6.7 The Fundamental Theorem of Calculus and Definite Integrals

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.

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:

1. Find an antiderivative of the integrand.
2. Plug in the upper bound.
3. Plug in the lower bound.
4. 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.

1. Calculate the integral of $f(x) = 3x^2$ from $a = 1$ to $b = 4$.

2. Evaluate the integral $\int_{0}^{2} (5x - 2)\, dx$.

3. Find the definite integral of $f(x)=2e^x$ from $a=0$ to $b=1$.

4. 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

• Misconception: Integration is just differentiation in reverse that you can do automatically.
  Actually: 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.
• Misconception: The derivative of $\int_{32}^{2x} 5t^4\, dt$ is just $5(2x)^4$.
  Actually: 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 $5(2x)^4$.
• Misconception: You can apply the evaluation rule even if the function has a discontinuity inside the interval.
  Actually: The evaluation rule needs $f$ continuous on $[a,b]$. If the function has a discontinuity inside the interval, you cannot blindly apply $F(b)-F(a)$.
• Misconception: You add a constant of integration when evaluating a definite integral.
  Actually: 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.
• Misconception: $F(b)-F(a)$ means lower minus upper.
  Actually: $F(b)-F(a)$ means upper minus lower. Reversing the order flips the sign of your answer.

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
• 6.4 The Fundamental Theorem of Calculus and Accumulation Functions