2.5.3 The fundamental theorem of calculus

The area function

Fix a starting point $a$ and let the right edge move. The area under $f$ from $a$ up to $x$ is then a function of $x$ : the area function.

Rule: Area function

A(x) = \int_a^x f(t)\,dt

The signed area under $f$ from $a$ to $x$ . Inside the integral the letter $t$ runs along the axis, so that $x$ can name the right edge. At the start there is no width yet, so $A(a) = 0$ .

Take $f(t) = 2t$ from 0. The region up to $x$ is a triangle with base $x$ and height $2x$ , so $A(x) = \frac{1}{2} \times x \times 2x = x^2$ .

right edge $x$	1	2	3
height $f(x) = 2x$	2	4	6
area $A(x) = x^2$	1	4	9

Now differentiate: the slope of $A(x) = x^2$ is $2x$ , the function we started from. The area function is an antiderivative of $f$ . A constant height gives the same pattern: under $f(t) = 3$ , $A(x) = 3x$ , whose slope is 3.

Why the slope of the area is the height

Push the right edge from $x$ to $x + h$ . The area grows by a thin strip of width $h$ under the curve. Its height is almost $f(x)$ , so the extra area is about $f(x) \times h$ .

So the growth rate $\frac{A(x + h) - A(x)}{h}$ is about $f(x)$ , and it gets closer as $h$ shrinks. In the limit (lesson 2.2.1) the slope of $A$ at $x$ is exactly $f(x)$ .

Rule: Slope of the area function

\frac{d}{dx}\int_a^x f(t)\,dt = f(x)

The area grows fastest where the curve is tallest. It stops growing where $f(x) = 0$ , and it shrinks where $f(x) < 0$ , since area below the axis counts as negative. This holds for every continuous $f$ ; the Proof Gym has the proof.

The area under $f(t) = 4 - t^2$ from 0 up to $x$ , and the graph of that area. At $x = 1$ the area grows at rate $f(1) = 3$ ; it peaks at $x = 2$ , where $f$ crosses zero, and falls once $f$ is negative.

Numbers show the same thing for $f(t) = 4 - t^2$ at $x = 1$ , where the height is $f(1) = 3$ . Each strip ratio below comes from Riemann sums, with no formula for $A$ .

strip width $h$	0.1	0.01	0.001
$\frac{A(1 + h) - A(1)}{h}$	2.896667	2.989967	2.999

4 - t^2

import numpy as np def f(t): return 4 - t**2 def area(x, n=10_000): dt = x / n return np.sum(f((np.arange(n) + 0.5) * dt)) * dt for h in [0.1, 0.01, 0.001]: print(h, round((area(1 + h) - area(1)) / h, 6))

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Drag the right edge yourself. Watch the area function's tangent: its slope matches the height of the curve at every $x$ , and it turns flat at $x = 2$ , where $4 - t^2$ crosses zero. Then try $\frac{1}{t}$ , whose area function from 1 is $\ln x$ (lesson 2.5.2).

Explore

Drag the right edge x. The upper graph shades the area under the curve from 0 to x; the lower graph plots that area for every x. Its slope at x is always the height f(x).

The area function A(x) = 4x − x³/3, with its tangent at x

Right edge x = 1

Show the next strip (width 0.1)

Area from 0 to x: A(1) = 3.6667
Height there: f(1) = 3
Next strip ÷ its width: (A(1.1) − A(1)) ÷ 0.1 = 2.8967, close to the height
So the area is: growing, at rate f(x)

The fundamental theorem

To find $\int_a^b f(x)\,dx$ , find any antiderivative $F$ and subtract its value at $a$ from its value at $b$ .

Here is why. The area function $A$ is an antiderivative of $f$ , and any other antiderivative $F$ differs from it by a constant (lesson 2.5.2), so $A(x) = F(x) + C$ . At $x = a$ the area is 0, which forces $C = -F(a)$ , so $A(b) = F(b) - F(a)$ .

Rule: Fundamental theorem of calculus

\begin{gathered} \int_a^b f(x)\,dx = F(b) - F(a) \\ \text{where } F' = f \end{gathered}

For $f$ continuous on $[a, b]$ . Any antiderivative works: a $+ C$ would appear in both terms and cancel. The bracket $\big[F(x)\big]_a^b$ is shorthand for $F(b) - F(a)$ .

Four examples

$\int_0^1 x^2\,dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}$ . That is the number the Riemann sums of lesson 2.5.1 approached: 0.21875, 0.328125 and 0.46875 with 4 strips, and a midpoint sum of 0.333333 with 10,000.
$\int_1^2 \frac{1}{x}\,dx = \big[\ln x\big]_1^2 = \ln 2 - \ln 1 = \ln 2 \approx 0.693147$ . It lies between that lesson's left and right sums with 4 strips, 0.759524 and 0.634524.
$\int_0^1 e^x\,dx = \big[e^x\big]_0^1 = e - e^0 = e - 1 \approx 1.718282$ . The lower limit gives $e^0 = 1$ , not 0.
$\int_1^2 (3x^2 + 2x)\,dx = \big[x^3 + x^2\big]_1^2 = (8 + 4) - (1 + 1) = 10$ .

The theorem also explains the table above. For $4 - t^2$ it gives $A(x) = 4x - \frac{x^3}{3}$ , and then the strip ratio is exactly $3 - h - \frac{h^2}{3}$ : 2.896667 for $h = 0.1$ .

Where it breaks

The theorem needs $f$ continuous on the whole interval. $\frac{1}{x^2}$ is positive everywhere, yet the formula would give $\int_{-1}^{1} \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_{-1}^{1} = -1 - 1 = -2$ .

That is impossible for a curve above the axis. The trouble is at 0, where $\frac{1}{x^2}$ has no value and blows up. Its midpoint sums grow without limit (47 with 10 strips, 491 with 100, 4,933 with 1,000), and lesson 2.6.3 deals with such integrals.

\ln 2

import math import sympy as sp from scipy.integrate import quad x = sp.symbols("x") print(sp.integrate(x**2, (x, 0, 1)), quad(lambda t: t**2, 0, 1)[0]) print(sp.integrate(1 / x, (x, 1, 2)), quad(lambda t: 1 / t, 1, 2)[0]) print(sp.integrate(sp.exp(x), (x, 0, 1)), quad(math.exp, 0, 1)[0]) print(sp.integrate(3 * x**2 + 2 * x, (x, 1, 2))) print(sp.integrate(1 / x**2, (x, -1, 1)))

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Net and total area

$F(b) - F(a)$ is the net area: parts below the axis count as negative (lesson 2.5.1). For the total area, split the interval where $f$ crosses zero and add the pieces as positive numbers.

Take $f(x) = 4 - x^2$ on $[0, 3]$ , the curve of the figure. With $F(x) = 4x - \frac{x^3}{3}$ , the net area is $F(3) - F(0) = 12 - 9 = 3$ .

For the total, split at $x = 2$ , where $f$ crosses zero. The piece above the axis is $F(2) - F(0) = \frac{16}{3} \approx 5.333$ , and the piece below is $F(3) - F(2) = -\frac{7}{3} \approx -2.333$ . So the total area is $\frac{16}{3} + \frac{7}{3} = \frac{23}{3} \approx 7.667$ .

These are values of the area function in the figure. $A(2) = \frac{16}{3}$ is its peak, and $A(3) = 3$ is lower because the last piece counts as negative.

Reversed and equal limits

The formula $F(b) - F(a)$ works for any two limits, in either order. That gives three rules.

Rule: Limits in any order

\begin{gathered} \int_a^a f\,dx = 0 \\[4pt] \int_b^a f\,dx = -\int_a^b f\,dx \\[4pt] \int_a^b f\,dx + \int_b^c f\,dx = \int_a^c f\,dx \end{gathered}

Equal limits give no width and no area. Swapping the limits flips the sign: in a Riemann sum, going from right to left makes $\Delta x$ negative. Areas side by side add.

So $\int_2^0 x^2\,dx = -\frac{8}{3}$ , and $\int_5^5 e^{x^2}\,dx = 0$ , even though $e^{x^2}$ has no antiderivative formula. Splitting $[0, 2]$ at 1 gives $\int_0^1 x^2\,dx + \int_1^2 x^2\,dx = \frac{1}{3} + \frac{7}{3} = \frac{8}{3}$ , the whole of $\int_0^2 x^2\,dx$ .

4 - x^2

import sympy as sp from scipy.integrate import quad x = sp.symbols("x") f = 4 - x**2 print(sp.integrate(f, (x, 0, 2)), sp.integrate(f, (x, 2, 3)), sp.integrate(f, (x, 0, 3))) print(round(quad(lambda t: abs(4 - t**2), 0, 3)[0], 6)) print(sp.integrate(x**2, (x, 2, 0)), round(quad(lambda t: t**2, 2, 0)[0], 6)) print(sp.integrate(sp.exp(x**2), (x, 5, 5)))

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The integral of a rate is a total change

If $F' = f$ , then $f$ is the rate at which $F$ changes. Integrating the rate over $[a, b]$ gives the total change in $F$ .

Rule: Total change

\int_a^b F'(t)\,dt = F(b) - F(a)

The units are the rate's units times the input's units. A rate in cm per year, integrated over years, gives cm.

A tree trunk widens at $2e^{-0.1t}$ cm per year, $t$ years after planting. Over the first 10 years it widens by

\begin{aligned} \int_0^{10} 2e^{-0.1t}\,dt &= \left[-20e^{-0.1t}\right]_0^{10} \\ &= 20\left(1 - e^{-1}\right) \approx 12.64 \text{ cm} \end{aligned}

The antiderivative divides by $k = -0.1$ , as in lesson 2.5.2. The rate and the time must use the same unit: ten years is $t = 10$ , not 120 months.

A velocity can be negative, and then the net and total areas mean different things. At $4 - t^2$ metres per second for 3 seconds, an object ends 3 m from where it started. It went forward $\frac{16}{3}$ m, then back $\frac{7}{3}$ m, so it travelled $\frac{23}{3} \approx 7.67$ m in all.

Case study 2.5.5 integrates a cash-flow rate in dollars per year. Lesson 2.6.1 adds substitution, which finds many more antiderivatives.

v

import math from scipy.integrate import quad print(round(20 * (1 - math.exp(-1)), 4), round(quad(lambda t: 2 * math.exp(-0.1 * t), 0, 10)[0], 4)) def v(t): return 4 - t**2 print(round(quad(v, 0, 3)[0], 4), round(quad(lambda t: abs(v(t)), 0, 3)[0], 4))

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Common mistakes

Upper limit first. The answer is $F(b) - F(a)$ , not $F(a) - F(b)$ .
The lower limit counts too. $F(a)$ need not be 0: $\int_0^1 e^x\,dx = e - 1$ , not $e$ .
Net area is not total area. For the total, split where $f$ crosses zero and add the pieces as positive numbers.
Swapping the limits flips the sign. $\int_2^0 x^2\,dx = -\frac{8}{3}$ .
The curve must be continuous on the whole interval. $\int_{-1}^{1} \frac{1}{x^2}\,dx$ is not $-2$ .
Match the units. A rate per year needs the time in years.

The fundamental theorem

0 of 12 correct

1.
Find $\int_0^3 2x\,dx$
2.
Find $\int_1^2 x^3\,dx$
3.
Find $\int_0^1 e^x\,dx$
4.
Find $\int_1^4 \frac{1}{x}\,dx$
5.
Find $\int_0^4 \sqrt{x}\,dx$
6.
$A(x) = \int_0^x (t^2 + 1)\,dt$ . Find the slope $A'(2)$
7.
Find the net area $\int_0^3 (x - 1)\,dx$
8.
Find the total area between $y = x - 1$ and the $x$ -axis from 0 to 3
9.
Find $\int_3^1 x\,dx$
10.
Find $\int_5^5 e^{x^2}\,dx$
11.
If $\int_0^2 f(x)\,dx = 5$ and $\int_2^6 f(x)\,dx = -3$ , find $\int_0^6 f(x)\,dx$
12.
Water flows into a tank at $3t^2$ litres per minute. How many litres arrive between $t = 0$ and $t = 2$ minutes?

How to type answers

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A(x) = \int_a^x f(t)\,dt

The fundamental theorem of calculus

The area function

Rule: Area function

Why the slope of the area is the height

Rule: Slope of the area function

The fundamental theorem

Rule: Fundamental theorem of calculus

Four examples

Where it breaks

Net and total area

Reversed and equal limits

Rule: Limits in any order

The integral of a rate is a total change

Rule: Total change

Common mistakes

The fundamental theorem

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