Fundamental theorem of calculus | Integration

What you’ll learn:

Understand how definite integrals can be computed using antiderivatives
How to apply the Fundamental theorem of calculus part 2 to evaluate definite integrals
Use the reverse power rule to find basic antiderivatives

In a previous section, it was shown that derivatives and integrals are opposite processes - if $F (x) = \int_{a}^{x} f (t) d t$ , then $F^{'} (x) = f (x)$ .

Since integration can “undo” a derivative, then given a function $f (x)$ , we can look for an antiderivative - a function $F (x)$ such that its derivative is $f (x)$ .

This idea leads to the 2nd part of the Fundamental theorem of calculus, which gives a practical way to compute definite integrals using antiderivatives.

Fundamental Theorem of Calculus (FTC), Part 2

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

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

To evaluate a definite integral, just find an antiderivative $F$ , and then calculate the difference between the function values.

The vertical bar after $F (x)$ means "evaluate $F (x)$ from $a$ to $b$ . It’s a notation you’ll see and can write as an intermediate step when evaluating definite integrals.

Now the question is, how do we find antiderivatives?

Just like with derivatives, there are a set of rules you can use, such as the:

Reverse power rule

Recall that the power rule for derivatives is

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

For integrals, add 1 to the exponent and divide by the new value in the exponent:

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

Note that there is an additional limitation of $n \neq = - 1$ , otherwise we’d be dividing by $0$ . Don’t worry about that case for now - more on that, as well as the $+ C$ , will be covered in the next section along with other antiderivative rules.

Examples

Evaluate

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

Solution

(spoiler)

Graphically, the region under the function $f (x) = 2 x + 3$ from $x = - 1$ to $x = 4$ is a trapezoid with a height of $5$ and base lengths of $f (- 1) = 1$ and $f (4) = 11$ , so we should expect the area of the trapezoid as the answer:

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

Here’s how it’s done using the FTC:

Let’s rewrite function $f (x) = 2 x + 3$ with explicit exponents to make it clear what to apply the reverse power rule on.

$f (x) = 2 x^{1} + 3 x^{0}$

Then using the reverse power rule, antiderivative $F (x)$ is

$F (x) = \frac{2 x ^{1 + 1}}{1 + 1} + \frac{3 x ^{0 + 1}}{0 + 1} + C$

$= x^{2} + 3 x + C$

Using the FTC,

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

$= [4^{2} + 3 (4)] - [(- 1)^{2} + 3 (- 1)]$

$= 28 - (- 2)$

$= 30$

Evaluate

$\int_{0}^{2} (4 x^{2} - x) d x$

(spoiler)

Rewrite $f (x)$ so to make it clear what the exponents are:

$f (x) = 4 x^{2} - x^{1/2}$

Using the reverse power rule,

$F (x) = \frac{4 x ^{2 + 1}}{2 + 1} - \frac{x ^{(1/2) + 1}}{( 1/2 ) + 1} + C$

$= \frac{4}{3} x^{3} - \frac{2}{3} x^{3/2} + C$

Then applying the FTC,

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

$= [\frac{4}{3} (2^{3}) - \frac{2}{3} (2^{3/2})] - [\frac{4}{3} (0^{3}) - \frac{2}{3} (0^{3/2})]$

$= (\frac{32}{3} - \frac{2 8}{3}) - (0 - 0)$

$= \frac{32 - 4 2}{3}$

Let $f$ be the function whose graph is given below and let $g$ be an antiderivative of $f$ .

Figure 6.6.1 Graph of f

a) If $g (5) = - 2$ , find $g (1)$ .
b) Find $g^{'} (3)$ .

Solutions

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

(spoiler)

$g$ is an antiderivative of $f$ and by the FTC,

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

Rearranging,

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

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

To evaluate the definite integral, find 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
- $A = \frac{1}{2} (2) (1 + 3) = 4$

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

Then

$g (1) = - 2 - 6.5$

$= - 8.5$

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

(spoiler)

Since $g$ is an antiderivative of $f$ , then $g^{'} (x) = f (x)$ .

So $g^{'} (3) = f (3)$ . From the graph, $f (3) = 2$ .

The table below gives values of a function $g$ at selected points:

$x$ $g (x)$

$0$ $2$

$2$ $1$

$3$ $5$

$5$ $6$

$x$	$g (x)$
$0$	$2$
$2$	$1$
$3$	$5$
$5$	$6$

Suppose that $G$ is an antiderivative of $g$ such that $G (0) = 1$ . Use a right Riemann sum with the values in the table to estimate $G (3)$ .

(spoiler)

$G$ is an antiderivative of $g$ and by the FTC,

$\int_{0}^{3} g (x) d x = G (3) - G (0)$

Rearranging,

$G (3) = G (0) + \int_{0}^{3} g (x) d x$

We can only get an estimate of the definite integral, or area, using a right Riemann sum from $x = 0$ to $x = 3$ .

It may help to plot the points on a graph and draw the rectangles. Here is how it looks:

Using right endpoints to build rectangles that approximate the area under G

Starting from the right endpoint $x = 3$ , the height of the 1st rectangle is $g (3) = 5$ . The width is the distance to the next point of $x = 2$ , or width = $1$ . Then the area of the rectangle is $5 \times 1 = 5$ .

The 2nd rectangle starts at $x = 2$ and has height $g (2) = 1$ . Its width, however, is the distance to the last point of $x = 0$ , or width = $2$ . Then the area of the rectangle is $1 \times 2 = 2$ .

The right Riemann sum is $5 + 2 = 7$ , which is the estimate for $\int_{0}^{3} g (x) d x$ .

Then

$G (3) = G (0) + \int_{0}^{3} g (x) d x$

$= 1 + 7 = 8$

Here’s a problem that appeared on the multiple-choice question portion of the 1998 AB exam (#11):

If $f$ is a linear function and $0 < a < b$ , then $\int_{a}^{b} f^{''} (x) d x =$

a) $0$
b) $1$
c) $\frac{ab}{2}$
d) $b - a$
e) $\frac{b ^{2} - a ^{2}}{2}$

Solution

(spoiler)

Using the FTC,

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

If $f$ is a linear function, then it has equation

$f (x) = m x + b$

Differentiating,

$f^{'} (x) = m$

The slope of the tangent line is $m$ for any input $x$ . So $f^{'} (a) = m$ and $f^{'} (b) = m$ .

Then $f^{'} (b) - f^{'} (a) = 0$ .

The correct answer choice is $a) 0$

What you’ll learn:

Understand how definite integrals can be computed using antiderivatives
How to apply the Fundamental theorem of calculus part 2 to evaluate definite integrals
Use the reverse power rule to find basic antiderivatives

In a previous section, it was shown that derivatives and integrals are opposite processes - if $F (x) = \int_{a}^{x} f (t) d t$ , then $F^{'} (x) = f (x)$ .

Since integration can “undo” a derivative, then given a function $f (x)$ , we can look for an antiderivative - a function $F (x)$ such that its derivative is $f (x)$ .

This idea leads to the 2nd part of the Fundamental theorem of calculus, which gives a practical way to compute definite integrals using antiderivatives.

Fundamental Theorem of Calculus (FTC), Part 2

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

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

To evaluate a definite integral, just find an antiderivative $F$ , and then calculate the difference between the function values.

The vertical bar after $F (x)$ means "evaluate $F (x)$ from $a$ to $b$ . It’s a notation you’ll see and can write as an intermediate step when evaluating definite integrals.

Now the question is, how do we find antiderivatives?

Just like with derivatives, there are a set of rules you can use, such as the:

Reverse power rule

Recall that the power rule for derivatives is

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

For integrals, add 1 to the exponent and divide by the new value in the exponent:

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

Examples

Evaluate

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

Solution

(spoiler)

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

Here’s how it’s done using the FTC:

Let’s rewrite function $f (x) = 2 x + 3$ with explicit exponents to make it clear what to apply the reverse power rule on.

$f (x) = 2 x^{1} + 3 x^{0}$

Then using the reverse power rule, antiderivative $F (x)$ is

$F (x) = \frac{2 x ^{1 + 1}}{1 + 1} + \frac{3 x ^{0 + 1}}{0 + 1} + C$

$= x^{2} + 3 x + C$

Using the FTC,

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

$= [4^{2} + 3 (4)] - [(- 1)^{2} + 3 (- 1)]$

$= 28 - (- 2)$

$= 30$

Evaluate

$\int_{0}^{2} (4 x^{2} - x) d x$

(spoiler)

Rewrite $f (x)$ so to make it clear what the exponents are:

$f (x) = 4 x^{2} - x^{1/2}$

Using the reverse power rule,

$F (x) = \frac{4 x ^{2 + 1}}{2 + 1} - \frac{x ^{(1/2) + 1}}{( 1/2 ) + 1} + C$

$= \frac{4}{3} x^{3} - \frac{2}{3} x^{3/2} + C$

Then applying the FTC,

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

$= [\frac{4}{3} (2^{3}) - \frac{2}{3} (2^{3/2})] - [\frac{4}{3} (0^{3}) - \frac{2}{3} (0^{3/2})]$

$= (\frac{32}{3} - \frac{2 8}{3}) - (0 - 0)$

$= \frac{32 - 4 2}{3}$

Let $f$ be the function whose graph is given below and let $g$ be an antiderivative of $f$ .

Figure 6.6.1 Graph of f

a) If $g (5) = - 2$ , find $g (1)$ .
b) Find $g^{'} (3)$ .

Solutions

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

(spoiler)

$g$ is an antiderivative of $f$ and by the FTC,

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

Rearranging,

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

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

To evaluate the definite integral, find 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
- $A = \frac{1}{2} (2) (1 + 3) = 4$

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

Then

$g (1) = - 2 - 6.5$

$= - 8.5$

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

(spoiler)

Since $g$ is an antiderivative of $f$ , then $g^{'} (x) = f (x)$ .

So $g^{'} (3) = f (3)$ . From the graph, $f (3) = 2$ .

The table below gives values of a function $g$ at selected points:

$x$ $g (x)$

$0$ $2$

$2$ $1$

$3$ $5$

$5$ $6$

$x$	$g (x)$
$0$	$2$
$2$	$1$
$3$	$5$
$5$	$6$

Suppose that $G$ is an antiderivative of $g$ such that $G (0) = 1$ . Use a right Riemann sum with the values in the table to estimate $G (3)$ .

(spoiler)

$G$ is an antiderivative of $g$ and by the FTC,

$\int_{0}^{3} g (x) d x = G (3) - G (0)$

Rearranging,

$G (3) = G (0) + \int_{0}^{3} g (x) d x$

We can only get an estimate of the definite integral, or area, using a right Riemann sum from $x = 0$ to $x = 3$ .

It may help to plot the points on a graph and draw the rectangles. Here is how it looks:

The right Riemann sum is $5 + 2 = 7$ , which is the estimate for $\int_{0}^{3} g (x) d x$ .

Then

$G (3) = G (0) + \int_{0}^{3} g (x) d x$

$= 1 + 7 = 8$

Here’s a problem that appeared on the multiple-choice question portion of the 1998 AB exam (#11):

If $f$ is a linear function and $0 < a < b$ , then $\int_{a}^{b} f^{''} (x) d x =$

a) $0$
b) $1$
c) $\frac{ab}{2}$
d) $b - a$
e) $\frac{b ^{2} - a ^{2}}{2}$

Solution

(spoiler)

Using the FTC,

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

If $f$ is a linear function, then it has equation

$f (x) = m x + b$

Differentiating,

$f^{'} (x) = m$

The slope of the tangent line is $m$ for any input $x$ . So $f^{'} (a) = m$ and $f^{'} (b) = m$ .

Then $f^{'} (b) - f^{'} (a) = 0$ .

The correct answer choice is $a) 0$