Topic 6.7 Notes – The Fundamental Theorem of Calculus and Definite Integrals

The Fundamental Theorem of Calculus

Before using the theorem, lock in one definition:

An antiderivative of $f$ is a function $F$ such that
$$F'(x) = f(x).$$

So if you differentiate $F$, you get back $f$. Integration asks the reverse question: “What function has this derivative?”

The Fundamental Theorem of Calculus explains how this connects to definite integrals.

Part 1 - Accumulation Functions

Suppose you define a function by accumulating area:

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

If $f$ is continuous, then

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

That’s huge. It says: the derivative of accumulated area is the original function.

What this means visually

Here’s the idea. Focus on the shaded region from $a$ to the moving right endpoint $x$.

Signed area under $y = f(x)$

• The integral represents signed area from $a$ to $x$.
• As $x$ moves, the accumulated area changes.
• The instantaneous rate at which that area changes equals the height of the curve.

So if the graph is high above the x-axis, area is increasing quickly. If the graph is below the axis, accumulated area decreases.

This shows up often on FRQs where they define something like
$G(x) = \int_2^x f(t)\,dt$
and ask for $G'(x)$. The answer is just $f(x)$. No integration needed.

Part 2 - Evaluating a Definite Integral

Now the computational tool you’ll use constantly.

If:

• $f$ is continuous on $[a,b]$, and
• $F$ is any antiderivative of $f$,

then

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

This replaces Riemann sums with simple substitution.

Why this works

The definite integral measures total signed area. The antiderivative keeps track of accumulation. Evaluating at endpoints gives the net change.

The $+C$ disappears because constants cancel when you subtract.

Evaluating a Definite Integral Step-by-Step

Let’s walk through one:

$$\int_{1}^{3} (4x^3 - 2x)\,dx$$

1. Find an antiderivative.
   $$F(x) = x^4 - x^2$$

2. Plug in the upper bound:
   $F(3) = 81 - 9 = 72$

3. Plug in the lower bound:
   $F(1) = 1 - 1 = 0$

4. Subtract:
   $$72 - 0 = 72$$

That’s it.

On no-calculator multiple choice, algebra mistakes here are common. Keep parentheses when plugging in negative bounds.

Geometry and Properties That Save Time

Not every integral needs power rule work.

Signed Area

• Above x-axis → positive contribution
• Below x-axis → negative contribution

If a graph forms simple shapes, use geometry. In the example below, the region from $0$ to $4$ is a triangle above the x-axis, and from $4$ to $8$ is a semicircle below the x-axis.

You might compute:

• Triangle area using $\frac{1}{2}bh$
• Semicircle using $\frac{1}{2}\pi r^2$

Then subtract the negative portion.

Useful Integral Properties

$$\int_a^a f(x)\,dx = 0$$

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

$$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$$

Splitting intervals helps when:

• A function changes sign
• A graph is piecewise
• You’re given table values

Recognizing structure is part of what this topic is testing.