Constructing Functions from Derivatives using Definite Integrals

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Constructing Functions from Derivatives using Definite Integrals

TL;DR

If you know a derivative, you can recover the original function by integrating. A definite integral from a fixed point to a variable point gives you back the function itself (up to an unknown constant). This is the Fundamental Theorem of Calculus at work, and it's how we "undo" differentiation.

1. The Mental Model

When you differentiate, you lose information—specifically, any constant term vanishes. So if you only have a derivative, you know the shape of the original function but not its exact vertical position. Definite integrals let you pin down that position: integrate from a known starting point (where you know the function's value) to a variable endpoint, and you've reconstructed the function. The definite integral literally accumulates the rate of change from start to finish.

2. The Core Material

The Fundamental Theorem in Action

The Fundamental Theorem of Calculus (Part 1) says that if $f$ is continuous on $[a, b]$, then
$$F(x) = \int_a^x f(t) \, dt$$
is a differentiable function, and $\frac{dF}{dx} = f(x)$.

In English: if you integrate a function $f$ from a fixed point $a$ to a variable point $x$, what you get is a new function $F(x)$ whose derivative is exactly $f$. This is the core tool.

Why does this work? The definite integral accumulates area (or, more generally, net signed change) as $x$ moves. Each tiny step $dx$ adds a slice of height $f(x)$, so the rate of change of that accumulated total is $f(x)$ itself.

Recovering a Function from Its Derivative

Suppose you're given:
- A derivative: $\frac{dy}{dx} = 3x^2$
- A starting condition: $y(1) = 5$

You want to find $y(x)$ for any $x$.

Here's the recipe:
1. Integrate the derivative from the starting point to $x$:
$$y(x) = y(1) + \int_1^x 3t^2 \, dt$$

2. Evaluate the definite integral:
   $$\int_1^x 3t^2 \, dt = \left[ t^3 \right]_1^x = x^3 - 1$$

3. Substitute back:
   $$y(x) = 5 + (x^3 - 1) = x^3 + 4$$

Notice: we never wrote $+ C$. The definite integral from a fixed point automatically captures the constant. That's the whole point.

Why Use a Definite Integral Instead of Antiderivatives?

When you learn integration, you write $\int f(x) \, dx = F(x) + C$ and then solve for $C$ using an initial condition. That works, but it's two steps. The definite integral does both at once:

$$y(x) = y_0 + \int_{x_0}^x f(t) \, dt$$

The lower limit $x_0$ pins down the "anchor point" where we know the value ($y_0$), and the upper limit $x$ is our variable. As $x$ changes, the definite integral measures the accumulated change from the anchor to the new point. It's cleaner and less error-prone.

A Word on the Dummy Variable

Notice that inside the integral, we write $\int_{x_0}^x f(t) \, dt$, not $\int_{x_0}^x f(x) \, dt$. The reason: the integration variable ($t$) must be different from the upper limit ($x$). If you use $x$ for both, you create notational confusion—the $x$ inside the integral is not the same as the $x$ you're varying. Always use a dummy variable like $t$ or $u$ inside the integral.

graph LR
    A["Given: f'(x) = derivative<br/>and initial condition y(x₀) = y₀"] --> B["Set up definite integral<br/>y(x) = y₀ + ∫(x₀ to x) f'(t) dt"]
    B --> C["Evaluate the integral<br/>using antiderivative"]
    C --> D["Substitute limits<br/>to get explicit y(x)"]
    D --> E["Result: function fully<br/>reconstructed"]
    style A fill:#e1f5ff
    style E fill:#c8e6c9

3. Worked Example

Problem: You measure that a ball's velocity is $v(t) = 4t + 1$ m/s. At time $t = 0$, the ball is at position $s = 2$ m. Find the position function $s(t)$.

Step 1: Recognise the setup.
- $v(t) = \frac{ds}{dt} = 4t + 1$ (velocity is the derivative of position)
- Initial condition: $s(0) = 2$

Step 2: Set up the definite integral.
$$s(t) = s(0) + \int_0^t v(\tau) \, d\tau = 2 + \int_0^t (4\tau + 1) \, d\tau$$

(I used $\tau$ as the dummy variable to avoid confusion with the upper limit $t$.)

Step 3: Evaluate the integral.
$$\int_0^t (4\tau + 1) \, d\tau = \left[ 2\tau^2 + \tau \right]_0^t = 2t^2 + t - 0 = 2t^2 + t$$

Step 4: Write the final answer.
$$s(t) = 2 + 2t^2 + t$$

Check: $\frac{ds}{dt} = 4t + 1$ ✓ and $s(0) = 2 + 0 + 0 = 2$ ✓

If you had instead used the antiderivative method: $s(t) = 2t^2 + t + C$, then solved for $C$ using $s(0) = 2$, you'd get $2 = 0 + 0 + C$, so $C = 2$, giving $s(t) = 2t^2 + t + 2$. Same answer—but the definite integral approach gives it directly.

4. Key Takeaways

4.1 Most Important Concepts

1. Definite integral from a fixed anchor reconstructs the function: $y(x) = y_0 + \int_{x_0}^x f'(t) \, dt$ recovers $y$ from its derivative and starting value without needing to solve for $C$ separately.

2. The Fundamental Theorem ties derivatives and integrals: If $F(x) = \int_a^x f(t) \, dt$, then $F'(x) = f(x)$—integration and differentiation are (almost) inverses.

3. The dummy variable must differ from the limit: Write $\int_a^x f(t) \, dt$, never $\int_a^x f(x) \, dx$ when $x$ is the upper limit, to avoid notational mixups.

4. The lower limit acts as the anchor: Changing $x_0$ shifts the entire function vertically by a constant, but doesn't change the derivative (since constants differentiate to zero).

5. Definite integrals encode initial conditions automatically: You never write "$+ C$" because the lower limit already specifies which particular antiderivative you want.

6. Physical interpretation: The integral accumulates the rate of change from a starting point, so the result is the total change in the quantity.

4.2 Common Misconceptions

4.3 Compare & Contrast

5. Now Try It

Exercise: construct the function from its derivative.

A population of bacteria has growth rate $\frac{dP}{dt} = 5e^{t}$ (individuals per day), and at time $t = 0$ there are $P(0) = 100$ bacteria.

(a) Write the integral formula for $P(t)$.

(b) Evaluate the definite integral.

(c) Write the explicit formula for $P(t)$.

(d) Check your answer by differentiating it.

What a complete answer must include:

• The setup: $P(t) = 100 + \int_0^t 5e^\tau \, d\tau$ (using a dummy variable inside).
• The antiderivative of $5e^\tau$ is $5e^\tau$; evaluate from $0$ to $t$ to get $5e^t - 5$.
• The final formula: $P(t) = 95 + 5e^t$.
• Verification: $\frac{dP}{dt} = 5e^t$ ✓ and $P(0) = 95 + 5 = 100$ ✓.

Success looks like: you write down the three components (initial condition, integral setup, and final formula) and verify that your answer differentiates back to the original rate.

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