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

1. Antiderivatives and What the Fundamental Theorem Connects

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

If one antiderivative works, then infinitely many do:

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

That $+C$ is there because the derivative of a constant is zero.

Now here’s the big connection.

If $f$ is continuous and you define a new function

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

then something amazing happens:

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

So the function defined by “area from $a$ to $x$” is automatically an antiderivative of $f$.

That’s FTC Part 1.

FTC Part 2 goes the other direction. If $F'(x) = f(x)$, then

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

So:

• Differentiate an integral → original function
• Integrate a derivative over an interval → net change

The slope of the accumulation function equals the height of the original function.

2. FTC Part 1 and Differentiating Definite Integrals

If

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

then

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

You don’t evaluate the integral. You don’t find an antiderivative. You just remove the integral and plug in $x$.

When the upper bound isn’t just $x$

If

$$G(x) = \int_a^{h(x)} f(t)\,dt,$$

then

$$G'(x) = f(h(x)) \cdot h'(x).$$

You:

1. Plug the upper bound into the integrand.
2. Multiply by the derivative of that upper bound.

Example idea:
If $G(x) = \int_1^{x^3} \sqrt{t}\,dt$, then

$$G'(x) = \sqrt{x^3} \cdot 3x^2.$$

Students almost always forget the $3x^2$. That chain rule factor is where points disappear on FRQs.

If the lower bound is a function, the derivative picks up a negative sign. If both bounds are functions, treat it as:

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

and differentiate both pieces.

3. FTC Part 2 and Evaluating Definite Integrals

If $F'(x) = f(x)$, then

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

This is what you use constantly on quizzes and the no-calculator section.

The flow

1. Find any antiderivative $F(x)$.
2. Plug in the upper bound.
3. Plug in the lower bound.
4. Subtract.

No $+C$. It cancels anyway.

Quick example:

$$\int_0^2 (4x^3 - 1)\,dx$$

An antiderivative is $F(x) = x^4 - x$.

Evaluate:

$$F(2) - F(0) = (16 - 2) - (0 - 0) = 14.$$

That number represents net signed area.

4. Net Area, Geometry, and Common Traps

A definite integral gives net accumulation, not total area.

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

If a graph crosses the axis and the question asks for total area, you must:

• Split at intercepts
• Make each piece positive
• Add

Here’s the sign idea visually.

Positive and negative signed area on a graph

The middle shaded region is above the x-axis, so it contributes positive area. The shaded regions on the left and right are below the axis, so they count as negative when you evaluate a definite integral.

Other properties you should know instantly:

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

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

If you’re given a graph made of rectangles or triangles, use geometry. The exam loves simple shapes because it tests whether you understand accumulation, not algebra stamina.