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 lets you turn an accumulation function—defined by an integral—into a derivative. When you differentiate an integral with a variable upper limit, you get back the integrand. This is one of the most powerful shortcuts in calculus and lets you solve problems without ever computing the antiderivative explicitly.

1. The Mental Model

You're building up area under a curve as you slide your right endpoint to the right. The rate at which that area grows is exactly the height of the curve at your current position. FTC 1 formalizes this intuition: if you define $F(x) = \int_a^x f(t) \, dt$, then $F'(x) = f(x)$. You're not finding an antiderivative by guessing—you're reading off the instantaneous rate of change of accumulated area.

2. The Core Material

The Core Theorem

FTC 1 states: if $f$ is continuous on $[a, b]$ and you define
$$F(x) = \int_a^x f(t) \, dt,$$
then $F'(x) = f(x)$ for all $x \in (a, b)$.

The key insight is that differentiation and integration are inverse operations. When you differentiate an accumulation function, the integral symbol cancels and you recover the original function.

Why This Matters

In applications, you often know a rate (like velocity, revenue, or population growth) and want to find the total amount (distance, profit, population size). You build that total as an integral. Then a question asks: "What's the rate of change of the total at time $t = 5$?" Instead of finding a closed-form antiderivative, you just apply FTC 1—the answer is the integrand evaluated at $t = 5$.

How to Apply It: Three Patterns

Pattern 1: Variable upper limit, constant lower limit

If $F(x) = \int_a^x f(t) \, dt$, then $F'(x) = f(x)$.

Pattern 2: Composite upper limit (chain rule)

If $F(x) = \int_a^{g(x)} f(t) \, dt$, then
$$F'(x) = f(g(x)) \cdot g'(x).$$

You differentiate the upper limit and multiply by the derivative of that limit.

Pattern 3: Both limits variable (add a negative)

If $F(x) = \int_{h(x)}^{g(x)} f(t) \, dt$, split it:
$$F(x) = \int_a^{g(x)} f(t) \, dt - \int_a^{h(x)} f(t) \, dt,$$
then
$$F'(x) = f(g(x)) \cdot g'(x) - f(h(x)) \cdot h'(x).$$

This is just the chain rule applied twice.

A Concrete Walkthrough

Let's compute $\frac{d}{dx} \int_1^{x^2} \sqrt{t^3 + 1} \, dt$.

Here $f(t) = \sqrt{t^3 + 1}$, $g(x) = x^2$, and $g'(x) = 2x$.

By Pattern 2:
$$\frac{d}{dx} \int_1^{x^2} \sqrt{t^3 + 1} \, dt = \sqrt{(x^2)^3 + 1} \cdot 2x = 2x\sqrt{x^6 + 1}.$$

Notice: we never computed the antiderivative of $\sqrt{t^3 + 1}$. We don't need to. FTC 1 gives us the derivative directly.

3. Worked Example

Problem: A chemical tank starts with 100 litres. Water flows in at a rate of $r(t) = 5 + 2\sin(t)$ litres per minute, where $t$ is time in minutes. Find the rate of change of the total volume at $t = 3$ minutes.

Solution:

The total volume at time $t$ is
$$V(t) = 100 + \int_0^t (5 + 2\sin(s)) \, ds.$$

We want $V'(3)$. By FTC 1, the derivative of the integral is the integrand:
$$V'(t) = 5 + 2\sin(t).$$

At $t = 3$:
$$V'(3) = 5 + 2\sin(3) \approx 5 + 2(0.1411) \approx 5.28 \text{ litres/min}.$$

This tells us that at exactly $t = 3$ minutes, water is flowing in at about 5.28 litres per minute. FTC 1 bypassed the need to actually integrate $5 + 2\sin(t)$; we read the answer straight from the rate function.

4. Key Takeaways

4.1 Most Important Concepts

1. FTC 1 is the inverse of integration. Differentiate an accumulation integral and you recover the original integrand.
2. Chain rule applies to the limits. If the upper limit is $g(x)$ instead of $x$, multiply by $g'(x)$.
3. You don't need a closed-form antiderivative. Even if $\int f(t) \, dt$ can't be written in elementary functions, you can still find the derivative of $F(x) = \int_a^x f(t) \, dt$.
4. Both limits can vary. When both limits depend on $x$, treat the lower limit as a negative contribution.
5. Continuity matters. FTC 1 requires $f$ to be continuous; if $f$ has a jump discontinuity inside the interval, FTC 1 doesn't apply at that point.
6. This unlocks optimisation problems. Find where $F'(x) = 0$ to locate critical points of accumulation functions without ever computing $F(x)$ in closed form.
7. Real-world rates become integrals. Velocity → displacement, flow rate → total volume, marginal cost → total cost. FTC 1 lets you swap between rate and total effortlessly.

4.2 Common Misconceptions

4.3 Compare & Contrast

5. Now Try It

Exercise: A factory's cumulative energy consumption (in megawatt-hours) after $t$ hours is given by
$$E(t) = \int_0^t (12 + 3\sqrt{s}) \, ds.$$

(a) Find $E'(t)$.

(b) What is $E'(4)$? Interpret this value in context.

(c) Without computing $E(t)$ in closed form, explain why $E'(t)$ is always positive. What does this tell you about the cumulative energy consumption?

What to check:
- (a): You should get $E'(t) = 12 + 3\sqrt{t}$ using FTC 1 directly.
- (b): $E'(4) = 12 + 3\sqrt{4} = 12 + 6 = 18$ megawatt-hours per hour. This is the instantaneous power consumption at $t = 4$ hours.
- (c): Since $12 + 3\sqrt{t} > 0$ for all $t \geq 0$, the factory is always consuming energy at a positive rate, so cumulative consumption never decreases.

Success looks like: You answer (a) and (b) correctly without computing an antiderivative, and you explain (c) by reasoning about the sign of $E'(t)$ and what it means physically.

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