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

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

The Fundamental Theorem of Calculus (FTC) establishes a profound and powerful connection between the two major branches of calculus: differential calculus (the study of rates of change and derivatives) and integral calculus (the study of accumulation and area). This topic focuses on the "Evaluation Part" of the theorem, which provides a direct method for calculating the exact value of a definite integral without resorting to the lengthy process of finding the limit of a Riemann sum.

The core idea is that the definite integral of a continuous function $f$ from $a$ to $b$, which represents the net accumulation of $f$ over the interval $[a,b]$, can be found simply by finding any antiderivative $F$ of $f$ and calculating the change in $F$ from $a$ to $b$. This transforms the problem of finding an area or total accumulation into a much simpler problem of antidifferentiation and algebraic substitution.

The Fundamental Theorem of Calculus (Evaluation Part)

The primary rule for this topic is the Evaluation Part of the Fundamental Theorem of Calculus.

The Theorem: 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:

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

Vocabulary and Notation:

In the expression ${\int }_a^{b}f(x)dx$:

• $f(x)$ is the integrand. This is the function whose net accumulation is being calculated.

• $a$ is the lower limit of integration. This is the starting point of the interval.

• $b$ is the upper limit of integration. This is the ending point of the interval.

A common shorthand notation for $F(b)-F(a)$ is:

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

Understanding the Condition of Continuity

The Fundamental Theorem of Calculus as stated here comes with a critical condition: the integrand $f(x)$must be continuous on the entire closed interval $[a,b]$.

This means that for the formula ${\int }_a^{b}f(x)dx=F(b)-F(a)$ to be valid, the graph of $f(x)$ cannot have any discontinuities—such as holes, jumps, or vertical asymptotes—anywhere between $x=a$ and $x=b$, inclusive. If the function is not continuous on the interval, this part of the Fundamental Theorem of Calculus cannot be directly applied. For the AP Calculus AB exam, you can generally assume that functions given in problems requiring the use of the FTC will meet this condition unless otherwise specified.

Core Concepts & Rules

• The Bridge Between Derivatives and Integrals: The FTC provides the essential link between differentiation and integration, showing they are inverse processes.

• The Evaluation Formula: The definite integral of a continuous function $f$ from $a$ to $b$ is the value of its antiderivative at the upper limit $b$ minus the value of the same antiderivative at the lower limit $a$. The formula is ${\int }_a^{b}f(x)dx=F(b)-F(a)$.

• The Role of the Antiderivative: To use the FTC, you must first find a function $F(x)$ such that $F^{\prime }(x)=f(x)$.

• The Constant of Integration is Not Needed: When finding the antiderivative $F(x)$ for a definite integral, you do not need to include the constant of integration $"+C"$. This is because it would cancel out during the subtraction: $(F(b)+C)-(F(a)+C)=F(b)-F(a)$.

• Continuity is Required: The function $f(x)$ must be continuous on the interval of integration $[a,b]$ for the theorem to apply.

Step-by-Step Example 1: Basic Application

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

Step 1: Identify the integrand and check for continuity.

The integrand is $f(x)=6x^2-2x$. As a polynomial function, it is continuous everywhere, so it is continuous on the interval $[1,2]$. We can proceed with the FTC.

Step 2: Find an antiderivative, $F(x)$.

Using the power rule for antiderivatives, $\int x^{n}dx=\frac{x^{n+1}}{n+1}$:

The antiderivative of $6x^2$ is $6\cdot \frac{x^3}{3}=2x^3$.

The antiderivative of $-2x$ is $-2\cdot \frac{x^2}{2}=-x^2$.

So, an antiderivative is $F(x)=2x^3-x^2$.

Step 3: Apply the FTC formula, $F(b)-F(a)$.

Using the standard notation:

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

This means we will evaluate the expression at $x=2$ and subtract the value of the expression at $x=1$.

Step 4: Evaluate $F(2)$ and $F(1)$.

• $F(2)=2(2{)}^3-(2{)}^2=2(8)-4=16-4=12$

• $F(1)=2(1{)}^3-(1{)}^2=2(1)-1=2-1=1$

Step 5: Compute the final result.

$$F(2)-F(1)=12-1=11$$

Therefore, ${\int }_1^2(6x^2-2x)dx=11$.

Step-by-Step Example 2: Exam-Style Application

Problem: Evaluate the definite integral ${\int }_0^{\pi }\sin (x)dx$.

Step 1: Identify the integrand and check for continuity.

The integrand is $f(x)=\sin (x)$. The sine function is continuous everywhere, so it is continuous on the interval $[0,\pi ]$. The FTC is applicable.

Step 2: Find an antiderivative, $F(x)$.

We need a function whose derivative is $\sin (x)$. Recall that $\frac{d}{dx}(\cos (x))=-\sin (x)$. To get a positive $\sin (x)$, we must start with $-\cos (x)$.

So, an antiderivative is $F(x)=-\cos (x)$.

Step 3: Apply the FTC formula, $F(b)-F(a)$.

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

Step 4: Evaluate $F(\pi )$ and $F(0)$.

Be careful with signs and trigonometric values.

• $F(\pi )=-\cos (\pi )=-(-1)=1$

• $F(0)=-\cos (0)=-(1)=-1$

Step 5: Compute the final result.

$$F(\pi )-F(0)=(1)-(-1)=1+1=2$$

Therefore, ${\int }_0^{\pi }\sin (x)dx=2$. This result means the area under the curve of $y=\sin (x)$ from $x=0$ to $x=\pi$ is exactly 2.

Using Your Calculator

A graphing calculator is an essential tool for evaluating definite integrals, especially on the calculator-active section of the AP exam. The primary function for this is fnInt.

Problem: Evaluate \int_{0}^{\pi} \sin(x) \,dx` using a TI-84 style calculator. **Steps:** 1. Press the `MATH` key. 2. Scroll down to option `9: fnInt(` and press `ENTER`. 3. The screen will show a template for the definite integral. * Enter the lower limit: $0

*   Enter the upper limit: $\pi$ (use the $\pi$ key)

*   Enter the integrand: $sin(X)$ (use the `X,T,\theta,n` key for the variable)

*   Enter the variable of integration: $X$

4. The input should look like: $\int (sin(X),X,0,\pi )$ or a more visual template depending on your calculator's operating system.

5. Press ENTER. The calculator will return the numerical value, which is $2$.

This feature is extremely useful for checking your by-hand calculations or for solving problems where the integral is complex or the question is on the calculator-active section.

AP Exam Quick Hit

Common Question Types

• Direct Evaluation (No Calculator): You will be asked to evaluate a definite integral where the antiderivative is one you are expected to know (e.g., power rule, basic trig, $e^x$, $1/x$).

  • Example: "Evaluate ${\int }_1^4\frac{1}{\sqrt{x}}dx$."

• Total Change from a Rate (Calculator or No Calculator): You will be given a function representing a rate of change ($f(x)$) and asked to find the net change in the original quantity over an interval $[a,b]$. This is a direct application of the FTC.

  • Example: "The velocity of a particle is given by $v(t)=t^2-4$ m/s. Find the particle's displacement from $t=0$ to $t=3$ seconds." (This requires calculating ${\int }_0^3(t^2-4)dt$).

Common Mistakes

• Antiderivative vs. Derivative: Accidentally taking the derivative of the integrand instead of the antiderivative. For example, for $\int x^{2}dx$, writing $2x$ instead of $\frac{x^3}{3}$.

• Reversing the Limits: Calculating $F(a)-F(b)$ instead of the correct $F(b)-F(a)$. This will give you the opposite sign of the correct answer.

• Sign Errors in Subtraction: A very common arithmetic error occurs when subtracting a negative value. For example, $F(b)-F(a)$ where $F(a)=-5$ might be incorrectly calculated as $F(b)-5$ instead of the correct $F(b)-(-5)=F(b)+5$. Always use parentheses when substituting: $(F(b))-(F(a))$.

• Incorrect Antiderivative Rules: Misremembering rules for specific functions. A common mistake is thinking the antiderivative of $\cos (x)$ is $-\sin (x)$ (it is $\sin (x)$) or that the antiderivative of $e^x$ is $x{e}^{x-1}$.

• Forgetting to Evaluate at Both Limits: Only calculating $F(b)$ and forgetting to subtract $F(a)$. This is especially common when $F(a)=0$, but it must still be formally shown and subtracted.