Topic 6.4 Notes – The Fundamental Theorem of Calculus and Accumulation Functions

The Fundamental Theorem of Calculus and Accumulation Functions

Suppose $f$ is continuous, and you define a new function:

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

This is an accumulation function. It measures the net (signed) area under $f$ from a fixed starting point $a$ to a moving endpoint $x$.

Here’s the key result:

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

That’s Fundamental Theorem of Calculus Part 1.

What this means

• Integration builds area.
• Differentiation measures rate of change.
• When you differentiate accumulated area, you get back the original function.

The variable inside the integral (usually $t$) is a dummy variable. It disappears when you differentiate. Only the upper bound matters.

What Accumulation Functions Represent

Think of $F(x) = \int_a^x f(t)\,dt$ as “total change so far.”

• If $f(x) > 0$, area is added → $F$ increases.
• If $f(x) < 0$, area is subtracted → $F$ decreases.
• If $f(x) = 0$, then $F'(x) = 0$ → horizontal tangent.
• If $f$ changes sign, $F$ can have a local max or min.

The integrand $f(x)$ controls the slope of the accumulation function.

Here’s what that relationship looks like visually. In the graph below, the red curve is the accumulation function and its slope at any point matches the value of $f(x)$.

Accumulation function g and its derivative f on the same axes

You should be able to look at a graph of $f$ and describe where $F$ is increasing, decreasing, or has extrema. That’s a common no-calculator multiple choice setup.

How to Differentiate Accumulation Functions

1. Upper bound is just $x$

If

$$F(x)=\int_2^x (3t^2+1)\,dt$$

Then

$$F'(x)=3x^2+1$$

You replace $t$ with $x$. Nothing else.

If asked for $F'(4)$, compute $3(4)^2+1=49$.
You never needed to evaluate the integral.

2. Upper bound is a function of $x$

If

$$F(x)=\int_1^{x^3} \sqrt{t+5}\,dt$$

Now the upper bound is $x^3$, so the chain rule joins the party:

$$F'(x)=\sqrt{x^3+5}\cdot 3x^2$$

What happened?

1. Plug in the upper bound.
2. Multiply by the derivative of the upper bound.

This shows up a lot on FRQs. Students often forget the extra factor.

3. Variable in the lower bound

If

$$F(x)=\int_{x^2}^4 \cos(t)\,dt$$

Rewrite first by switching bounds:

$$F(x)=-\int_4^{x^2}\cos(t)\,dt$$

Now differentiate:

$$F'(x)=-\cos(x^2)\cdot 2x$$

Switching bounds introduces a negative sign. Missing that sign costs easy points.

Representing Functions with Definite Integrals

A definite integral can define a function.

Example:

$$H(x)=\int_0^x e^{-t^2}\,dt$$

You cannot find an elementary antiderivative for $e^{-t^2}$. That’s fine.

By FTC:

$$H'(x)=e^{-x^2}$$

This is powerful. You can:

• Find slopes
• Determine increasing/decreasing intervals
• Evaluate derivatives at specific points

All without ever computing the integral.

That’s the point of this topic. The integral defines the function. The derivative reveals its behavior.