The Fundamental Theorem of Calculus connects derivatives and integrals: if you build a function by integrating from a fixed point, its derivative gives you back the original function, and any definite integral can be evaluated by finding an antiderivative and subtracting. In practice, $\int_a^b f(x)\,dx=F(b)-F(a)$ , where $F$ is an antiderivative of $f$ . For AP Calculus, check continuity before applying the theorem.

Why This Matters for the AP Calculus Exam

This theorem is the bridge between the two main ideas of AP Calculus: differentiation and integration. It shows up constantly because it lets you evaluate definite integrals exactly instead of estimating with Riemann sums. You will use it to find accumulated change, compute exact areas, and differentiate functions that are defined by integrals. Getting comfortable with both directions of this theorem, evaluating integrals and differentiating integral-defined functions, sets up later work on areas, volumes, and motion.

more resources to help you study

practice multiple choice FRQ practice & scoring cheatsheets score calculator key terms

Key Takeaways

An antiderivative of $f$ is a function whose derivative equals $f$ .
If $f$ is continuous near $a$ , then $F(x)=\int_{a}^{x} f(t)\,dt$ is an antiderivative of $f$ , so $\frac{d}{dx}\int_{a}^{x} f(t)\,dt = f(x)$ .
To evaluate a definite integral, use $\int_{a}^{b} f(x)\,dx = F(b) - F(a)$ for any antiderivative $F$ .
When the upper limit is a function of $x$ (like $2x$ ), apply the chain rule after substituting.
The function $f$ must be continuous on $[a,b]$ for the evaluation rule to apply.
The constant of integration cancels in a definite integral, so you do not need it when evaluating $F(b)-F(a)$ .

Connecting Antiderivatives and Definite Integrals

The Fundamental Theorem of Calculus has two parts that work together. The first part tells you what happens when you differentiate an integral-defined function. The second part tells you how to evaluate a definite integral using an antiderivative.

Part 1: Differentiating an Integral-Defined Function

If $f$ is continuous on an interval containing $a$ , then the function

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

is an antiderivative of $f$ , which means

g'(x)=f(x)

Taking the derivative of an integral hands you back the inside function. The bounds still matter here because this works for a definite integral with a variable upper limit.

If the upper bound is not just $x$ , substitute the upper bound for $t$ and apply the chain rule.

Example 1: Standard Form

Find $g'(x)$ .

g(x)=\int_{32}^{x} 5t^4\, dt

By the theorem, differentiating both sides gives $g'(x)=5x^4$ .

Example 2: Variable Upper Bound

Find $g'(x)$ .

g(x)=\int_{32}^{2x} 5t^4\, dt

Here the upper bound is $2x$ , so substitute $2x$ for $t$ and multiply by the derivative of $2x$ (the chain rule). That gives $g'(x)=5(2x)^4 \cdot 2 = 160x^4$ .

This first part connects directly to accumulation functions. For more detail and examples, see The Fundamental Theorem of Calculus and Accumulation Functions.

Part 2: Evaluating a Definite Integral

The second part is the one you will reach for most often. If $f$ is continuous on $[a,b]$ and $F$ is an antiderivative of $f$ , then

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

So to find a definite integral, evaluate the antiderivative at the upper limit and subtract its value at the lower limit. No Riemann sums needed.

How to Use This on the AP Calculus Exam

Problem Solving

Use these steps to evaluate a definite integral with the theorem:

Find an antiderivative of the integrand.
Plug in the upper bound.
Plug in the lower bound.
Subtract: upper result minus lower result.

Walkthrough

Evaluate the following integral.

\int_{0}^{5}e^x\, dx

The antiderivative of $e^x$ is $e^x$ , so evaluate it at $5$ and subtract its value at $0$ :

\int_{0}^{5}e^x\, dx=e^5-e^0=e^5-1

The answer is $e^5-1$ because $e^0 = 1$ .

Free Response

When a free-response question lets you use a calculator to evaluate a definite integral, write out the full integral expression, including the limits of integration and the differential, before giving the numerical answer. When working without a calculator, show an appropriate antiderivative. Clear notation and careful parentheses make your work easy to follow, which is important for clear exam work. Avoid stringing together equal signs between expressions that are not actually equal.

Common Trap

You can abbreviate the name of the theorem and you do not need to state which part you are using. Spend your time on the math, not on labeling.

Practice Problems

Try each of these before checking the solutions.

Calculate the integral of $f(x) = 3x^2$ from $a = 1$ to $b = 4$ .
Evaluate the integral $\int_{0}^{2} (5x - 2)\, dx$ .
Find the definite integral of $f(x)=2e^x$ from $a=0$ to $b=1$ .
Evaluate $\int_{2}^{3} \frac{1}{x}\, dx$ .

Question 1 Solution

An antiderivative of $3x^2$ is $F(x) = x^3$ . Then:

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

4^3 - 1^3 = 64 - 1 = \boxed{63}

Question 2 Solution

An antiderivative of $5x-2$ is $F(x)=\frac{5}{2}x^2 - 2x$ . Then:

\int_{0}^{2} (5x - 2)\, dx = F(2) - F(0)

=\left( \frac{5}{2} \cdot 2^2 - 2 \cdot 2 \right) - \left( \frac{5}{2} \cdot 0^2 - 2 \cdot 0 \right)

=10 - 4 = \boxed{6}

Question 3 Solution

An antiderivative of $2e^x$ is $F(x)=2e^x$ . Apply the theorem:

\int_{0}^{1} 2e^x\, dx = F(1) - F(0)

= 2e^1 - 2e^0 = \boxed{2e - 2}

Question 4 Solution

An antiderivative of $\frac{1}{x}$ is $\ln|x|$ . So:

\int_{2}^{3} \frac{1}{x}\, dx = \ln|3| - \ln|2|

Common Misconceptions

Integration is not just "differentiation in reverse" that you can do automatically. You have to recognize patterns and choose the right approach, which gets more involved as integrands get harder.
When the upper limit is a function like $2x$ , forgetting the chain rule is a common error. The derivative of $\int_{32}^{2x} 5t^4\, dt$ is $5(2x)^4 \cdot 2$ , not just $5(2x)^4$ .
The evaluation rule needs $f$ continuous on $[a,b]$ . If the function has a discontinuity inside the interval, you cannot blindly apply $F(b)-F(a)$ .
You do not add a constant of integration when evaluating a definite integral; it cancels in the subtraction $F(b)-F(a)$ . Save the $+C$ for indefinite integrals.
$F(b)-F(a)$ means upper minus lower. Reversing the order flips the sign of your answer.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

Term	Definition
antiderivative	Functions whose derivative equals a given function; the reverse process of differentiation.
continuous	A function that has no breaks, jumps, or holes in its graph over a given interval.
definite integral	The integral of a function over a specific interval [a, b], representing the net signed area between the curve and the x-axis.
Fundamental Theorem of Calculus	The central theorem linking differentiation and integration, stating that if f is continuous on [a, b] and F is an antiderivative of f, then ∫(a to b) f(x) dx = F(b) - F(a).

Frequently Asked Questions

What is the Fundamental Theorem of Calculus in AP Calc?

The Fundamental Theorem of Calculus connects derivatives and integrals. It lets you evaluate a definite integral by using an antiderivative instead of only relying on area geometry or numerical approximation.

How do you evaluate a definite integral using the Fundamental Theorem of Calculus?

Find an antiderivative F of the integrand f, then compute F(b) - F(a). This gives the value of the definite integral from a to b when f is continuous on the interval.

What does F(x) = integral from a to x of f(t) dt mean?

If f is continuous, F(x) = integral from a to x of f(t) dt defines an antiderivative of f. That means F'(x) = f(x).

When do you use the chain rule with the Fundamental Theorem of Calculus?

Use the chain rule when the upper or lower limit is a function of x. For example, the derivative of integral from a to g(x) of f(t) dt is f(g(x))g'(x).

Why is there no + C when evaluating a definite integral?

The + C is not needed because definite integrals use a difference of antiderivative values. Any constant added to the antiderivative cancels out in F(b) - F(a).

How is AP Calc 6.7 tested?

AP Calc 6.7 is tested through definite integral evaluation, integral-defined functions, continuity conditions, and FRQ work where clear notation for F(b) - F(a) matters.

2,589 studying →