Applications of FTC 1: Differentiating Partial Area Functions

TL;DR

You'll use the First Fundamental Theorem of Calculus (FTC 1) to find derivatives of functions defined by integrals, especially when the upper limit of integration is itself a function of $x$. The key is to remember the chain rule when the upper limit isn't just $x$. This lets you calculate how the accumulated area changes with respect to a variable.

1. The Mental Model

Imagine an integral as a "net area accumulator." When you differentiate this accumulated area with respect to a variable, you're essentially asking: "How fast is this accumulated area changing as that variable changes?"

2. The Core Material

The First Fundamental Theorem of Calculus (FTC 1) tells us that if $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$. This means the derivative of an integral with a constant lower limit and a variable upper limit is just the integrand evaluated at that upper limit.

However, things get a bit more interesting when the upper limit of integration isn't just $x$, but a function of $x$, say $g(x)$. Or even when both limits are functions of $x$. This is where the chain rule becomes your best friend.

Differentiating with a Function as an Upper Limit

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If you have a function like $G(x) = \int_a^{g(x)} f(t) dt$, to find $G'(x)$, you combine FTC 1 with the chain rule.
1. Evaluate the integrand $f(t)$ at the upper limit $g(x)$, giving you $f(g(x))$.
2. Multiply this by the derivative of the upper limit, $g'(x)$.

So, $G'(x) = f(g(x)) \cdot g'(x)$.

Differentiating with a Function as a Lower Limit

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What if the lower limit is a function, say $h(x)$, and the upper limit is a constant, $b$?
$H(x) = \int_{h(x)}^b f(t) dt$.
You can rewrite this using the property $\int_a^b f(t) dt = -\int_b^a f(t) dt$:
$H(x) = -\int_b^{h(x)} f(t) dt$.
Now, apply the same rule as above:
$H'(x) = -f(h(x)) \cdot h'(x)$.

Differentiating with Functions as Both Limits

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This is the most general case: $K(x) = \int_{h(x)}^{g(x)} f(t) dt$.
You can split this integral into two using an arbitrary constant $c$ between $h(x)$ and $g(x)$:
$K(x) = \int_{h(x)}^c f(t) dt + \int_c^{g(x)} f(t) dt$.
Then rewrite the first part:
$K(x) = -\int_c^{h(x)} f(t) dt + \int_c^{g(x)} f(t) dt$.
Now differentiate each part separately:
$K'(x) = -f(h(x)) \cdot h'(x) + f(g(x)) \cdot g'(x)$.

Let's visualize the process:

graph TD
    A["Start: Define H(x) = ∫[h(x) to g(x)] f(t) dt"] --> B{Are both limits functions of x?};
    B -- Yes --> C["Split integral using a constant 'c': -∫[c to h(x)] f(t) dt + ∫[c to g(x)] f(t) dt"];
    B -- No, only upper limit g(x) --> D["Apply Chain Rule: f(g(x)) * g'(x)"];
    B -- No, only lower limit h(x) --> E["Rewrite: -∫[constant to h(x)] f(t) dt"];
    E --> F["Apply Chain Rule: -f(h(x)) * h'(x)"];
    C --> G["Differentiate each part separately"];
    G --> H["First part: -f(h(x)) * h'(x)"];
    G --> I["Second part: +f(g(x)) * g'(x)"];
    H & I --> J["Combine: f(g(x)) * g'(x) - f(h(x)) * h'(x)"];
    D --> K["Result: Derivative of partial area function"];
    F --> K;
    J --> K;

3. Worked Example

Let's find the derivative of $F(x) = \int_{x^2}^{e^x} \sin(t^3) dt$.

Here, $f(t) = \sin(t^3)$, the lower limit is $h(x) = x^2$, and the upper limit is $g(x) = e^x$.

1. Differentiate with respect to the upper limit:

   • Evaluate $f(t)$ at $g(x)$: $f(e^x) = \sin((e^x)^3) = \sin(e^{3x})$.
   • Find the derivative of the upper limit: $g'(x) = \frac{d}{dx}(e^x) = e^x$.
   • The first part is $f(g(x)) \cdot g'(x) = \sin(e^{3x}) \cdot e^x$.

2. Differentiate with respect to the lower limit:

   • Evaluate $f(t)$ at $h(x)$: $f(x^2) = \sin((x^2)^3) = \sin(x^6)$.
   • Find the derivative of the lower limit: $h'(x) = \frac{d}{dx}(x^2) = 2x$.
   • The second part, which gets subtracted, is $f(h(x)) \cdot h'(x) = \sin(x^6) \cdot 2x$.

3. Combine the results:
   $F'(x) = f(g(x)) \cdot g'(x) - f(h(x)) \cdot h'(x)$
   $F'(x) = \sin(e^{3x}) \cdot e^x - \sin(x^6) \cdot 2x$.

4. Key Takeaways

• FTC 1 lets you find the derivative of an integral: $ \frac{d}{dx} \int_a^x f(t) dt = f(x)$.
• When the upper limit is a function $g(x)$, apply the chain rule: $ \frac{d}{dx} \int_a^{g(x)} f(t) dt = f(g(x)) \cdot g'(x)$.
• When the lower limit is a function $h(x)$, use the negative sign from flipping limits: $ \frac{d}{dx} \int_{h(x)}^b f(t) dt = -f(h(x)) \cdot h'(x)$.
• For both limits as functions, subtract the derivative from the lower limit part from the upper limit part.
• The integrand $f(t)$ itself isn't differentiated, only evaluated at the limits.
• Always remember to multiply by the derivative of the limit function (the chain rule part).

Common mistakes to avoid:
- Forgetting the chain rule part (multiplying by $g'(x)$ or $h'(x)$).
- Incorrectly handling the negative sign when the lower limit is a function.
- Differentiating the integrand $f(t)$ instead of just evaluating it.
- Not recognizing when the upper or lower limit is a constant (in which case its derivative is zero).

5. Now Try It

Find the derivative of $G(x) = \int_{\sqrt{x}}^{x^3} \cos(t^2) dt$.

What to do:
1. Identify $f(t)$, $h(x)$, and $g(x)$.
2. Calculate $g'(x)$ and $h'(x)$.
3. Evaluate $f(g(x))$ and $f(h(x))$.
4. Combine these pieces using the formula for differentiating with functions at both limits.

What success looks like:
You should arrive at an expression for $G'(x)$ that involves $\cos$ terms, multiplied by derivatives of $x^3$ and $\sqrt{x}$.