Constructing Functions from Derivatives using Definite Integrals

TL;DR

You can build a function $f(x)$ if you know its derivative $f'(x)$ and one specific point on $f(x)$. Definite integrals help you find the total change from that known point to any other $x$. This process essentially reverses differentiation, adding up tiny changes to get the original function.

1. The Mental Model

Think of your car's speedometer. It tells you your instantaneous speed ($f'(x)$). If you want to know your exact position ($f(x)$), just knowing your speed isn't enough; you also need to know where you started. Integrating your speed tells you how much you've changed position from that start.

2. The Core Material

When you're given a derivative, $f'(x)$, and you want to find the original function, $f(x)$, you're essentially performing the reverse operation of differentiation. Indefinite integration gives you $F(x) + C$, where $C$ is an unknown constant. This $C$ represents the "starting point" or initial value that indefinite integrals can't capture alone.

That's where definite integrals come in handy. If you know a specific value of the function, say $f(a) = y_a$ for some point $x=a$, you can use this "initial condition" to pinpoint the exact function.

The fundamental theorem of calculus connects derivatives and definite integrals:
$\int_a^x f'(t) dt = f(x) - f(a)$

Rearranging this, you can express $f(x)$ like this:
$f(x) = f(a) + \int_a^x f'(t) dt$

Here, $f(a)$ is your known starting value (your initial condition). The integral $\int_a^x f'(t) dt$ represents the total change in the function's value as $t$ goes from $a$ to $x$. By adding this total change to your starting value $f(a)$, you get the exact value of $f(x)$ at any point $x$.

Notice we use $t$ as the dummy variable inside the integral. This is because $x$ is already being used as one of the limits of integration, and it helps avoid confusion.

How it works step-by-step:

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1. Identify $f'(x)$ and the initial condition $(a, f(a))$: You need both the derivative function and one point on the original function.
2. Set up the definite integral: Use the formula $f(x) = f(a) + \int_a^x f'(t) dt$. The lower limit of the integral will be $a$ (from your initial condition), and the upper limit will be $x$.
3. Evaluate the integral: Find the antiderivative of $f'(t)$, let's call it $F(t)$. Then evaluate it at the limits: $F(x) - F(a)$.
4. Add the initial condition: Add the result of the integral to $f(a)$ to get your final $f(x)$.

graph TD
    A["Know f'(x) (the derivative)"] --> B["Know f(a) (an initial condition)"]
    B --> C{"Use the formula:
    f(x) = f(a) + ∫[a to x] f'(t) dt"}
    C --> D["Find the antiderivative of f'(t)"]
    D --> E["Evaluate the antiderivative from 'a' to 'x'"]
    E --> F["Add the initial condition f(a)"]
    F --> G["Result: The specific function f(x)"]

3. Worked Example

Let's say you're given $f'(x) = 2x - 3$ and you know that $f(1) = 4$. We want to find the specific function $f(x)$.

1. Identify $f'(x)$ and the initial condition:
   $f'(x) = 2x - 3$
   $a = 1$, $f(a) = f(1) = 4$

2. Set up the definite integral:
   $f(x) = f(1) + \int_1^x (2t - 3) dt$
   $f(x) = 4 + \int_1^x (2t - 3) dt$

3. Evaluate the integral:
   First, find the antiderivative of $2t - 3$:
   $\int (2t - 3) dt = t^2 - 3t + C$ (we don't need the $C$ for definite integrals as it cancels out, but it's good to remember it's there for indefinite ones).

   Now, evaluate from $1$ to $x$:
   $[t^2 - 3t]_1^x = (x^2 - 3x) - (1^2 - 3 \cdot 1)$
   $= (x^2 - 3x) - (1 - 3)$
   $= (x^2 - 3x) - (-2)$
   $= x^2 - 3x + 2$

4. Add the initial condition:
   $f(x) = 4 + (x^2 - 3x + 2)$
   $f(x) = x^2 - 3x + 6$

So, the specific function is $f(x) = x^2 - 3x + 6$. You can check this by differentiating $f(x)$ to get $2x-3$ and checking $f(1) = 1^2 - 3(1) + 6 = 1 - 3 + 6 = 4$. It matches!

4. Key Takeaways

• You need both the derivative $f'(x)$ and an initial condition $f(a)$ to find a unique $f(x)$.
• The definite integral $\int_a^x f'(t) dt$ calculates the net change in $f(x)$ from $x=a$ to the current $x$.
• The formula $f(x) = f(a) + \int_a^x f'(t) dt$ essentially reconstructs the function by adding its starting value to its total change.
• Using a dummy variable like $t$ inside the integral helps avoid confusion when $x$ is also a limit of integration.
• This method is more robust than just finding an indefinite integral and solving for $C$, especially in applications where the initial condition is naturally a starting point.

Common Mistakes to Avoid:

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• Forgetting the initial condition $f(a)$: Your final answer will just be an indefinite integral plus a general $C$.
• Mixing up integration limits: The lower limit must match the $a$ from $f(a)$, and the upper limit should be $x$.
• Incorrectly evaluating the definite integral: Double-check your antiderivative and your subtraction of the evaluated limits.
• Not using a dummy variable: While it might work out, it's bad practice and can lead to errors in more complex problems or when $x$ has other meanings.

5. Now Try It

Given $f'(x) = \cos(x) + 2$ and $f(0) = 5$, find the function $f(x)$. Your final answer should be a unique function, not one with an arbitrary constant $C$.