Fundamental Theorem of Calculus (Part 1)

TL;DR

The Fundamental Theorem of Calculus (Part 1), often called FTC1, connects differentiation and integration, showing they're inverse processes. It states that differentiating an integral with a variable upper limit essentially "undoes" the integration. This theorem is crucial for evaluating definite integrals and understanding the relationship between a function and its accumulation function.

1. The Mental Model

Imagine you have a function that calculates the accumulated area under another curve up to a certain point. FTC1 tells you that if you then find the rate of change of this accumulated area function, you'll get back the original curve itself at that point.

2. The Core Material

FTC1 is a foundational concept in calculus because it establishes the relationship between the two main branches: differential calculus (rates of change) and integral calculus (accumulation).

Let's say you have a continuous function $f(t)$. If you define a new function $F(x)$ as the accumulated area under $f(t)$ from a constant $a$ to a variable $x$, like this:

$F(x) = \int_{a}^{x} f(t) dt$

FTC1 states that if you differentiate $F(x)$ with respect to $x$, you'll get the original function $f(x)$ back:

$\frac{d}{dx} \left[ \int_{a}^{x} f(t) dt \right] = f(x)$

The variable $t$ inside the integral is a "dummy" variable; it just helps define the area as you move from $a$ to $x$. When you differentiate with respect to $x$, that's what makes the magic happen.

How to apply FTC1

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1. Direct Application: If your upper limit is just $x$, and the lower limit is a constant, simply replace the dummy variable in the integrand with $x$.

   Example: $\frac{d}{dx} \left[ \int_{2}^{x} \sin(t) dt \right] = \sin(x)$

2. Chain Rule with FTC1: If your upper limit is a function of $x$, say $g(x)$, you need to use the chain rule.

   $\frac{d}{dx} \left[ \int_{a}^{g(x)} f(t) dt \right] = f(g(x)) \cdot g'(x)$

   You substitute the upper limit function into the integrand and then multiply by the derivative of that upper limit function.

   Example: $\frac{d}{dx} \left[ \int_{1}^{x^2} e^t dt \right]$
   Here, $f(t) = e^t$ and $g(x) = x^2$. So $g'(x) = 2x$.
   Applying the rule: $e^{x^2} \cdot 2x$

3. Variable in the Lower Limit: If the variable is in the lower limit, you can use a property of integrals to flip the limits and introduce a negative sign: $\int_{x}^{a} f(t) dt = - \int_{a}^{x} f(t) dt$. Then apply FTC1.

   Example: $\frac{d}{dx} \left[ \int_{x}^{5} \sqrt{t} dt \right] = \frac{d}{dx} \left[ - \int_{5}^{x} \sqrt{t} dt \right] = - \sqrt{x}$

4. Variables in Both Limits: If both the upper and lower limits are functions of $x$, you can split the integral into two parts using an arbitrary constant $c$:

   $\int_{h(x)}^{g(x)} f(t) dt = \int_{h(x)}^{c} f(t) dt + \int_{c}^{g(x)} f(t) dt$
   $= - \int_{c}^{h(x)} f(t) dt + \int_{c}^{g(x)} f(t) dt$

   Then apply the chain rule version of FTC1 to each part.

   $\frac{d}{dx} \left[ \int_{h(x)}^{g(x)} f(t) dt \right] = f(g(x)) \cdot g'(x) - f(h(x)) \cdot h'(x)$

Here's how the process generally flows:

graph TD
    A["Start: You have d/dx [∫(from a to g(x)) f(t) dt]"] --> B{Is upper limit g(x) = x?};
    B -- Yes --> C["Substitute x for t in f(t)"] --> D["Result: f(x)"];
    B -- No (g(x) is a function of x) --> E["Substitute g(x) for t in f(t)"] --> F["Multiply by derivative of g(x) (g'(x))"] --> G["Result: f(g(x)) * g'(x)"];
    A --> H{Is lower limit h(x) a variable?};
    H -- Yes, and upper is constant --> I["Rewrite as -∫(from constant to h(x)) f(t) dt"] --> J["Apply FTC1 (Chain Rule) to -f(h(x)) * h'(x)"];
    H -- Yes, and upper is variable --> K["Split into -∫(from C to h(x)) f(t) dt + ∫(from C to g(x)) f(t) dt"] --> L["Apply FTC1 (Chain Rule) to both parts and combine"];

3. Worked Example

Let's find the derivative of $G(x) = \int_{x^3}^{5} \cos(t^2) dt$.

First, notice that the variable $x^3$ is in the lower limit, and the upper limit is a constant (5).
We can rewrite this integral by flipping the limits and changing the sign:
$G(x) = - \int_{5}^{x^3} \cos(t^2) dt$

Now, this looks like the form we can use FTC1 with the chain rule.
Here, our original function is $f(t) = \cos(t^2)$, and our upper limit function is $g(x) = x^3$.

According to the chain rule extension of FTC1:
$\frac{d}{dx} \left[ - \int_{a}^{g(x)} f(t) dt \right] = - f(g(x)) \cdot g'(x)$

1. Identify $f(t)$: $f(t) = \cos(t^2)$
2. Identify $g(x)$: $g(x) = x^3$
3. Find $g'(x)$: $g'(x) = \frac{d}{dx}(x^3) = 3x^2$
4. Substitute $g(x)$ into $f(t)$: $f(g(x)) = \cos((x^3)^2) = \cos(x^6)$
5. Combine them with the negative sign: $- \cos(x^6) \cdot (3x^2)$

So, $\frac{d}{dx} \left[ \int_{x^3}^{5} \cos(t^2) dt \right] = -3x^2 \cos(x^6)$.

4. Key Takeaways

• FTC1 states that differentiating an accumulation function $\int_{a}^{x} f(t) dt$ returns the original function $f(x)$.
• The key idea is that differentiation and integration are inverse operations, undoing each other.
• When the upper limit is a function of $x$ (e.g., $g(x)$), remember to apply the chain rule: $f(g(x)) \cdot g'(x)$.
• If the variable is in the lower limit, first use the integral property $\int_{b}^{a} f(t) dt = - \int_{a}^{b} f(t) dt$ to move it to the upper limit.
• If both limits are functions of $x$, split the integral using an arbitrary constant and apply FTC1 to both parts.

Common Mistakes to Avoid:

• Forgetting the Chain Rule: This is the most frequent error when the upper limit is a function of $x$ other than just $x$. Always multiply by the derivative of the upper limit.
• Not Handling Lower Limit Variables Correctly: If the variable is in the lower limit, don't just substitute; remember to introduce a negative sign after flipping the limits.
• Confusing the Dummy Variable: The $t$ inside the integral is a placeholder. When you apply FTC1, it's replaced by the upper limit variable (or function of $x$).
• Differentiating the Integrand: You don't differentiate $f(t)$ itself; you substitute the upper limit into $f(t)$.

5. Now Try It

Find the derivative of the function $H(x) = \int_{2x}^{x^4} \sqrt{1 + t^2} dt$.

What to do: Apply the rule for when both limits are functions of $x$. Identify $f(t)$, $g(x)$, and $h(x)$, and their derivatives. Then substitute and combine.

What success looks like: Your final answer should be in terms of $x$, involving the square root term, multiplied by the derivatives of the upper and lower limits. It should be a single expression.