The Fundamental Theorem of Calculus and Definite Integrals - AP Calculus BC Study Guide

The Core Idea: The Fundamental Theorem of Calculus and Definite Integrals

The Fundamental Theorem of Calculus (FTC) provides the essential, powerful connection between differentiation and integration. Specifically, this part of the theorem (often called Part 2) gives us a direct method for calculating the exact value of a definite integral without resorting to the approximations of Riemann sums or geometric formulas. It establishes that if we can find an antiderivative of a function, we can find the value of the definite integral of that function over an interval by simply evaluating the antiderivative at the endpoints of the interval and subtracting the results.

This theorem transforms the problem of finding the value of an integral, ${\int }_a^{b}f(x)dx$, into a problem of finding an antiderivative, $F(x)$. The result of this calculation is not a family of functions (like an indefinite integral), but a single, specific real number that represents the value of the definite integral. The process hinges on the function being continuous over the interval of integration.

Key Formulas/Rules/Theorems

The primary theorem for this topic is the Fundamental Theorem of Calculus, Part 2.

The Fundamental Theorem of Calculus, Part 2

If a function $f$ is continuous on the closed interval $[a,b]$ and $F$ is any antiderivative of $f$ (meaning $F^{\prime }(x)=f(x)$), then:

$${\int }_a^{b}f(x)dx=F(b)-F(a)$$

Evaluation Notation

To streamline the process of applying the theorem, a specific notation is used to represent the evaluation step $F(b)-F(a)$. This is written as:

$$F(x){|}_a^b\text{or}[F(x){]}_a^b$$

Both notations mean exactly the same thing: evaluate the function $F(x)$ at the upper limit $b$, and from that result, subtract the value of the function $F(x)$ at the lower limit $a$.

Understanding the Conditions and the Result

There are two critical conceptual points to understand when applying the Fundamental Theorem of Calculus.

1. The Condition of Continuity: The theorem begins with the condition, "If a function $f$ is continuous on the interval $[a,b]$...". This is not an optional suggestion; it is a mandatory requirement for the theorem to apply. If the function $f(x)$ has a discontinuity (such as a jump, hole, or vertical asymptote) within the interval $[a,b]$, the FTC cannot be directly applied as stated. You must always mentally verify that the function you are integrating is continuous on the given closed interval.

2. The Nature of the Result: An indefinite integral, $\int f(x)dx$, results in a family of functions, $F(x)+C$. In contrast, a definite integral, ${\int }_a^{b}f(x)dx$, results in a single, specific real number. This number is the value of the integral. The FTC provides the mechanism to find this exact numerical value. The constant of integration $C$ is not necessary when evaluating definite integrals because it would cancel out in the subtraction: $(F(b)+C)-(F(a)+C)=F(b)-F(a)$.

Core Concepts & Rules

• The Purpose of the FTC: The Fundamental Theorem of Calculus, Part 2, provides a direct, analytical method for evaluating a definite integral.

• The Process: The method involves two main steps: first, finding an antiderivative $F(x)$ of the integrand $f(x)$, and second, calculating the difference $F(b)-F(a)$.

• The Requirement: The function $f(x)$ must be continuous over the entire closed interval of integration $[a,b]$.

• The Result: The value of a definite integral ${\int }_a^{b}f(x)dx$ is always a single real number.

• The Notation: The expression $F(b)-F(a)$ is commonly written using the evaluation notation $F(x){|}_a^b$.

Step-by-Step Example 1: Evaluating a Polynomial Integral

Problem: Evaluate the definite integral ${\int }_1^2(6x^2-4x)dx$.

Step 1: Verify Continuity

The integrand is $f(x)=6x^2-4x$. This is a polynomial function, which is continuous for all real numbers. Therefore, it is continuous on the closed interval $[1,2]$. The FTC can be applied.

Step 2: Find an Antiderivative

Find an antiderivative $F(x)$ for $f(x)$ using the power rule for integration.

$$F(x)=\int (6x^2-4x)dx=6\left(\frac{x^3}{3}\right)-4\left(\frac{x^2}{2}\right)=2x^3-2x^2$$

We do not need the constant of integration, $C$.

Step 3: Apply the FTC Notation

Set up the evaluation using the notation from the theorem.

$${\int }_1^2(6x^2-4x)dx=[2x^3-2x^2{]}_1^2$$

Step 4: Evaluate $F(b)-F(a)$

Substitute the upper limit ($b=2$) and the lower limit ($a=1$) into the antiderivative $F(x)$ and subtract.

$$F(2)-F(1)=\left(2(2{)}^3-2(2{)}^2\right)-\left(2(1{)}^3-2(1{)}^2\right)$$

Step 5: Simplify

Perform the arithmetic to find the final numerical value.

$$(2(8)-2(4))-(2(1)-2(1))$$

$$(16-8)-(2-2)$$

$$8-0=8$$

The value of the definite integral is $8$.

Step-by-Step Example 2: An Integral with a Trigonometric Function

Problem: Find the value of ${\int }_0^{\pi }\sin (x)dx$.

Step 1: Verify Continuity

The function $f(x)=\sin (x)$ is continuous for all real numbers, so it is continuous on the interval $[0,\pi ]$. The FTC is applicable.

Step 2: Find an Antiderivative

Find an antiderivative $F(x)$ for $f(x)=\sin (x)$. Recall that the derivative of $\cos (x)$ is $-\sin (x)$, so the antiderivative of $\sin (x)$ is $-\cos (x)$.

$$F(x)=-\cos (x)$$

Step 3: Apply the FTC Notation

Set up the evaluation.

$${\int }_0^{\pi }\sin (x)dx=[-\cos (x){]}_0^{\pi }$$

Step 4: Evaluate $F(b)-F(a)$

Substitute the upper and lower limits into the antiderivative. Be careful with signs.

$$F(\pi )-F(0)=(-\cos (\pi ))-(-\cos (0))$$

Step 5: Simplify

Use unit circle values to find the result. Recall $\cos (\pi )=-1$ and $\cos (0)=1$.

$$(-(-1))-(-(1))$$

$$(1)-(-1)$$

$$1+1=2$$

The value of the definite integral is $2$.

Using Your Calculator

The Fundamental Theorem of Calculus outlines an analytical procedure. You are expected to be able to evaluate definite integrals by hand. A calculator is a tool for verifying your answer, not for replacing the analytical steps.

To check the result of a definite integral on a TI-84 style calculator:

1. Press the MATH key.

2. Select option 9: fnInt(.

3. The calculator will display a template. Fill in the values as follows: $fnInt(expression,variable,lowerlimit,upperlimit)$.

4. For Example 1, you would enter: fnInt(6X^2 - 4X, X, 1, 2).

5. Pressing ENTER will yield the result $8$, confirming the hand calculation.

6. For Example 2, you would enter: fnInt(sin(X), X, 0, π).

7. Pressing ENTER will yield the result $2$.

AP Exam Quick Hit

Common Question Types

• Direct Evaluation: You will be asked to evaluate a definite integral where the antiderivative is based on standard rules (power rule, trig, exponential, logarithmic).

  • Example: Evaluate ${\int }_0^1{e}^{x}dx$.

• Solving for a Limit of Integration: You will be given the value of a definite integral and asked to find an unknown limit of integration.

  • Example: If ${\int }_1^k\frac{1}{x^2}dx=\frac{1}{2}$, find the value of $k$.

Common Mistakes

• Antiderivative Errors: The most frequent source of error is an incorrect antiderivative. For example, finding the antiderivative of $\frac{1}{x}$ to be $\ln |x|$ but forgetting the absolute value, or incorrectly applying the power rule, such as $\int x^{2}dx=\frac{x^2}{2}$.

• Arithmetic Errors in Evaluation: Simple arithmetic mistakes when calculating $F(b)-F(a)$ are very common. This is especially true when dealing with negative signs or fractions. For example, $(F(b))-(F(a))$ where $F(a)$ is negative, such as $(10)-(-4)$, is often incorrectly computed as $10-4=6$ instead of $10+4=14$.

• Swapping Limits: Accidentally calculating $F(a)-F(b)$ instead of $F(b)-F(a)$. This will result in the correct magnitude but the opposite sign.

• Evaluating the Original Function: Plugging the limits of integration $a$ and $b$ into the original function $f(x)$ instead of the antiderivative $F(x)$. Remember, the theorem requires you to use the antiderivative: $F(b)-F(a)$, not $f(b)-f(a)$.