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FTC Part 1: Differentiating Integrals

Reference: Stewart §5.3 • Chapter: 4 • Section: 3

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Why Area and Slope Are Connected

Imagine filling a bathtub. The water level rises over time, and how fast it rises depends on the flow rate from the faucet. If you know the flow rate at every moment, you can figure out the total water accumulated. Conversely, if you know how much water has accumulated over time, you can figure out the current flow rate by looking at how fast the total is changing.

This is exactly what the Fundamental Theorem of Calculus, Part 1 tells us: the rate at which area accumulates under a curve equals the height of the curve. It's the mathematical version of "the flow rate determines how fast the total grows."

Prerequisite Map

Quick Reference

Property	Value
Concept	Fundamental Theorem of Calculus
Course	MATH161 (Calculus I)
Difficulty	Intermediate
Time	~15 minutes

Key Concepts

The Accumulation Function

When we write

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

we're defining a new function $g$ that measures accumulated area. As $x$ increases, $g(x)$ captures more and more area under the curve $f$.

    y
    │     ┌───f(t)───┐
    │    /           \
    │   /  shaded    \
    │  /   area       \
    │ /    = g(x)      \
    └─────┬─────────┬───────── t
          a         x

Think of $g$ as the "area so far" function.

The Fundamental Theorem, Part 1

If $f$ is continuous on $[a, b]$, then the function

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

is differentiable on $(a, b)$, and

$$\boxed{g'(x) = f(x)}$$

Or in Leibniz notation:

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

The key insight: The derivative of accumulated area equals the height of the curve.

Why This Works

Consider what happens when $x$ increases by a tiny amount $h$:

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

This is the area of a thin strip. For small $h$, this strip is approximately a rectangle with height $f(x)$ and width $h$:

$$g(x+h) - g(x) \approx f(x) \cdot h$$

Dividing by $h$:

$$\frac{g(x+h) - g(x)}{h} \approx f(x)$$

Taking the limit as $h \to 0$ gives $g'(x) = f(x)$.

Physical Interpretation

If $f(t)$ represents:

Velocity → $g(x)$ is displacement (position change)
Flow rate → $g(x)$ is total volume accumulated
Power → $g(x)$ is total energy consumed

The derivative brings you back to the original rate.

Important Note on Variable Names

Notice that we use different variable names: $t$ inside the integral and $x$ as the upper limit. The variable $t$ is a "dummy variable" (it gets integrated away). The function $g$ depends only on $x$.

Practice Problems

Level 1 Direct Application

Find $g'(x)$ if $g(x) = \displaystyle\int_0^x \cos(t)\, dt$.

Thought Process

This is a direct application of FTC Part 1. The integrand is $f(t) = \cos(t)$, which is continuous everywhere. The lower limit is constant (0) and the upper limit is just $x$.

According to FTC1: $\frac{d}{dx}\int_a^x f(t)\, dt = f(x)$

So we simply replace $t$ with $x$ in the integrand.

Show Answer

By FTC Part 1:

$$g'(x) = \cos(x)$$

Level 2 More Complex Integrand

Find the derivative: $\displaystyle\frac{d}{dx}\int_1^x \sqrt{1 + t^3}\, dt$

Thought Process

Check the conditions:

Is the integrand continuous? Yes, $\sqrt{1 + t^3}$ is continuous for all $t$ (since $1 + t^3 > 0$ for reasonable values)
Is the upper limit just $x$? Yes
Is the lower limit constant? Yes (it's 1)

This fits the FTC1 pattern exactly. The fact that the integrand looks complicated doesn't matter. We don't need to actually integrate it!

Show Answer

By FTC Part 1, we simply substitute $x$ for $t$ in the integrand:

$$\frac{d}{dx}\int_1^x \sqrt{1 + t^3}\, dt = \sqrt{1 + x^3}$$

Level 3 Reversed Limits

Find $h'(x)$ if $h(x) = \displaystyle\int_x^5 \sin(t^2)\, dt$.

Thought Process

Here's the catch: the variable $x$ is in the lower limit, not the upper limit. FTC1 as stated has $x$ in the upper limit.

Use the property of integrals: $\int_x^5 f(t)\, dt = -\int_5^x f(t)\, dt$

Now the $x$ is in the upper limit, and we can apply FTC1.

Show Answer

First, flip the limits (which introduces a negative sign):

$$h(x) = \int_x^5 \sin(t^2)\, dt = -\int_5^x \sin(t^2)\, dt$$

Now apply FTC Part 1:

$$h'(x) = -\sin(x^2)$$

Level 4 Graphical Analysis

Let $g(x) = \displaystyle\int_0^x f(t)\, dt$, where $f$ is a continuous function.

If $f(t) > 0$ for $0 < t < 3$ and $f(t) < 0$ for $t > 3$, determine:

On what interval is $g$ increasing?
Where does $g$ have a maximum?
Is $g(5)$ greater than, less than, or equal to $g(3)$?

Thought Process

By FTC1, $g'(x) = f(x)$. This connects the sign of $f$ to whether $g$ is increasing or decreasing:

When $f(x) > 0$, we have $g'(x) > 0$, so $g$ is increasing
When $f(x) < 0$, we have $g'(x) < 0$, so $g$ is decreasing

For the maximum: $g$ is increasing before $x = 3$ and decreasing after, so $x = 3$ is where the max occurs.

For part (c): Since $g$ is decreasing from $x = 3$ to $x = 5$, the value at $x = 5$ must be less than at $x = 3$.

Show Answer

(a) By FTC1, $g'(x) = f(x)$.

$g'(x) > 0$ when $f(x) > 0$, which is $0 < x < 3$
Therefore, $g$ is increasing on $(0, 3)$

(b) At $x = 3$:

$g'(3) = f(3) = 0$ (transition from positive to negative)
$g'$ changes from positive to negative at $x = 3$
By the First Derivative Test, $g$ has a maximum at $x = 3$

(c) Since $g$ is decreasing for $x > 3$, we have $g(5) < g(3)$.

Level 5 Finding Concavity

Let $g(x) = \displaystyle\int_0^x \frac{t^2}{t^2 + t + 2}\, dt$.

Find $g'(x)$.
Find $g''(x)$.
Determine where $g$ is concave upward and where it is concave downward.

Thought Process

For part (a), FTC1 applies directly.

For part (b), we need to differentiate $g'(x)$: this is ordinary differentiation using the quotient rule.

For part (c), concavity depends on the sign of $g''(x)$:

$g'' > 0$ means concave up
$g'' < 0$ means concave down

We'll need to analyze when the numerator of $g''$ is positive or negative.

Show Answer

(a) By FTC Part 1: $$g'(x) = \frac{x^2}{x^2 + x + 2}$$

(b) Use the quotient rule to differentiate $g'(x)$: $$g''(x) = \frac{(2x)(x^2 + x + 2) - x^2(2x + 1)}{(x^2 + x + 2)^2}$$

Expand the numerator: $$= \frac{2x^3 + 2x^2 + 4x - 2x^3 - x^2}{(x^2 + x + 2)^2} = \frac{x^2 + 4x}{(x^2 + x + 2)^2} = \frac{x(x + 4)}{(x^2 + x + 2)^2}$$

(c) The denominator $(x^2 + x + 2)^2$ is always positive.

The numerator $x(x + 4)$ is:

Positive when $x < -4$ or $x > 0$
Negative when $-4 < x < 0$
Zero when $x = -4$ or $x = 0$

Therefore:

$g$ is concave upward on $(-\infty, -4)$ and $(0, \infty)$
$g$ is concave downward on $(-4, 0)$

Mastery Checklist

[ ] I can state FTC Part 1 precisely
[ ] I can differentiate $\int_a^x f(t)\, dt$ when the integrand is continuous
[ ] I understand that I don't need to actually compute the integral to find its derivative
[ ] I can handle the case when $x$ appears in the lower limit (by flipping and negating)
[ ] I can connect the sign of $f$ to increasing/decreasing behavior of the accumulation function
[ ] I understand that FTC1 only works when $f$ is continuous

Mental Model

The Bathtub Analogy: Think of $f(t)$ as the rate water flows into a bathtub and $g(x)$ as the total water in the tub at time $x$. The FTC says: the rate of change of total water equals the current flow rate. This is almost obvious when you think about it. Of course the water level rises at the rate water flows in!

Connections

Looking back:

This builds on the definition of the derivative (limit of difference quotient)
Uses the interpretation of definite integrals as accumulated area

Looking ahead:

FTC Part 1 with Chain Rule handles when the upper limit is a function of $x$
FTC Part 2 shows how to evaluate definite integrals using antiderivatives
Together, they reveal that differentiation and integration are inverse operations

Real-world connections:

In physics: velocity integrates to displacement; FTC1 says differentiating displacement gives velocity
In economics: marginal cost is the derivative of total cost (the inverse operation of integration)

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Definite Integrals	Skills Index	FTC Part 1 + Chain Rule

Last updated: 2026-01-22