6.4 The Fundamental Theorem of Calculus and Accumulation Functions

AP Calculus 6.4 Accumulation Functions Summary

The Fundamental Theorem of Calculus connects derivatives and integrals: if you build a new function from a definite integral with a variable upper limit, its derivative is the function inside the integral. In AP Calculus, this lets you define accumulation functions like $g(x)=\int_a^x f(t)\,dt$ and differentiate them, even when the upper limit is itself a function of $x$.

Why This Matters for the AP Calculus Exam

This topic ties together the two halves of calculus. Differentiation finds an instantaneous rate of change at a point, and integration finds accumulated change over an interval. The Fundamental Theorem of Calculus shows they are inverse processes.

On the AP Calculus exam, you will see accumulation functions in both multiple-choice and free-response questions. You may need to:

• Differentiate a function defined by an integral.
• Use a graph of $f$ to describe the behavior of $g(x)=\int_a^x f(t)\,dt$.
• Interpret accumulated change in context, including units.

Recognizing when a function is built from a definite integral, and knowing that its derivative is the integrand, saves time and prevents errors on both sections.

Key Takeaways

• An accumulation function has the form $g(x)=\int_a^x f(t)\,dt$, where the variable upper limit makes $g$ a function of $x$.
• If $f$ is continuous on an interval containing $a$, then $\dfrac{d}{dx}\left(\int_a^x f(t)\,dt\right)=f(x)$.
• At $x=a$, the accumulation starts at zero: $g(a)=\int_a^a f(t)\,dt=0$.
• When the upper limit is a function like $h(x)$, use the chain rule: $\dfrac{d}{dx}\int_a^{h(x)} f(t)\,dt = f(h(x))\cdot h'(x)$.
• Reversing the limits flips the sign: $\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$, which helps when the variable is on the lower limit.
• Graphical behavior connects $f$ and $g$: where $f$ is positive, $g$ increases; where $f$ is negative, $g$ decreases.

Fundamental Theorem of Calculus and Accumulation Functions

The Fundamental Theorem of Calculus links differentiation and integration. The key idea: the derivative of an integral is the function inside the integral.

You can use a definite integral to define a new function. Let $F$ be defined so that:

$$F(x)=\int_{a}^{x}f(t)\,dt$$

If $f$ is continuous on an interval containing $a$, then for every $x$ in the interval:

$$\frac{d}{dx}\left[\int_{a}^{x}f(t)\,dt\right]=F'(x)=f(x)$$

This means you can find the derivative of an accumulation function without ever computing the integral itself. A classic example of a function defined this way is $f(x)=\int_{0}^{x}e^{-t^2}\,dt$, which has no simple closed-form antiderivative but is still a perfectly good function.

Example 1: Variable Upper Limit Is Just x

Find $g'(16)$ if:

$$g(x)=\int_{5}^{x}\sqrt[4]{t}\,dt$$

Functions defined by definite integrals are called accumulation functions. To find $g'(16)$, first find $g'(x)$, then substitute.

By the Fundamental Theorem of Calculus, the derivative of the integral is the integrand:

$$g'(x)=\sqrt[4]{x}$$

Now substitute $16$:

$$g'(16)=\sqrt[4]{16}=2$$

Example 2: Upper Limit Is a Function of x

When the upper limit is something other than $x$, the chain rule comes in. Find $F'(x)$ for:

$$F(x)=\int_{3}^{x^2}(t+4)\,dt$$

The upper limit is $x^2$, so substitute it into the integrand and multiply by its derivative:

$$F'(x)=\frac{d}{dx}\int_{3}^{x^2}(t+4)\,dt$$ $$F'(x)=(x^2+4)\cdot\frac{d}{dx}x^2$$ $$F'(x)=(x^2+4)(2x)$$

So when the upper limit of the definite integral is a function of $x$, multiply the integrand (evaluated at that function) by the derivative of the upper limit. Keep an eye on the upper limit.

How to Use This on the AP Calculus Exam

MCQ

• Spot when a function is defined by an integral with a variable upper limit. The derivative is the integrand, no integration needed.
• Check the upper limit. If it is $x$, the derivative is just $f(x)$. If it is $h(x)$, multiply by $h'(x)$.
• If the variable is on the lower limit, flip the limits and the sign first.

Free Response

• Show the chain rule step clearly when the upper limit is a function of $x$. Write $f(h(x))\cdot h'(x)$ before simplifying.
• When you describe the behavior of $g(x)=\int_a^x f(t)\,dt$ from a graph of $f$, connect the sign of $f$ to whether $g$ is increasing or decreasing.
• In context problems, include units. The accumulated quantity uses the units of the rate multiplied by the units of the independent variable.

Common Trap

When the variable appears on the lower limit, such as $F(x)=\int_{3x}^{1}\sec^2(t)\,dt$, you cannot apply the rule directly. Reverse the limits first:

$$\int_{a}^{b}f(x)\, dx = -\int_{b}^{a}f(x)\, dx$$

Then differentiate with the chain rule.

Practice Problems

Problems

Question 1

Let $g(x)=\int_{0}^{x}\sqrt{8+\cos(t)}\,dt$. Find $g'(0)$.

Question 2

Let $g(x)=\int_{1}^{x}(5t^2+2t)\,dt$. Find $g'(3)$.

Question 3

Let $g(x)=\int_{0}^{x}\sqrt{\sin\left(t\right)+15}\,dt$. Find $g'\left(\frac{\pi}{2}\right)$.

Question 4

Let $F(x)=\int_{3x}^{1}\sec^2(t)\,dt$. Find $F'(x)$.

Answers and Solutions

Question 1

The upper limit is already $x$, so find $g'(x)$ and evaluate at $x=0$.

By the Fundamental Theorem of Calculus, $g'(x)=\sqrt{8+\cos(x)}$.

Therefore, $g'(0)=\sqrt{8+\cos(0)}=\sqrt{8+1}=\sqrt{9}=3$.

Question 2

Find $g'(x)$ and evaluate at $x=3$.

By the Fundamental Theorem of Calculus, $g'(x)=5x^2+2x$.

Therefore, $g'(3)=5\cdot3^2+2\cdot 3=45+6=51$.

Question 3

Find $g'(x)$ and evaluate at $x=\frac{\pi}{2}$.

By the Fundamental Theorem of Calculus, $g'(x)=\sqrt{\sin(x)+15}$.

Therefore, $g'\left(\frac{\pi}{2}\right)=\sqrt{15+\sin\left(\frac{\pi}{2}\right)}=\sqrt{15+1}=\sqrt{16}=4$.

Question 4

Notice the variable is on the lower limit, not the upper limit. Reverse the limits and flip the sign first.

This uses a property of definite integrals covered in a later topic, 6.6 Applying Properties of Definite Integrals:

$$\int_{a}^{b}f(x)\, dx = -\int_{b}^{a}f(x)\, dx$$

Apply it here:

$$F'(x)=\frac{d}{dx}\left(-\int_{1}^{3x}\sec^2(t)\,dt\right)$$

Now differentiate using the chain rule:

$$F'(x)=-\sec^2(3x)\cdot\frac{d}{dx}3x$$ $$=-3\sec^2(3x)$$

Common Misconceptions

• Integration is not just "reverse differentiation" in a mechanical sense. The Fundamental Theorem of Calculus tells you the derivative of an accumulation function is the integrand, but applying it correctly still requires watching the limits and using the chain rule.
• The rule only gives $f(x)$ directly when the upper limit is exactly $x$. If the upper limit is a function like $x^2$ or $3x$, you must multiply by the derivative of that function.
• A variable on the lower limit changes the sign. $\frac{d}{dx}\int_x^b f(t)\,dt=-f(x)$, not $f(x)$. Reverse the limits before applying the rule.
• The starting point $a$ does not change the derivative. Whether the integral starts at $0$, $5$, or any constant $a$, the derivative is still the integrand. The constant lower limit only sets where the accumulation equals zero.
• The variable inside the integral is a dummy variable. Using $t$ in $\int_a^x f(t)\,dt$ keeps it separate from the limit $x$. Do not confuse the integration variable with the variable upper limit.

Related AP Calculus Guides

• Unit 6 Overview: Integration and Accumulation of Change
• 6.11 Integrating Using Integration by Parts
• 6.1 Integration and Accumulation of Change
• 6.12 Integrating Using Linear Partial Fractions
• 6.3 Riemann Sums, Summation Notation, and Definite Integral Notation
• 6.5 Interpreting the Behavior of Accumulation Functions Involving Area