Applications of FTC 1: Differentiating Partial Area Functions

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Applications of FTC 1: Differentiating Partial Area Functions

TL;DR

The Fundamental Theorem of Calculus Part 1 tells you that if you define a function as the area under a curve from a fixed point to a variable point, then differentiating that function gives you back the original curve. This lets you turn accumulation problems into derivative problems and is the foundation for understanding why integration and differentiation are inverses.

1. The Mental Model

You know how taking the derivative of a position function gives velocity—the rate of change? The FTC says something similar works for areas: if you build a function by accumulating area under a curve, differentiating that accumulation function recovers the curve you were accumulating. It's the formal reason integration and differentiation undo each other. The core insight: a partial area function measures how much total accumulation has happened so far, and its derivative tells you the rate at which that accumulation is happening right now.

2. The Core Material

What is a Partial Area Function?

A partial area function A(x) is defined like this:

$$A(x) = \int_a^x f(t) \, dt$$

Here, $a$ is a fixed starting point, and $x$ is the variable upper limit. As $x$ changes, the area between the curve $f(t)$ and the t-axis accumulates. $A(x)$ keeps track of the total accumulated area from $a$ to wherever you currently are.

Think of it this way: if $f(t)$ is the rate of water flowing into a tank at time $t$, then $A(x)$ is the total volume of water in the tank by time $x$.

The Fundamental Theorem of Calculus Part 1

FTC 1 states:

$$\frac{d}{dx} \int_a^x f(t) \, dt = f(x)$$

In words: if you differentiate a partial area function, you get back the original function $f(x)$. This works because area accumulates at a rate equal to the height of the curve at that point. At position $x$, the "infinitesimal slice" of area being added is $f(x) \cdot dx$, so the rate of area accumulation is $\frac{dA}{dx} = f(x)$.

Why the dummy variable? Notice we use $t$ inside the integral and $x$ as the upper limit. This is crucial. The variable $t$ is just a placeholder that lets us integrate; $x$ is the variable we care about. You can't use $x$ for both or you'll confuse yourself.

Step-by-Step Application

Step 1: Identify the partial area function. Look for an integral with a constant lower limit and a variable upper limit. That's your $A(x)$.

Step 2: Apply FTC 1 directly. Just differentiate: $\frac{d}{dx} A(x) = f(x)$, where $f$ is the integrand.

Step 3: Handle variable limits. If the upper limit isn't just $x$ but a function like $g(x)$, use the chain rule:

$$\frac{d}{dx} \int_a^{g(x)} f(t) \, dt = f(g(x)) \cdot g'(x)$$

Step 4: If both limits are variable, use the linearity of the integral:

$$\frac{d}{dx} \int_{u(x)}^{v(x)} f(t) \, dt = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x)$$

This is because you're subtracting the "accumulated area up to $u(x)$" from the "accumulated area up to $v(x)$."

graph LR
    A["Partial Area Function<br/>A(x) = ∫ₐˣ f(t) dt"] --> B["Differentiate<br/>Apply FTC 1"]
    B --> C["Get f(x) back<br/>dA/dx = f(x)"]
    D["Variable upper limit g(x)?<br/>Use chain rule"] --> C
    C --> E["Result: rate of<br/>area accumulation"]
    style A fill:#e1f5ff
    style C fill:#c8e6c9
    style E fill:#fff9c4

3. Worked Example

Let's say $A(x) = \int_1^x (3t^2 + 2) \, dt$ and you're asked to find $A'(x)$.

Application of FTC 1:
$$A'(x) = 3x^2 + 2$$

That's it. You don't evaluate the integral; you just replace the variable in the integrand.

Now a slightly harder one: $A(x) = \int_0^{x^2} \sin(t) \, dt$.

Here the upper limit is $g(x) = x^2$, so use the chain rule:
$$A'(x) = \sin(x^2) \cdot \frac{d}{dx}(x^2) = \sin(x^2) \cdot 2x = 2x \sin(x^2)$$

One more with both limits variable:
$$B(x) = \int_{x}^{x^2} (4t - 1) \, dt$$

Using the two-limit rule:
$$B'(x) = (4 \cdot x^2 - 1) \cdot 2x - (4 \cdot x - 1) \cdot 1$$
$$= 2x(4x^2 - 1) - (4x - 1)$$
$$= 8x^3 - 2x - 4x + 1$$
$$= 8x^3 - 6x + 1$$

Notice you evaluate the integrand at each limit, then multiply by the derivative of that limit, and subtract.

4. Key Takeaways

4.1 Most Important Concepts

1. FTC 1 inverts integration. Differentiating a partial area function recovers the integrand—they're inverse operations.

2. Dummy variables matter. Use $t$ inside the integral and $x$ as your variable upper limit so you don't mix them up.

3. Chain rule applies to variable limits. If the upper limit is $g(x)$, multiply by $g'(x)$; this captures how fast the limit itself is changing.

4. Accumulation rate equals height. The derivative of total accumulated area is the current rate of accumulation, which equals the function value at that point.

5. Two-limit rule uses subtraction. When both limits vary, treat them as separate partial areas and subtract: $F(v) - F(u)$ approach.

6. No evaluation needed. FTC 1 gives you the derivative directly; you don't compute the antiderivative first.

7. Works only if the integrand is continuous. If $f$ has a jump discontinuity, FTC 1 doesn't apply at that point.

4.2 Common Misconceptions

4.3 Compare & Contrast

5. Now Try It

Problem 1 (Warm-up): Differentiate $F(x) = \int_2^x (5t^3 - 4t) \, dt$ using FTC 1.

Problem 2 (Chain rule): Differentiate $G(x) = \int_0^{\sqrt{x}} e^{-t^2} \, dt$.

Problem 3 (Both limits variable): Differentiate $H(x) = \int_{x}^{2x} \cos(t) \, dt$.

Problem 4 (Application): The amount of charge (in coulombs) that has flowed into a battery by time $x$ (in hours) is $Q(x) = \int_0^x (12 - 0.5t) \, dt$. What is the current (rate of charge flow) at $x = 3$ hours? (Hint: current is $\frac{dQ}{dx}$.)

Expected answers:
1. $F'(x) = 5x^3 - 4x$
2. $G'(x) = \frac{e^{-x}}{2\sqrt{x}}$
3. $H'(x) = 2\cos(2x) - \cos(x)$
4. Current at $x = 3$ is $12 - 0.5(3) = 10.5$ coulombs/hour

Success looks like: You can apply FTC 1 directly to constant-limit integrals, correctly use the chain rule when the limit is a function, and handle two-variable limits by subtracting the two derivatives.

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