: $\int_{t_i}^{t_f} a(t)\,dt = v(t_f) - v(t_i).$

It does not give $v(t_f)$ by itself — to recover the actual velocity you still need the initial velocity $v(t_i)$ : $v(t_f) = v(t_i) + \int_{t_i}^{t_f} a(t)\,dt.$

The indefinite integral (not the definite one) is what returns a function plus a constant $C$ .

Question 3

Consider $a(t) = t - 2$ (the curve from Visualization 2). What is the net change in velocity over the interval from $t_i = 0$ to $t_f = 4$ , i.e. $\displaystyle\int_{0}^{4} (t - 2)\,dt$ ?

✅ Correct! The negative area on $[0,2]$ cancels the positive area on $[2,4]$ , giving a net change of $0$ .

❌ Not quite. Use the antiderivative $F(t) = t^2/2 - 2t$ and remember the integral is signed area — the part below the axis is negative and cancels the part above.

Show solution

Solution:

An antiderivative of $a(t) = t - 2$ is $F(t) = \dfrac{t^2}{2} - 2t$ . Evaluate from $0$ to $4$ : $\int_{0}^{4} (t - 2)\,dt = F(4) - F(0) = \left(\frac{16}{2} - 8\right) - 0 = (8 - 8) - 0 = 0.$

Geometrically, the triangle of negative (signed) area on $[0, 2]$ exactly cancels the triangle of positive area on $[2, 4]$ — both have area $2$ . The signed areas sum to zero.

Question 4

True or False: Because the derivative of position is velocity, you can recover position by integrating velocity: $x(t) = x_i + \int_{t_i}^{t} v(t')\,dt'.$

The full pipeline is therefore: integrate acceleration to get velocity, then integrate velocity to get position, supplying one initial condition at each step.

✅ Correct! Integrating velocity recovers position, with $x_i$ fixing the constant — the same logic that recovers velocity from acceleration.

❌ Not quite. Since $dx/dt = v(t)$ , integrating velocity does give displacement, and adding $x_i$ recovers position. The statement is true.

Show solution

Solution: True.

Since $\dfrac{dx}{dt} = v(t)$ , the definite integral of velocity is the net change in position (displacement): $x(t_f) - x(t_i) = \int_{t_i}^{t_f} v(t')\,dt'.$

Replacing the fixed upper limit with a variable $t$ gives position as a function of time: $x(t) = x_i + \int_{t_i}^{t} v(t')\,dt'.$

This is the second link in the two-step chain $a(t) \xrightarrow{\int dt} v(t) \xrightarrow{\int dt} x(t)$ , where the initial velocity $v_i$ fixes the constant in the first step and the initial position $x_i$ fixes it in the second.

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