Constructing Functions from Derivatives using Definite Integrals

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Constructing Functions from Derivatives using Definite Integrals

TL;DR

If you know a function's derivative and one value of the original function, you can rebuild the whole thing using a definite integral. The fundamental theorem of calculus says: if you integrate the derivative from a fixed point to any other point, you recover the original function's behaviour. This is the reverse of differentiation.

1. The Mental Model

You have a speedometer (the derivative) and you know your starting position. By adding up all the tiny displacements the speedometer records, you can reconstruct your entire journey. That's what integration does: it glues the derivative back into a function. The definite integral measures the total accumulation from one point to another, and that accumulation IS the change in the original function.

2. The Core Material

Why This Matters

Differentiation breaks things apart—it finds rates of change. Integration puts them back together. In practice, you often measure rates (velocity, profit margin, population growth) and need to recover the total quantity (distance, revenue, population). This is how you do it.

The Fundamental Theorem: Part 2

The second part of the fundamental theorem of calculus says:

If $F'(x) = f(x)$ (so $f$ is the derivative of $F$), then:
$$\int_a^b f(x) \, dx = F(b) - F(a)$$

In words: the definite integral of the derivative from $a$ to $b$ equals the change in the original function.

But here's the reconstruction angle: if you know $f(x)$ (the derivative) and you know $F(a)$ (one starting value), then:
$$F(x) = F(a) + \int_a^x f(t) \, dt$$

Read this carefully. $F(x)$ (what you're building) equals your starting value plus the accumulated derivative from $a$ out to wherever you are now. The variable of integration is $t$ (not $x$) because $x$ is the upper limit—it's the "where you are now" that changes.

Building a Function: The Three Ingredients

1. The derivative $f(x)$ — you're given this or can find it.
2. An initial condition $F(a) = c$ — one known value of the original function.
3. The reconstruction formula: $F(x) = F(a) + \int_a^x f(t) \, dt$.

That's it. Once you have these three, you have $F$ completely.

Why the Dummy Variable Matters

Notice the integral uses $t$ as the dummy variable, not $x$. This is crucial. $x$ is the upper limit—it's the input to your new function $F(x)$. $t$ is just a placeholder inside the integral that disappears when you evaluate it. Think of $t$ as "past time" and $x$ as "current time." You integrate all the changes from the start up to now.

Practical Reconstruction Steps

1. Check you have the derivative $f(x)$.
2. Find the antiderivative $\int f(x) \, dx$ symbolically (or identify a pattern).
3. Apply the limits using the formula $F(x) = F(a) + [F_{\text{antiderivative}}(x) - F_{\text{antiderivative}}(a)]$.
4. Simplify to get $F(x)$ as an explicit formula if possible.

graph TD
    A["Given: derivative f(x)<br/>and initial value F(a) = c"] --> B["Find the antiderivative<br/>of f(x)"]
    B --> C["Write F(x) = F(a) + ∫_a^x f(t) dt"]
    C --> D["Evaluate the definite integral<br/>using limits a and x"]
    D --> E["Simplify to get<br/>explicit formula for F(x)"]
    E --> F["Check: verify F'(x) = f(x)<br/>and F(a) = c"]

3. Worked Example

Problem: The derivative is $f(x) = 3x^2 + 2x$. We know $F(1) = 5$. Find $F(x)$.

Step 1: Apply the formula.
$$F(x) = F(1) + \int_1^x (3t^2 + 2t) \, dt$$

Step 2: Find the antiderivative.
$$\int (3t^2 + 2t) \, dt = t^3 + t^2 + C$$

(The $C$ disappears in definite integrals, but we'll use the antiderivative to evaluate the limits.)

Step 3: Evaluate the definite integral.
$$\int_1^x (3t^2 + 2t) \, dt = [t^3 + t^2]_1^x = (x^3 + x^2) - (1^3 + 1^2) = x^3 + x^2 - 2$$

Step 4: Plug back in.
$$F(x) = 5 + (x^3 + x^2 - 2) = x^3 + x^2 + 3$$

Verification:
- $F'(x) = 3x^2 + 2x$ ✓
- $F(1) = 1 + 1 + 3 = 5$ ✓

Done. You've reconstructed the entire function from its derivative and one starting point.

4. Key Takeaways

4.1 Most Important Concepts

1. The reconstruction formula: $F(x) = F(a) + \int_a^x f(t) \, dt$ is your workhorse. It says the function at $x$ equals its starting value plus the accumulated rate of change.

2. Dummy variables: The letter inside the integral (here $t$) is a dummy—it doesn't appear in the answer. Use it to avoid confusion with the upper limit $x$.

3. Definite integrals give changes, not totals: $\int_a^b f(t) \, dt$ is the change in $F$ from $a$ to $b$, not an absolute quantity.

4. Initial conditions are essential: Without knowing $F(a)$, you can only find $F$ up to an unknown constant. The definite integral alone gives you the shape, but you need one anchor point.

5. The fundamental theorem connects the two parts of calculus: Differentiation and integration are inverse operations. This is the deepest idea in calculus.

6. Different starting points give the same function: If you use $F(x) = F(0) + \int_0^x f(t) \, dt$ or $F(x) = F(5) + \int_5^x f(t) \, dt$, you get the same $F(x)$ in the end (though the intermediate steps differ).

7. This works for any continuous derivative: As long as $f(x)$ is continuous, you can always reconstruct $F(x)$ from it and one known value.

4.2 Common Misconceptions

4.3 Compare & Contrast

5. Now Try It

Exercise: You measure that a population is growing at a rate $f(t) = 100 + 10t$ people per year (where $t$ is years from now). Right now ($t = 0$), the population is 5000. Write a formula for the population $P(t)$ at any time $t$, then use it to predict the population after 2 years.

What you need to do:
1. Identify the derivative (rate), the initial condition, and set up the reconstruction formula.
2. Integrate $f(t) = 100 + 10t$ from 0 to $t$.
3. Use the initial condition $P(0) = 5000$ to find the constant.
4. Write the final formula for $P(t)$.
5. Plug in $t = 2$ to find the population at year 2.

Hints:
- The antiderivative of $100 + 10t$ is $100t + 5t^2$.
- Remember: $P(t) = P(0) + \int_0^t f(s) \, ds$.

Success looks like: You write $P(t) = 5000 + 100t + 5t^2$ and calculate $P(2) = 5000 + 200 + 20 = 5220$ people.

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