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Linearity, additivity over subintervals, the order property and the size estimate, each proved from the Darboux sums.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to use linearity and additivity to combine integrals you already know, bound an integral by the size estimate when no formula is available, and prove that the size of an integral is at most the integral of the size. You will also be able to say why a non-negative function with integral zero need not vanish, and what hypothesis repairs that statement.
The integral as the common value of the upper and lower integrals, and the criterion that establishes it. Every property here is proved from the Darboux sums, and each proof is short — which is the payment for having built the object carefully.
Linearity is the rule that constants come out of an integral and sums split; additivity joins two adjacent intervals into one. Orientation is the convention that reversing the two limits changes the sign, which is what makes additivity hold for three points in any order.
Throughout, $f$ and $g$ are integrable on $[a,b]$.
Linearity. $\int (f + g) = \int f + \int g$ and $\int cf = c\int f$. The second is immediate from the sums; the first is not quite, because the supremum of a sum can be less than the sum of the suprema — so the proof compares the sums for $f+g$ with those for $f$ and $g$ on a common refinement.
Additivity. For $a < c < b$, $\int_a^b f = \int_a^c f + \int_c^b f$, and each side's existence follows from the other's. Adding $c$ to a partition is a refinement, which is the whole proof.
Order. If $f \le g$ pointwise then $\int f \le \int g$. Not strictly: a function positive at one point only has the same integral as one that is zero.
Size. $\left|\int f\right| \le \int |f| \le M(b-a)$ where $M$ bounds $|f|$. The first follows from the order property applied to $-|f| \le f \le |f|$; the second from comparing with a constant.
Orientation. $\int_b^a f$ is defined as $-\int_a^b f$, which makes additivity hold for any three points in any order.
The size estimate is the one used most. Nearly every argument that needs an integral to be small is this inequality with a small bound or a narrow interval.
Another way: picture
Additivity is two adjacent regions whose areas add. The order property is one region sitting inside another. The size estimate is a region inside a rectangle as tall as the function ever gets. Each picture is exactly what the proof from the sums says, which is unusual and worth enjoying.
Another way: steps
To evaluate or bound an integral without a formula:
Most estimates in the rest of the course are exactly this, with step 3 doing the work.
For a single partition, $\sup(f+g) \le \sup f + \sup g$ on each piece, and the inequality can be strict: on $[0,1]$ take $f(x) = x$ and $g(x) = 1 - x$, whose suprema are $1$ and $1$ while the sum is constantly $1$.
So $U(f+g, P) \le U(f,P) + U(g,P)$, with no equality available. The proof takes $\varepsilon$, chooses $P_1$ good for $f$ and $P_2$ good for $g$, and works on their common refinement $P$, where both are good at once. Then $$\overline{\int}(f+g) \le U(f+g,P) \le U(f,P) + U(g,P) < \int f + \int g + \varepsilon,$$ and the matching lower bound comes from the lower sums. True for every $\varepsilon$, so the upper and lower integrals of $f + g$ both equal $\int f + \int g$.
The move — take a common refinement so that two partitions can be good simultaneously — is the integral's version of taking the larger of two indices in a sequence proof.
Non-negative with integral zero. If $f \ge 0$ and $\int_a^b f = 0$, it does not follow that $f = 0$; it follows that $f = 0$ at every point of continuity. With $f$ continuous, a value $f(c) > 0$ would be at least $f(c)/2$ on an interval about $c$, contributing at least $f(c)\delta/2 > 0$ — so then $f$ is identically zero.
The mean value theorem for integrals. If $f$ is continuous on $[a,b]$, then $\int_a^b f = f(p)(b-a)$ for some $p \in [a,b]$. It follows from the size estimate and the intermediate value theorem: the average value lies between the minimum and the maximum, both of which are attained, so it is attained too. Continuity is needed — a step function's average need not be one of its values.
Both are places where the integral's insensitivity to small sets shows through, and both are proved by going back to continuity at a point.
Expecting strict inequalities to survive. $f < g$ at every point gives only $\int f \le \int g$ in general, though with continuity and a strict inequality somewhere the integrals do separate.
Reading integral zero as function zero. True for continuous non-negative functions and false in general; a single non-zero value is invisible to the integral.
Assuming the integral of a product is the product of the integrals. It is not, and there is no rule that repairs it. $fg$ is integrable, and its integral is a new quantity.
Forgetting orientation. $\int_b^a$ is minus $\int_a^b$ by definition, and additivity holds for points in any order only because of that convention.
Bound $\left|\int_0^1 \dfrac{\sin(x^3)}{1 + x}\,dx\right|$, for which no antiderivative is available.
No formula, so estimate instead.
The size of the integrand is at most $1$, since the numerator is at most $1$ in size and the denominator at least $1$.
Bound the function.
By the size estimate the integral is at most $1 \times 1 = 1$ in size. Crude, exact enough for most purposes, and obtained in two lines.
Bound times width.
Compute $\int_0^2 f$ where $f(x) = x$ below $1$ and $f(x) = 2 - x$ from $1$ on.
One formula per piece.
Split at $1$: $\int_0^1 x\,dx + \int_1^2 (2-x)\,dx$.
Additivity, at the point where the rule changes.
Each piece is a triangle of area $1/2$, so the total is $1$.
Two easy integrals instead of one awkward one.
Suppose $f(c) > 0$ for some $c$. Continuity gives an interval about $c$ where $f > f(c)/2$.
The sign-preservation lemma from unit 3.
Splitting the integral at the ends of that interval, the middle piece is at least $f(c)\delta/2 > 0$, and the outer pieces are non-negative.
Additivity, then the order property.
So the whole integral is positive, contradicting the hypothesis. Hence no such $c$ exists. Dropping continuity breaks the first step, and the function that is $1$ at one point survives as a counterexample.
Suppose $\int_0^1 f = 7$, $\int_0^1 g = 7$ and $\int_1^2 f = 7$. Give the three integrals below.
| value | |
|---|---|
| the integral of $3f$ from $0$ to $1$ | |
| the integral of $f + g$ from $0$ to $1$ | |
| the integral of $f - g$ from $0$ to $1$ |
Compute $\displaystyle\int_0^1 (3x + 8)\,dx$ from the properties of the integral alone.
Answer:
Match each statement about integrals to its name.
| linearity | additivity over subintervals | the order property | the basic size estimate | |
|---|---|---|---|---|
| the integral of a sum is the sum of the integrals, and constants come out | ||||
| the integral over a whole interval is the sum over two pieces meeting at $6$ | ||||
| if one function is at most another at every point, so are their integrals | ||||
| the size of the integral is at most the largest size times the width |
An integrable $f$ satisfies $2 \le f(x) \le 5$ on the interval from $0$ to $3$. Give the set of values its integral over that interval must lie in.
This task has no paper form; do it on a device.
Put the steps that prove the integral of a function is at most the integral of its size into order.
Number the steps in order (write the number in the box):
Here is an argument about a non-negative integrable function whose integral over the interval from $0$ to $1$ is zero. Mark the one step that does not follow.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Suppose $\int_0^1 f = 2$, $\int_0^1 g = 4$ and $\int_1^2 f = 4$. Give the three integrals below.
| value | |
|---|---|
| the integral of $3f$ from $0$ to $1$ | |
| the integral of $f + g$ from $0$ to $1$ | |
| the integral of $f - g$ from $0$ to $1$ |
You can combine integrals by linearity and additivity, bound one by the size estimate, and say which order statements survive and which do not. Say in your own words why the proof of linearity needs a common refinement. Next: the theorem that connects the integral to the derivative.
10. Your turn: show that if $f$ is continuous and non-negative on $[0,1]$ with $\int_0^1 f = 0$, then $f$ is identically zero, step 3