Introduction to Fundamental Theorem of Calculus Part 1 (FTC 1)

TL;DR

FTC 1 connects differentiation and integration, showing they're inverse operations. It tells us how to differentiate an integral with a variable upper limit. This theorem is crucial for understanding the relationship between a function and its accumulation function.

1. The Mental Model

Think of FTC 1 as the formal handshake between two key calculus ideas: finding the rate of change (differentiation) and finding the total accumulation (integration). It basically says that if you accumulate something, and then look at the rate at which you're accumulating it, you'll get back to where you started.

2. The Core Material

The Fundamental Theorem of Calculus Part 1 (FTC 1) is a cornerstone of calculus. It formalizes the idea that differentiation and integration are inverse processes, much like addition and subtraction, or multiplication and division.

Let's break down what FTC 1 actually states.

If $f$ is a continuous function on an interval $[a, b]$, and we define a new function $F(x)$ as the definite integral of $f(t)$ from a constant $a$ to a variable $x$:

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

Then, FTC 1 states that the derivative of $F(x)$ with respect to $x$ is simply $f(x)$:

$F'(x) = \frac{d}{dx} \left[ \int_a^x f(t) \, dt \right] = f(x)$

What does this mean?

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1. Accumulation Function: The integral $\int_a^x f(t) \, dt$ represents an "accumulation function." It calculates the accumulated area under the curve of $f(t)$ starting from $a$ up to any point $x$.
2. Rate of Change of Accumulation: When you differentiate this accumulation function, you are asking: "At what rate is the accumulated area changing as I move the upper limit $x$?"
3. Back to the Original Function: FTC 1 tells us that this rate of change is precisely the value of the original function $f(x)$ at that point $x$.

The variable $t$ inside the integral is a "dummy variable." It's just a placeholder for the integration process. When we differentiate with respect to $x$, the $t$ gets replaced by $x$.

Handling Different Limits

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Case 1: Variable in the Upper Limit (as shown above)
If $G(x) = \int_a^x f(t) \, dt$, then $G'(x) = f(x)$. This is the most direct application.

Case 2: Variable in the Lower Limit
What if the variable is in the lower limit?
Let $H(x) = \int_x^a f(t) \, dt$.
We know that $\int_x^a f(t) \, dt = - \int_a^x f(t) \, dt$.
So, $H'(x) = \frac{d}{dx} \left[ - \int_a^x f(t) \, dt \right] = - f(x)$.
Just add a negative sign!

Case 3: Variable in Both Limits (using the Chain Rule)
What if both limits are functions of $x$? For example, $J(x) = \int_{g(x)}^{h(x)} f(t) \, dt$.
We can split this integral:
$\int_{g(x)}^{h(x)} f(t) \, dt = \int_{g(x)}^c f(t) \, dt + \int_c^{h(x)} f(t) \, dt$ (where $c$ is any constant)
$= - \int_c^{g(x)} f(t) \, dt + \int_c^{h(x)} f(t) \, dt$.

Now, let's differentiate using FTC 1 and the Chain Rule:
$J'(x) = \frac{d}{dx} \left[ \int_c^{h(x)} f(t) \, dt \right] - \frac{d}{dx} \left[ \int_c^{g(x)} f(t) \, dt \right]$
This gives us: $J'(x) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)$.

This is a very powerful generalization!

graph TD
    A["Start with a continuous function f(t)"] --> B["Define Accumulation Function F(x) = ∫_a^x f(t) dt"]
    B --> C{"What is the upper limit?"}
    C -- "It's x" --> D["Apply FTC 1: d/dx [F(x)] = f(x)"]
    C -- "It's a function of x, h(x)" --> E["Apply Chain Rule: d/dx [∫_a^(h(x)) f(t) dt] = f(h(x)) * h'(x)"]
    C -- "It's a constant, 'a', and lower limit is x" --> F["Rewrite: -∫_a^x f(t) dt"]
    F --> G["Apply FTC 1 & Negative: -f(x)"]
    C -- "Both limits are functions, g(x) and h(x)" --> H["Split the integral & apply Chain Rule for both: f(h(x))h'(x) - f(g(x))g'(x)"]

3. Worked Example

Let's find the derivative of $G(x) = \int_2^{x^3} \cos(t^2) \, dt$.

Here, our $f(t) = \cos(t^2)$, and our upper limit is $h(x) = x^3$. The lower limit is a constant, $a=2$.

Using the generalized FTC 1 (Case 3, but with a constant lower limit, so $g(x)$ would effectively be a constant whose derivative is 0):

$G'(x) = f(h(x)) \cdot h'(x)$

First, identify $f(t)$ and $h(x)$:
$f(t) = \cos(t^2)$
$h(x) = x^3$

Next, find the derivative of the upper limit:
$h'(x) = \frac{d}{dx}(x^3) = 3x^2$

Now, substitute $h(x)$ into $f(t)$ to get $f(h(x))$:
$f(h(x)) = \cos((x^3)^2) = \cos(x^6)$

Finally, multiply $f(h(x))$ by $h'(x)$:
$G'(x) = \cos(x^6) \cdot 3x^2$

So, $G'(x) = 3x^2 \cos(x^6)$.

4. Key Takeaways

• FTC 1 formally states that differentiation "undoes" integration of an accumulation function.
• When differentiating $\int_a^x f(t) \, dt$, the result is simply $f(x)$.
• If the upper limit is a function of $x$, say $h(x)$, you need to use the Chain Rule: $f(h(x)) \cdot h'(x)$.
• If the variable is in the lower limit, reverse the integral and add a negative sign before applying FTC 1.
• FTC 1 provides a powerful shortcut to differentiate certain integrals without actually evaluating the integral first.

Common Mistakes to Avoid:
- Forgetting the Chain Rule when the upper limit is a function of $x$ (e.g., $x^2$, $\sin(x)$).
- Not changing the dummy variable $t$ to $x$ (or $h(x)$) in the $f(t)$ part.
- Incorrectly handling variables in the lower limit (forgetting the negative sign).
- Trying to evaluate the integral first and then differentiate, which is often much harder or impossible.

5. Now Try It

Find the derivative of the function $K(x) = \int_{\sin(x)}^{x^2} \sqrt{t^3 + 1} \, dt$.
Think about how you would split this integral and apply the generalized FTC 1 for both limits. What should the two functions of $x$ for the limits be? What is $f(t)$? Success looks like a clear application of the formula $f(h(x))h'(x) - f(g(x))g'(x)$.