Math · Calculus I · Concept

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links derivatives and integrals. Part 1: if $f$ is continuous and $F (x) = \int_{a}^{x} f (t) d t$ , then $F^{'} (x) = f (x)$ . Part 2: if $F$ is any antiderivative of a continuous $f$ on $[a, b]$ , then the integral of $f$ from $a$ to $b$ is $F (b) - F (a)$ .

A concrete case

Distance two ways

A car pulls away from a stop sign and speeds up steadily. Its speed starts at $0 m / s$ and rises by $2 m / s$ every second, so after $3 seconds$ it is doing $6 m / s$ . How far has it gone?

One way reads the answer from the graph of speed against time. At a constant speed, distance is speed times time, which is the area of a rectangle under the speed graph. For a changing speed the distance is still the area under the graph. Here that area is a triangle $3 s$ wide and $6 m / s$ tall, so the car covers $\frac{1}{2} \cdot 3 \cdot 6 = 9 m$ .

The other way uses a distance formula. The speed after $t$ seconds is $2 t$ , and $t^{2}$ has derivative $2 t$ , so the distance is $t^{2}$ meters. That gives $1$ , $4$ and $9 m$ at $1$ , $2$ and $3 seconds$ . The table shows the two routes agreeing second by second.

Now ask for the distance between $t = 1$ and $t = 3$ . The area under the speed graph there is a trapezoid, $(2 + 6) / 2 \cdot 2 = 8 m$ . The formula gives $3^{2} - 1^{2} = 8 m$ . The area under a rate graph equals the change in the quantity the rate measures. That is the Fundamental Theorem of Calculus.

\frac{2 + 6}{2} \cdot 2 = 8 = 3^{2} - 1^{2}

The car’s distance, second by second
Second	Speed at its start and end $(m / s)$	Area under the speed graph $(m)$	Change in $t^{2}$ $(m)$
$0$ to $1$	$0$ and $2$	$(0 + 2) / 2 \cdot 1$ $= 1$	$1 - 0$ $= 1$
$1$ to $2$	$2$ and $4$	$(2 + 4) / 2 \cdot 1$ $= 3$	$4 - 1$ $= 3$
$2$ to $3$	$4$ and $6$	$(4 + 6) / 2 \cdot 1$ $= 5$	$9 - 4$ $= 5$

The rule

Part 1: the area function grows at rate $f$

Fix a starting point $a$ , and let $f$ be continuous. For each $x$ , let $A (x)$ be the signed area under the graph of $f$ from $a$ to $x$ , written $\int_{a}^{x} f (t) d t$ . $A$ is called an accumulation function. The letter $t$ is a placeholder for the variable of integration. The variable that matters is $x$ , the moving right edge.

Part 1 of the Fundamental Theorem says that $A$ has a derivative, and that its derivative is $f$ itself. The area grows exactly as fast as the curve is tall. Where $f$ is negative, new area counts as negative and $A$ falls.

A (x) = \int_{a}^{x} f (t) d t ⟹ A^{'} (x) = f (x)

\begin{array}{c} A (x) = \int_{a}^{x} f (t) d t \\ A^{'} (x) = f (x) \end{array}

Part 2: evaluate with any antiderivative

Part 2 turns the theorem into a way of computing. Suppose $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ there, meaning $F^{'} (x) = f (x)$ . Then the integral of $f$ from $a$ to $b$ is $F (b) - F (a)$ . Evaluate the antiderivative at the upper limit and subtract its value at the lower limit.

The difference is often written as $F (x)$ in square brackets, with $a$ and $b$ at the foot and top of the closing bracket. Any antiderivative gives the same answer. A constant $C$ added to $F$ appears in both terms and cancels.

Books do not agree on the numbering. Some swap Part 1 and Part 2, and some call them the First and Second Fundamental Theorems. One statement differentiates an area function, and the other evaluates an integral with an antiderivative.

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

\begin{array}{c} \int_{a}^{b} f (x) d x = [F (x)]_{a}^{b} \\ = F (b) - F (a) \end{array}

The integral of a rate is a net change

Part 2 has a reading that needs no area at all. If $Q (t)$ is any quantity and $Q^{'} (t)$ is its rate of change, integrating the rate gives the change in the quantity. Integrating a velocity gives a displacement, and integrating a flow rate gives the net amount added. The units multiply, so liters per minute times minutes gives liters.

\int_{a}^{b} Q^{'} (t) d t = Q (b) - Q (a)

Why it works

Why the area grows at rate $f$

Take $f (t) = t^{2}$ and $A (x) = \int_{0}^{x} t^{2} d t$ . As $x$ moves from $1$ to $1.01$ , $A$ gains a thin strip $0.01$ wide. Across that strip the curve is at least $f (1) = 1$ tall and at most $f (1.01) = 1.0201$ tall. The strip’s area therefore lies between $0.01$ and $0.010201$ , and its area divided by its width lies between $1$ and $1.0201$ . The table narrows the strip further.

The same squeeze works at any $x$ and for any small width $h$ . The gain $A (x + h) - A (x)$ , divided by $h$ , is the average height of $f$ across the strip. It lies between the lowest and highest values of $f$ there. As $h$ shrinks to $0$ , a continuous $f$ pins both of those values to $f (x)$ , and the average height becomes $A^{'} (x) = f (x)$ .

Part 2 follows from Part 1. $A$ and any antiderivative $F$ have the same derivative, $f$ . By the Mean Value Theorem, two functions with the same derivative on an interval differ by a constant, so $A (x) = F (x) + C$ . No area has built up at $x = a$ , so $0 = F (a) + C$ and $C = - F (a)$ . At $x = b$ this reads $A (b) = F (b) - F (a)$ .

\frac{A (x + h) - A (x)}{h} = \frac{1}{h} \int_{x}^{x + h} f (t) d t \underset{h \to 0}{→} f (x)

\begin{array}{c} \frac{A (x + h) - A (x)}{h} = \frac{1}{h} \int_{x}^{x + h} f (t) d t \\ \underset{h \to 0}{→} f (x) \end{array}

Squeezing the strip under $t^{2}$ at $x = 1$
Strip width $h$	Lowest height, $f (1)$	Highest height, $f (1 + h)$	$Strip area \div h$
$0.1$	$1$	$1.21$	$1.10333$
$0.01$	$1$	$1.0201$	$1.01003$
$0.001$	$1$	$1.002001$	$1.00100$

The method

Evaluating a definite integral, step by step

The table evaluates $\int_{1}^{2} (6 x^{2} - 4 x) d x$ . Leave out the $+ C$ , which would cancel in the subtraction anyway.

\int_{1}^{2} (6 x^{2} - 4 x) d x = [2 x^{3} - 2 x^{2}]_{1}^{2} = (16 - 8) - (2 - 2) = 8

\begin{aligned} \int_{1}^{2} (6 x^{2} - 4 x) d x \\ = [2 x^{3} - 2 x^{2}]_{1}^{2} \\ = (16 - 8) - (2 - 2) = 8 \end{aligned}

Part 2 on the integral of $6 x^{2} - 4 x$ from $1$ to $2$
Step	What to do	For $\int_{1}^{2} (6 x^{2} - 4 x) d x$
1	Check that $f$ is continuous on $[a, b]$	$6 x^{2} - 4 x$ is a polynomial, continuous everywhere
2	Find an antiderivative $F$	$F (x)$ $= 2 x^{3} - 2 x^{2}$
3	Check $F$ by differentiating	$F^{'} (x)$ $= 6 x^{2} - 4 x$
4	Evaluate $F$ at both limits	$F (2)$ $= 16 - 8$ $= 8$ and $F (1)$ $= 2 - 2$ $= 0$
5	Subtract, upper limit first	$8 - 0$ $= 8$
6	Read the sign and the units	Positive, since $6 x^{2} - 4 x$ $> 0$ on $[1, 2]$

Differentiating an accumulation function

Part 1 differentiates an accumulation function without integrating anything. Put the upper limit in place of $t$ . When the upper limit is a function $g (x)$ rather than $x$ , the chain rule also multiplies by $g^{'} (x)$ . When $x$ is the lower limit, swap the limits first, which changes the sign of the integral.

For $G (x) = \int_{0}^{x^{2}} cos t d t$ , the integrand at the upper limit is $cos (x^{2})$ , and the upper limit has derivative $2 x$ . $G^{'} (x) = 2 x cos (x^{2})$ . Integrating first gives the same answer. $G (x)$ $= sin (x^{2}) - sin 0$ $= sin (x^{2})$ , and the chain rule gives $2 x cos (x^{2})$ again.

\frac{d}{d x} \int_{0}^{x^{2}} cos t d t = cos (x^{2}) \cdot 2 x

Derivatives of accumulation functions
Function of $x$	Its derivative
$\int$ from $a$ to $x$ of $f (t) d t$	$f (x)$
$\int$ from $a$ to $g (x)$ of $f (t) d t$	$f (g (x)) \cdot g^{'} (x)$
$\int$ from $x$ to $b$ of $f (t) d t$	$- f (x)$

Average value

Dividing an integral by the length of its interval gives the average value of the function. It is the height of the rectangle on $[a, b]$ with the same area as the region under the curve. The Mean Value Theorem for integrals says that a continuous $f$ takes its average value at least once in the interval.

f_{avg} = \frac{1}{b - a} \int_{a}^{b} f (x) d x

Where it breaks

Part 2 needs $f$ to be continuous on the whole of $[a, b]$ , and $F$ to be an antiderivative on the whole of it. Take the integral of $1 / x^{2}$ from $- 1$ to $1$ . The function $- 1 / x$ has derivative $1 / x^{2}$ , and subtracting blindly gives $(- 1) - 1 = - 2$ . A positive function cannot have a negative integral. The trouble is at $x = 0$ , where $1 / x^{2}$ is undefined and grows without bound. This integral is improper, and it diverges.

Part 1 needs $f$ to be continuous at the point. Let $f (t)$ be $0$ for $t < 1$ and $1$ for $t \geq 1$ , and $A (x) = \int_{0}^{x} f (t) d t$ . $A$ is $0$ up to $x = 1$ and $x - 1$ after it. Its graph has a corner at $1$ , where $A$ has no derivative. Everywhere else $A^{'} (x) = f (x)$ .

The integral of a velocity is a displacement, not the distance traveled. Where the velocity is negative the object moves backward, and that stretch subtracts. For the distance, integrate the speed $∣ v (t) ∣$ instead.

A definite integral and an indefinite integral are different objects. $\int_{1}^{2} 2 x d x$ is the number $3$ , and $\int 2 x d x$ is the family $x^{2} + C$ . The theorem connects them without making them the same thing.

\int_{- 1}^{1} \frac{1}{x^{2}} d x \neq [- \frac{1}{x}]_{- 1}^{1} = - 2

Check your answer

An answer from the Fundamental Theorem can be checked backward, by its size and at its starting point.

Differentiate your antiderivative. $F^{'} (x)$ should give back the integrand exactly. For $2 x^{3} - 2 x^{2}$ it gives $6 x^{2} - 4 x$ .

Estimate the size. The integral lies between the smallest value of $f$ times $(b - a)$ and the largest value times $(b - a)$ . On $[1, 2]$ , $6 x^{2} - 4 x$ rises from $2$ to $16$ , so $\int_{1}^{2} (6 x^{2} - 4 x) d x$ lies between $2$ and $16$ . One midpoint rectangle, $f (1.5) \cdot 1 = 7.5$ , lands close to the exact $8$ .

Check the starting value. An accumulation function is $0$ at its lower limit, because no area has built up yet. $F (x) = x^{2} / 2 - x$ passes, since $F (0) = 0$ .

To have Chalk check the work, write it in a Math box. A definite integral can share one row with its evaluation, as in $\int_{1}^{2} (6 x^{2} - 4 x) d x$ $= (16 - 8) - (2 - 2)$ $= 8$ , and the Checker marks the row ✓ or ✗. The square bracket can share that row too, as in $\int_{1}^{2} (6 x^{2} - 4 x) d x$ $= {[2 x^{3} - 2 x^{2}]}_{1}^{2}$ $= 8$ , which reads ✓. A wrong bracket reads ✗, and so does a right bracket with a wrong value. The bracket also reads on a row of its own, below the integral.

For an accumulation function, write its definition $F (x) = \int \dots$ on the first row and the formula you found under it. The Checker marks that formula ✓ or ✗, and the derivative and the values you write under it are marked against it. Part 1 is checked directly too. Under $G (x) = \int_{0}^{x^{2}} cos t d t$ , the derivative $G^{'} (x) = 2 x cos (x^{2})$ reads ✓, and $G^{'} (x) = cos (x^{2})$ , without the chain rule factor, reads ✗.

A proof from sums, and functions with no formula

Part 2 can be proved straight from Riemann sums. Split $[a, b]$ at the points $x_{0}$ , $x_{1}$ , …, $x_{n}$ . The change $F (b) - F (a)$ is the sum of the small changes $F (x_{i}) - F (x_{i - 1})$ , because each middle value appears once with each sign and cancels. By the Mean Value Theorem, each small change equals $f (c_{i}) Δ x$ for some $c_{i}$ in its piece. $F (b) - F (a)$ is therefore a Riemann sum of $f$ , for every $n$ . A number equal to every one of the sums equals their limit, the integral.

Part 1 says something Part 2 cannot. Every function that is continuous on an interval has an antiderivative there, even when no formula for one exists. The function $e^{- t^{2}}$ has no antiderivative built from the usual functions, but $A (x) = \int_{0}^{x} e^{- t^{2}} d t$ is one all the same. Scaled by $2 / \sqrt{π}$ , it is the error function that statistics uses for the normal curve.

Some books define the natural logarithm this way, as $ln x = \int_{1}^{x} (1 / t) d t$ for $x > 0$ . Part 1 then gives its derivative, $1 / x$ , at once.

F (b) - F (a) = \sum_{i = 1}^{n} (F (x_{i}) - F (x_{i - 1})) = \sum_{i = 1}^{n} f (c_{i}) Δ x

\begin{aligned} F (b) - F (a) \\ = \sum_{i = 1}^{n} (F (x_{i}) - F (x_{i - 1})) \\ = \sum_{i = 1}^{n} f (c_{i}) Δ x \end{aligned}

Common mistakes

Subtracting in the wrong order. The integral from $a$ to $b$ is $F (b) - F (a)$ , upper limit first. $\int_{1}^{2} (6 x^{2} - 4 x) d x$ is $8 - 0 = 8$ , and $0 - 8$ has the right size and the wrong sign.
Forgetting the chain rule factor. The derivative of $\int_{0}^{x^{2}} cos t d t$ is $2 x cos (x^{2})$ , not $cos (x^{2})$ , because the upper limit $x^{2}$ moves $2 x$ times as fast as $x$ .
Using Part 2 across a point where $f$ blows up. The integral of $1 / x^{2}$ from $- 1$ to $1$ is not $- 2$ . The integrand is unbounded at $0$ , so the theorem does not apply, and the integral diverges.
Confusing the area function with its slope. For $F (x) = \int_{0}^{x} (t - 1) d t$ , $F (2) = 0$ is the area built up so far, and $F^{'} (2) = 1$ is how fast it is growing. A zero area can still be growing.
Reading a net change as a total. A velocity whose integral is $0$ can still cover ground. $v (t) = t^{2} - 4 t + 3$ has displacement $0$ on $[0, 3]$ , yet the particle travels $8 / 3 m$ , out and back.
Leaving the variable of integration in the answer. The derivative of $\int_{1}^{x} t^{2} d t$ is $x^{2}$ , not $t^{2}$ . The $t$ is a placeholder, and it disappears once the limit $x$ is put in its place.

Key terms

Fundamental theorem of calculus: The link between derivatives and integrals: if $F$ is an antiderivative of a continuous $f$ , then $\int_{a}^{b} f (x) d x$ $= F (b) - F (a)$ . It also says $d / d x \int_{a}^{x} f (t) d t = f (x)$ .
Accumulation function: A function defined by an integral with a variable upper limit, $F (x) = \int_{a}^{x} f (t) d t$ . It records the signed area from $a$ to $x$ , and where $f$ is continuous, $F^{'} (x) = f (x)$ .
Net change theorem: The integral of a rate of change over an interval equals the net change in the quantity: $\int_{a}^{b} F^{'} (t) d t$ $= F (b) - F (a)$ . Decreases count as negative, so net change can differ from the total amount of change.
Antiderivative: A function whose derivative is the given function, such as $x^{3}$ for $3 x^{2}$ . Any two antiderivatives on one interval differ by a constant, which is why answers carry $+ C$ .
Definite integral: $\int_{a}^{b} f (x) d x$ : the signed area between the graph of $f$ and the x-axis from $a$ to $b$ , defined as the limit of Riemann sums. Area below the axis counts as negative.
Average value of a function: The integral of $f$ over $[a, b]$ divided by $b - a$ : the height of the rectangle on $[a, b]$ with the same signed area as $f$ . A continuous function reaches its average value somewhere on the interval.

Work through an example

Let $F (x) = \int_{0}^{x} (t - 1) d t$ . Find a formula for $F$ and its derivative, and explain where $F$ decreases and why $F (2) = 0$ .

Find an accumulation function and its slope →

Find a net change from a rate →

Find the average value of a function →

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