Fundamental Theorem of Calculus

Antiderivatives and Indefinite Integrals

An antiderivative of $f(x)$ is a function $F(x)$ such that $F'(x) = f(x)$. The indefinite integral represents the family of all antiderivatives:

$\int f(x)\,dx = F(x) + C$

where $C$ is the constant of integration. For example, $\int 2x\,dx = x^2 + C$ because $\frac{d}{dx}[x^2 + C] = 2x$.

The Definite Integral as Area

The definite integral $\int_a^b f(x)\,dx$ represents the signed area between the graph of $f$ and the $x$-axis from $x = a$ to $x = b$. Area above the $x$-axis counts as positive; area below counts as negative.

This can be computed as a limit of Riemann sums:

$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x$

But the FTC gives us a much easier method.

FTC Part 1 (The Derivative of an Integral)

Statement: If $f$ is continuous on $[a, b]$ and we define:

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

then $F$ is differentiable on $(a, b)$ and:

$F'(x) = f(x)$

In plain language: if you integrate a continuous function and then differentiate the result, you get back the original function.

This is remarkable — it says that the accumulation function $F(x) = \int_a^x f(t)\,dt$ is always an antiderivative of $f$.

Key example:

$\frac{d}{dx} \int_0^x \cos(t)\,dt = \cos(x)$

FTC Part 1 with Chain Rule

When the upper limit is a function of $x$ (not just $x$ itself), combine FTC Part 1 with the chain rule:

$\frac{d}{dx} \int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)$

This is one of the most commonly tested variations on the AP exam.

Example: $\frac{d}{dx} \int_1^{x^2} \sqrt{t + 1}\,dt = \sqrt{x^2 + 1} \cdot 2x = 2x\sqrt{x^2 + 1}$

When both limits are functions of $x$, split the integral:

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

FTC Part 2 (The Evaluation Theorem)

Statement: If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ (meaning $F' = f$), then:

$\int_a^b f(x)\,dx = F(b) - F(a)$

This is the formula you use to evaluate every definite integral. The notation $F(x) \Big|_a^b$ means $F(b) - F(a)$.

Example:

$\int_1^4 (3x^2 + 2)\,dx = [x^3 + 2x]_1^4 = (64 + 8) - (1 + 2) = 72 - 3 = 69$

Accumulation Functions

An accumulation function has the form $F(x) = \int_a^x f(t)\,dt$. It tells you the net area under $f$ from $a$ to $x$.

Key properties:

• $F(a) = 0$ (the integral from $a$ to $a$ is always 0)
• $F'(x) = f(x)$ (by FTC Part 1)
• $F$ is increasing where $f > 0$ and decreasing where $f < 0$
• $F$ has a local max where $f$ changes from positive to negative
• $F$ has a local min where $f$ changes from negative to positive
• $F$ is concave up where $f$ is increasing ($f' > 0$) and concave down where $f$ is decreasing ($f' < 0$)

The AP exam frequently gives you the graph of $f$ and asks questions about $F$, like: "Where does $F$ have a maximum?" or "Where is $F$ concave up?"

Properties of Definite Integrals

These properties work alongside the FTC:

• $\int_a^a f(x)\,dx = 0$
• $\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$
• $\int_a^b [f(x) + g(x)]\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx$
• $\int_a^b cf(x)\,dx = c \int_a^b f(x)\,dx$
• $\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$ (splitting the interval)

The Mean Value Theorem for Integrals

If $f$ is continuous on $[a, b]$, then there exists a $c \in [a, b]$ such that:

$f(c) = \frac{1}{b - a} \int_a^b f(x)\,dx$

The quantity $\frac{1}{b-a} \int_a^b f(x)\,dx$ is the average value of $f$ on $[a, b]$.

Tip 1: Know Which Part of the FTC You Need

If the problem asks you to differentiate an integral → FTC Part 1. If the problem asks you to evaluate a definite integral → FTC Part 2. This distinction saves time.

Tip 2: Always Check the Upper Limit for Chain Rule

When differentiating $\int_a^{g(x)} f(t)\,dt$, the answer is NOT just $f(x)$. You must multiply by $g'(x)$. The AP exam tests this regularly.

Tip 3: For Accumulation Function Graphs, Focus on Signs

When $f(t) > 0$, the accumulation $F(x)$ is increasing. When $f(t) < 0$, $F(x)$ is decreasing. Zero crossings of $f$ correspond to extrema of $F$.

Tip 4: Compute the Average Value Using the Formula

Average value = $\frac{1}{b-a} \int_a^b f(x)\,dx$. This appears often on the AP exam, especially in free-response problems.

Tip 5: Use Properties to Simplify Before Evaluating

Splitting integrals, reversing limits, and using symmetry can simplify calculations. For even functions on symmetric intervals: $\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$.