6.6 Fundamental theorem of calculus

Achievable AP Calculus AB

6. Integration

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Fundamental theorem of calculus

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What you’ll learn

Evaluate definite integrals analytically using the Fundamental Theorem of Calculus (FTC), Part 2
Use the reverse power rule to calculate basic antiderivatives by hand
Use graphing calculator technology to compute definite integrals
Apply the FTC to problems with graphs and tables

The previous section showed that derivatives and integrals are opposite processes:

$If F (x) then F^{'} (x) = \int_{a}^{x} f (t) d t, = f (x) .$

Because integration can “undo” differentiation, when given a function $f (x)$ you can look for an antiderivative: a function $F (x)$ whose derivative is $f (x)$ .

To evaluate a definite integral using this relationship, follow these two steps:

Find an antiderivative $F (x)$ whose derivative is $f (x)$ .
Compute $F (b) - F (a)$ .

The 2nd part of the Fundamental Theorem of Calculus gives a way to evaluate definite integrals:

Fundamental Theorem of Calculus (FTC), Part 2:

If $f$ is continuous on $[a, b]$ , and $F$ is any antiderivative of $f$ , then

$\int_{a}^{b} f (x) d x = F (x)_{a}^{b} = F (b) - F (a)$

The notation $F (x)_{a}^{b}$ is a common intermediate step when working with definite integrals. It instructs you to evaluate $F (x)$ at the upper limit of integration $b$ and subtract its value at the lower limit $a$ .

Caution: The subtraction sign

When evaluating $F (b) - F (a)$ , always wrap the $F (a)$ expression in parentheses, especially if $F (x)$ has multiple terms or negative coefficients. A very common mistake is forgetting to distribute the subtraction sign to the entire second part of the evaluation:

$(F (b)) - (F (a))$

Reversing the power rule

Now the question is: how do you find antiderivatives?

Just like with derivatives, there are rules you can use. One of the most important is the reverse power rule. Recall the power rule for derivatives:

$\frac{d}{d x} (x^{n}) = n x^{n - 1}$

To reverse this process for integrals, add $1$ to the exponent and divide by the new exponent:

Reverse power rule:

$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$

This rule requires $n \neq = - 1$ (otherwise it would result in division by $0$ ). For now, we’ll avoid this case. It will be covered in the next section along with the meaning of $+ C$ .

Since the reverse power rule applies to powers of $x$ , algebraic manipulation - such as simplifying fractions or expanding polynomials - can be used to rewrite expressions in that form.

Example 1: Direct application

Evaluate the definite integral:

$\int_{- 1}^{4} (2 x + 3) d x$

Solution

(spoiler)

Graphically, the region under $f (x) = 2 x + 3$ from $x = - 1$ to $x = 4$ is a trapezoid with:

a height of $5$ and
parallel base lengths of $f (- 1) = 1$ and $f (4) = 11$ .

So we should expect the definite integral to be the area of the trapezoid:

$\frac{1}{2} (5) (1 + 11) = 30$

To compute the same value using the FTC, first rewrite $f (x)$ with explicit exponents so it’s clear how to apply the reverse power rule:

$\int_{- 1}^{4} f (x) d x = \int_{- 1}^{4} (2 x^{1} + 3 x^{0}) d x$

Apply the reverse power rule term-by-term:

$\int_{- 1}^{4} (2 x^{1} + 3 x^{0}) d x = (\frac{2 x ^{1 + 1}}{1 + 1} + \frac{3 x ^{0 + 1}}{0 + 1})_{- 1}^{4} = (x^{2} + 3 x)_{- 1}^{4}$

Now apply the FTC for definite integrals:

$(x^{2} + 3 x)_{- 1}^{4} = [(4)^{2} + 3 (4)] - [(- 1)^{2} + 3 (- 1)] = 28 - (- 2) = 30$

Using your calculator

While you could use the reverse power rule to evaluate an integral like

$\int_{1.5}^{4.2} (0.35 x^{5} - 2.4 x^{3} + 7) d x$

doing so by hand takes more time than necessary.

Instead, the calculation can be handled in Desmos with the integral tool:

Type “int” and an integral symbol will appear.
Plug in the boundaries.
Type the integrand.

Syntax in Desmos:

Make sure to put parentheses around the integrand expression before typing $d x$ . Otherwise, you’ll see an error message.

By entering the integral exactly as shown, Desmos gives the answer of $154.764$ .

Example 2: Using graphs

The AP Calculus exam frequently tests the FTC using a visual graph instead of a function. In these problems, remember that the definite integral represents the net area under the curve. Calculate these integrals by finding the areas of the geometric shapes formed by the graph.

Let $f$ be the function whose graph is given below and let $g$ be an antiderivative of $f$ such that $g (5) = - 2$ .

Figure 6.6.1 Graph of f

a) Find $g (1)$ .

b) Find $g^{'} (3)$ .

Solutions

a) Find $g (1)$ given $g (5) = - 2$ .

(spoiler)

Since $g$ is an antiderivative of $f$ , we can use the FTC part 2 to evaluate an integral of $f$ . The problem gives us $g (5) = - 2$ and asks for $g (1)$ , so setup the integral as follows:

$\int_{1}^{5} f (x) d x = g (5) - g (1)$

Rearrange to solve for $g (1)$ :

$g (1) = g (5) - \int_{1}^{5} f (x) d x$

Compute the definite integral by finding the net area under the graph from $x = 1$ to $x = 5$ . The region consists of:

A trapezoid on interval $[1, 2]$ with area
- $A = \frac{1}{2} (1) (3 + 2) = 2.5$
A trapezoid on interval $[2, 5]$ with area
- $A = \frac{1}{2} (2) (1 + 3) = 4$

Therefore,

$\int_{1}^{5} f (x) d x = 4 + 2.5 = 6.5$

Lastly, substitute this in to find $g (1)$ :

$g (1) = g (5) - \int_{1}^{5} f (x) d x = - 2 - 6.5 = - 8.5$

b) Find $g^{'} (3)$ .

(spoiler)

Since $g$ is an antiderivative of $f$ , we have $g^{'} (x) = f (x)$ by the FTC part 1.

Therefore, $g^{'} (3) = f (3)$ . Using the graph of $f$ given, locate the function value $f (3) = 2$ .

Example 3: Using tables

A tank contains $1$ gallon of water at time $t = 0$ . Water is poured into the tank at a rate of $R (t)$ gallons per minute, where $t$ is measured in minutes. The table below gives values of $R$ at selected points:

$t$ $R (t)$

$0$ $2$

$2$ $1$

$3$ $5$

$5$ $6$

Use a right Riemann sum with the values in the table to estimate the amount of water in the tank at $t = 3$ minutes.

$t$	$R (t)$
$0$	$2$
$2$	$1$
$3$	$5$
$5$	$6$

Solution

(spoiler)

Let $A (t)$ be the amount of water in the tank at time $t$ . The problem gives the initial amount of water in the tank, $A (0) = 1$ gallon.

Integrating the rate $R (t)$ over the time interval from $t = 0$ to $t = 3$ gives the net change in $A (t)$ , or how much water was added:

$\int_{0}^{3} R (t) d t = A (3) - A (0)$

Solve for $A (3)$ :

$A (3) = A (0) + \int_{0}^{3} R (t) d t$

Next, estimate $\int_{0}^{3} R (t) d t$ using a right Riemann sum on the subintervals determined by the given $t$ -values.

It may help to plot the points and draw the rectangles:

Using right endpoints:

On $[2, 3]$ , the right endpoint is $t = 3$ , so the height is $R (3) = 5$ and the width is $3 - 2 = 1$ .
- $A = 5 \cdot 1 = 5$ .
On $[0, 2]$ , the right endpoint is $t = 2$ , so the height is $R (2) = 1$ and the width is $2 - 0 = 2$ .
- $A = 1 \cdot 2 = 2$ .

So the right Riemann sum is $5 + 2 = 7$ , giving the estimate

$\int_{0}^{3} R (t) d t \approx 7.$

Then the amount of water in the tank at $t = 3$ minutes is

$A (3) = A (0) + \int_{0}^{3} R (t) d t \approx 1 + 7 = 8$

The FTC (part 2) for definite integrals:

To evaluate $\int_{a}^{b} f (x) d x$ , find the antiderivative $F (x)$ and compute $F (b) - F (a)$ . Always wrap $F (a)$ in parentheses to distribute negative signs correctly.

Reverse power rule:

Add 1 to the exponent and divide by the new exponent:

$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$

(where $n \neq = - 1$ ).

Desmos shortcut:

Type “int” for the integral symbol. You must wrap multi-term integrands in parentheses before typing $d x$ so the calculator evaluates it.

Graphs and tables:

A definite integral represents the net area under a curve. Use geometric formulas for graphs, or Riemann sums for tables, to calculate or estimate this area.

Fundamental theorem of calculus

What you’ll learn

Evaluate definite integrals analytically using the Fundamental Theorem of Calculus (FTC), Part 2
Use the reverse power rule to calculate basic antiderivatives by hand
Use graphing calculator technology to compute definite integrals
Apply the FTC to problems with graphs and tables

The previous section showed that derivatives and integrals are opposite processes:

$If F (x) then F^{'} (x) = \int_{a}^{x} f (t) d t, = f (x) .$

Because integration can “undo” differentiation, when given a function $f (x)$ you can look for an antiderivative: a function $F (x)$ whose derivative is $f (x)$ .

To evaluate a definite integral using this relationship, follow these two steps:

Find an antiderivative $F (x)$ whose derivative is $f (x)$ .
Compute $F (b) - F (a)$ .

The 2nd part of the Fundamental Theorem of Calculus gives a way to evaluate definite integrals:

Fundamental Theorem of Calculus (FTC), Part 2:

If $f$ is continuous on $[a, b]$ , and $F$ is any antiderivative of $f$ , then

$\int_{a}^{b} f (x) d x = F (x)_{a}^{b} = F (b) - F (a)$

Caution: The subtraction sign

$(F (b)) - (F (a))$

Reversing the power rule

Now the question is: how do you find antiderivatives?

Just like with derivatives, there are rules you can use. One of the most important is the reverse power rule. Recall the power rule for derivatives:

$\frac{d}{d x} (x^{n}) = n x^{n - 1}$

To reverse this process for integrals, add $1$ to the exponent and divide by the new exponent:

Reverse power rule:

$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$

This rule requires $n \neq = - 1$ (otherwise it would result in division by $0$ ). For now, we’ll avoid this case. It will be covered in the next section along with the meaning of $+ C$ .

Since the reverse power rule applies to powers of $x$ , algebraic manipulation - such as simplifying fractions or expanding polynomials - can be used to rewrite expressions in that form.

Example 1: Direct application

Evaluate the definite integral:

$\int_{- 1}^{4} (2 x + 3) d x$

Solution

(spoiler)

Graphically, the region under $f (x) = 2 x + 3$ from $x = - 1$ to $x = 4$ is a trapezoid with:

a height of $5$ and
parallel base lengths of $f (- 1) = 1$ and $f (4) = 11$ .

So we should expect the definite integral to be the area of the trapezoid:

$\frac{1}{2} (5) (1 + 11) = 30$

To compute the same value using the FTC, first rewrite $f (x)$ with explicit exponents so it’s clear how to apply the reverse power rule:

$\int_{- 1}^{4} f (x) d x = \int_{- 1}^{4} (2 x^{1} + 3 x^{0}) d x$

Apply the reverse power rule term-by-term:

$\int_{- 1}^{4} (2 x^{1} + 3 x^{0}) d x = (\frac{2 x ^{1 + 1}}{1 + 1} + \frac{3 x ^{0 + 1}}{0 + 1})_{- 1}^{4} = (x^{2} + 3 x)_{- 1}^{4}$

Now apply the FTC for definite integrals:

$(x^{2} + 3 x)_{- 1}^{4} = [(4)^{2} + 3 (4)] - [(- 1)^{2} + 3 (- 1)] = 28 - (- 2) = 30$

Using your calculator

While you could use the reverse power rule to evaluate an integral like

$\int_{1.5}^{4.2} (0.35 x^{5} - 2.4 x^{3} + 7) d x$

doing so by hand takes more time than necessary.

Instead, the calculation can be handled in Desmos with the integral tool:

Type “int” and an integral symbol will appear.
Plug in the boundaries.
Type the integrand.

Syntax in Desmos:

Make sure to put parentheses around the integrand expression before typing $d x$ . Otherwise, you’ll see an error message.

By entering the integral exactly as shown, Desmos gives the answer of $154.764$ .

Example 2: Using graphs

Let $f$ be the function whose graph is given below and let $g$ be an antiderivative of $f$ such that $g (5) = - 2$ .

a) Find $g (1)$ .

b) Find $g^{'} (3)$ .

Solutions

a) Find $g (1)$ given $g (5) = - 2$ .

(spoiler)

Since $g$ is an antiderivative of $f$ , we can use the FTC part 2 to evaluate an integral of $f$ . The problem gives us $g (5) = - 2$ and asks for $g (1)$ , so setup the integral as follows:

$\int_{1}^{5} f (x) d x = g (5) - g (1)$

Rearrange to solve for $g (1)$ :

$g (1) = g (5) - \int_{1}^{5} f (x) d x$

Compute the definite integral by finding the net area under the graph from $x = 1$ to $x = 5$ . The region consists of:

A trapezoid on interval $[1, 2]$ with area
- $A = \frac{1}{2} (1) (3 + 2) = 2.5$
A trapezoid on interval $[2, 5]$ with area
- $A = \frac{1}{2} (2) (1 + 3) = 4$

Therefore,

$\int_{1}^{5} f (x) d x = 4 + 2.5 = 6.5$

Lastly, substitute this in to find $g (1)$ :

$g (1) = g (5) - \int_{1}^{5} f (x) d x = - 2 - 6.5 = - 8.5$

b) Find $g^{'} (3)$ .

(spoiler)

Since $g$ is an antiderivative of $f$ , we have $g^{'} (x) = f (x)$ by the FTC part 1.

Therefore, $g^{'} (3) = f (3)$ . Using the graph of $f$ given, locate the function value $f (3) = 2$ .

Example 3: Using tables

A tank contains $1$ gallon of water at time $t = 0$ . Water is poured into the tank at a rate of $R (t)$ gallons per minute, where $t$ is measured in minutes. The table below gives values of $R$ at selected points:

$t$ $R (t)$

$0$ $2$

$2$ $1$

$3$ $5$

$5$ $6$

Use a right Riemann sum with the values in the table to estimate the amount of water in the tank at $t = 3$ minutes.

$t$	$R (t)$
$0$	$2$
$2$	$1$
$3$	$5$
$5$	$6$

Solution

(spoiler)

Let $A (t)$ be the amount of water in the tank at time $t$ . The problem gives the initial amount of water in the tank, $A (0) = 1$ gallon.

Integrating the rate $R (t)$ over the time interval from $t = 0$ to $t = 3$ gives the net change in $A (t)$ , or how much water was added:

$\int_{0}^{3} R (t) d t = A (3) - A (0)$

Solve for $A (3)$ :

$A (3) = A (0) + \int_{0}^{3} R (t) d t$

Next, estimate $\int_{0}^{3} R (t) d t$ using a right Riemann sum on the subintervals determined by the given $t$ -values.

It may help to plot the points and draw the rectangles:

Using right endpoints:

On $[2, 3]$ , the right endpoint is $t = 3$ , so the height is $R (3) = 5$ and the width is $3 - 2 = 1$ .
- $A = 5 \cdot 1 = 5$ .
On $[0, 2]$ , the right endpoint is $t = 2$ , so the height is $R (2) = 1$ and the width is $2 - 0 = 2$ .
- $A = 1 \cdot 2 = 2$ .

So the right Riemann sum is $5 + 2 = 7$ , giving the estimate

$\int_{0}^{3} R (t) d t \approx 7.$

Then the amount of water in the tank at $t = 3$ minutes is

$A (3) = A (0) + \int_{0}^{3} R (t) d t \approx 1 + 7 = 8$

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.