The Fundamental Theorem of Calculus and Accumulation Functions - AP Calculus AB Study Guide

The Core Idea: The Fundamental Theorem of Calculus and Accumulation Functions

The Fundamental Theorem of Calculus (FTC) establishes the profound, inverse relationship between differentiation and integration. It provides a powerful method for evaluating definite integrals without resorting to the tedious process of calculating limits of Riemann sums. The theorem essentially states that the definite integral of a function's rate of change ($f(x)$) over an interval gives the total net change in the original function ($F(x)$) over that same interval.

Furthermore, this topic introduces the concept of an "accumulation function," which is a function defined as a definite integral, such as $g(x)={\int }_a^{x}f(t)dt$. This function measures the accumulated "net area" under the curve of $f(t)$ from a fixed starting point $a$ to a variable endpoint $x$. The second part of the FTC reveals that the rate at which this area accumulates at $x$ is precisely the value of the function $f(x)$ itself. This connects the process of accumulation (integration) directly back to the rate of change (the derivative).

Key Theorems: The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is presented in two parts that connect the concepts of derivatives, antiderivatives, and definite integrals.

The Fundamental Theorem of Calculus, Part 1: The Evaluation Theorem

This part provides a method for calculating the exact value of a definite integral.

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:

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

This is often written using the notation $[F(x){]}_a^b$ or $F(x){|}_a^b$ to represent $F(b)-F(a)$.

The Fundamental Theorem of Calculus, Part 2: The Derivative of an Accumulation Function

This part explains how to differentiate a function that is defined as an integral.

If $f$ is a continuous function on an interval and $a$ is a constant in that interval, then the function $g(x)$ defined by

$$g(x)={\int }_a^{x}f(t)dt$$

is an antiderivative of $f$. Therefore, its derivative is:

$$\frac{d}{dx}\left[{\int }_a^{x}f(t)dt\right]=f(x)$$

The FTC Part 2 with the Chain Rule

A crucial extension of Part 2 involves cases where the upper limit of integration is a differentiable function of $x$, say $u(x)$.

If $h(x)={\int }_a^{u(x)}f(t)dt$, then the derivative of $h(x)$ is found by applying the Chain Rule:

$$h^{\prime }(x)=\frac{d}{dx}\left[{\int }_a^{u(x)}f(t)dt\right]=f(u(x))\cdot u^{\prime }(x)$$

Understanding Accumulation Functions

An accumulation function, $g(x)={\int }_a^{x}f(t)dt$, represents the net accumulation of a quantity whose rate of change is given by $f(t)$. Conceptually, $g(x)$ measures the net signed area under the curve of $f$ from the fixed starting point $t=a$ to the variable endpoint $t=x$.

• The integrand $f(t)$ is the rate of change. It tells you how fast the quantity is accumulating at any given moment $t$.

• The accumulation function $g(x)$ is the net change or total accumulated amount. It is the result of summing up all the small changes from $a$ to $x$.

The relationship $g^{\prime }(x)=f(x)$ is the cornerstone of this concept. It means that the rate of change of the accumulated amount at $x$ is simply the value of the integrand function at $x$. For example, if $f(t)$ represents the rate at which water flows into a reservoir in gallons per hour, then $g(x)={\int }_0^{x}f(t)dt$ represents the total number of gallons that have flowed into the reservoir in the first $x$ hours. The derivative, $g^{\prime }(x)=f(x)$, tells us that the instantaneous rate of change of the total water volume at time $x$ is equal to the flow rate $f(x)$ at that exact moment.

Core Concepts & Rules

• Integral of a Rate is Net Change: The definite integral of a rate of change, ${\int }_a^b{f}^{\prime }(t)dt$, calculates the net change in the original function, $f(b)-f(a)$, over the interval $[a,b]$.

• Evaluating Definite Integrals: To compute ${\int }_a^{b}f(x)dx$, first find an antiderivative $F(x)$ of $f(x)$. Then, evaluate $F(x)$ at the upper and lower limits of integration and subtract: $F(b)-F(a)$.

• Functions Defined by Integrals: A function can be defined using a definite integral with a variable upper limit, such as $g(x)={\int }_a^{x}f(t)dt$. This function $g(x)$ is called an accumulation function.

• The Accumulation Function is an Antiderivative: The function $g(x)={\int }_a^{x}f(t)dt$ is a specific antiderivative of the integrand function $f(x)$.

• Differentiating an Accumulation Function: The derivative of $g(x)={\int }_a^{x}f(t)dt$ with respect to $x$ is simply the integrand evaluated at $x$: $g^{\prime }(x)=f(x)$.

• The Chain Rule is Essential: When differentiating an accumulation function where the upper limit is a function of $x$, say $u(x)$, you must apply the chain rule. The derivative of ${\int }_a^{u(x)}f(t)dt$ is $f(u(x))\cdot u^{\prime }(x)$.

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

Problem: Evaluate the definite integral ${\int }_1^2(4x^3-6x)dx$.

This problem requires applying the Evaluation part of the Fundamental Theorem of Calculus (FTC Part 1).

Step 1: Find an antiderivative of the integrand.

The integrand is $f(x)=4x^3-6x$. We find an antiderivative, $F(x)$, by applying the power rule for integration to each term.

$$F(x)=4\left(\frac{x^{3+1}}{3+1}\right)-6\left(\frac{x^{1+1}}{1+1}\right)$$

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

$$F(x)=x^4-3x^2$$

(Note: We do not need the constant of integration $+C$ for definite integrals because it would cancel out in the next step: $(F(b)+C)-(F(a)+C)=F(b)-F(a)$.)

Step 2: Apply the FTC formula ${\int }_a^{b}f(x)dx=F(b)-F(a)$.

Here, $a=1$, $b=2$, and $F(x)=x^4-3x^2$.

First, calculate $F(b)=F(2)$:

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

Next, calculate $F(a)=F(1)$:

$$F(1)=(1{)}^4-3(1{)}^2=1-3(1)=1-3=-2$$

Step 3: Subtract the values.

$${\int }_1^2(4x^3-6x)dx=F(2)-F(1)=4-(-2)=4+2=6$$

Final Answer: The value of the definite integral is 6.

Step-by-Step Example 2: Differentiating an Accumulation Function with the Chain Rule

Problem: Let $h(x)={\int }_{\pi }^{x^2}\cos (t)dt$. Find $h^{\prime }(x)$.

This problem requires applying the second part of the Fundamental Theorem of Calculus combined with the Chain Rule, as the upper limit of integration is a function of $x$.

Step 1: Identify the components for the formula $\frac{d}{dx}{\int }_a^{u(x)}f(t)dt=f(u(x))\cdot u^{\prime }(x)$.

• The integrand function is $f(t)=\cos (t)$.

• The lower limit is a constant, $a=\pi$.

• The upper limit is a function of $x$, $u(x)=x^2$.

Step 2: Find the derivative of the upper limit, $u^{\prime }(x)$.

The upper limit is $u(x)=x^2$. Its derivative is:

$$u^{\prime }(x)=2x$$

Step 3: Substitute $u(x)$ into the integrand $f(t)$ to get $f(u(x))$.

Replace every $t$ in $f(t)=\cos (t)$ with $u(x)=x^2$.

$$f(u(x))=\cos (x^2)$$

Step 4: Assemble the final derivative using the formula $h^{\prime }(x)=f(u(x))\cdot u^{\prime }(x)$.

Multiply the results from Step 2 and Step 3.

$$h^{\prime }(x)=\cos (x^2)\cdot (2x)$$

Final Answer:$h^{\prime }(x)=2x\cos (x^2)$.

Using Your Calculator

While the Fundamental Theorem of Calculus provides the analytical tools for integrals and their derivatives, a graphing calculator is essential for numerical calculations, especially with functions that are difficult or impossible to antidifferentiate by hand.

Evaluating a Definite Integral

Your calculator can find the numerical value of a definite integral ${\int }_a^{b}f(x)dx$ using a numerical integration function.

Problem: Calculate ${\int }_0^{\pi }\sin (x^2)dx$.

TI-84 Style Steps:

1. Press the $[math]$ key.

2. Scroll down to option 9: fnInt( and press [ENTER].

3. The screen will display the integral template.

4. Enter the lower limit of integration: $0$.

5. Enter the upper limit of integration: $\pi$.

6. Enter the function (integrand): $\sin (X^2)$.

7. Enter the variable of integration: $X$.

8. Press [ENTER] to get the result.

The calculator will return an approximate value, such as $0.773$. This is extremely useful for checking answers or for problems on the calculator-active section of the exam where the antiderivative is not elementary.

Finding the Value of an Accumulation Function

If you are given $g(x)=5+{\int }_2^{x}f(t)dt$ and need to find $g(4)$, you can use fnInt to calculate the integral part.

$$g(4)=5+{\int }_2^{4}f(t)dt$$

You would use fnInt to calculate the value of the integral and then add 5 to the result.

AP Exam Quick Hit

Common Question Types

• FRQ with a Graphed Rate: A free-response question provides the graph of a function $f$ and defines a new function $g(x)={\int }_a^{x}f(t)dt$. You will be asked to:

  • Find values of $g(x)$ by calculating the geometric area under the graph of $f$.

  • Find values of $g^{\prime }(x)$ by reading the $y$-values directly from the graph of $f$ (since $g^{\prime }(x)=f(x)$).

  • Find values of $g^{\prime \prime }(x)$ by finding the slope of the graph of $f$ (since $g^{\prime \prime }(x)=f^{\prime }(x)$).

  • Find where $g$ is increasing/decreasing or concave up/down by analyzing the sign of $f$ and the slope of $f$, respectively.

• Differentiating an Integral (MCQ): A multiple-choice question will ask for the derivative of a function defined as an integral, often requiring the chain rule.

  • Example: Find $\frac{d}{dx}{\int }_1^{\tan (x)}\sqrt{t}dt$. The answer is $\sqrt{\tan (x)}\cdot {\sec }^2(x)$.

• Evaluating an Integral (MCQ/FRQ): A question will require you to find the value of a definite integral where the antiderivative can be found using basic rules (power rule, trig integrals, etc.).

  • Example: The value of ${\int }_0^4\frac{1}{\sqrt{x}}dx$ is $4$.

Common Mistakes

• Forgetting the Chain Rule: When differentiating ${\int }_a^{u(x)}f(t)dt$, the most common error is to write $f(u(x))$ but forget to multiply by the derivative of the upper bound, $u^{\prime }(x)$.

• Sign Errors in Evaluation: When calculating $F(b)-F(a)$, students often make simple arithmetic mistakes, especially when $F(a)$ is negative. For example, $10-(-2)$ incorrectly becomes $10-2=8$. Always use parentheses: $10-(-2)=10+2=12$.

• Confusing the Functions: In problems with $g(x)={\int }_a^{x}f(t)dt$, students confuse the properties of $g$ and $f$. For example, when asked for the value of $g^{\prime }(2)$, they might look for the slope of the graph of $f$ at $x=2$, when they should simply be reading the value of $f(2)$ from the graph.

• Incorrectly Handling Swapped Bounds: When differentiating an integral like ${\int }_x^{a}f(t)dt$, students forget to first flip the bounds and negate the integral: $-{\int }_a^{x}f(t)dt$. The derivative is therefore $-f(x)$.

• Plugging in the Bounds Before Differentiating: When asked to find $\frac{d}{dx}{\int }_a^{x}f(t)dt$, some students try to find the antiderivative of $f(t)$, plug in $x$ and $a$, and then take the derivative of the result. This is a long and error-prone process. The FTC Part 2 is a direct shortcut: the answer is simply $f(x)$.