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Linearity, additivity, orientation, comparison and average value — and why signed is not the same as area.
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 combine integrals using linearity and additivity without any integrand, apply the orientation convention and say what it buys, bound an integral by comparison without evaluating it, compute an average value and say what makes it an average, use odd and even symmetry to settle an integral at sight, and distinguish the signed integral from total area — knowing which one a question is asking for.
$\int_a^b f(x)\,dx$ is the number the Darboux sums close on. Everything below follows from the sums having the property and the limit preserving it — no new machinery, and no antiderivatives yet.
Linearity. $\displaystyle\int_a^b (cf + g) = c\int_a^b f + \int_a^b g$.
Additivity over intervals. $\displaystyle\int_a^c f = \int_a^b f + \int_b^c f$.
Orientation. $\displaystyle\int_b^a f = -\int_a^b f$, and $\displaystyle\int_a^a f = 0$.
The orientation rule is a definition, not a theorem — the Darboux construction assumes $a < b$. It is adopted because it makes additivity hold for every arrangement of $a$, $b$ and $c$ rather than only when $b$ lies between the other two, and one convention thereby removes a case analysis from every later proof.
Comparison. If $f \le g$ on $[a,b]$ then $\int_a^b f \le \int_a^b g$. In particular, if $m \le f \le M$ then $$m(b-a) \le \int_a^b f \le M(b-a),$$ which bounds an integral without evaluating it.
Average value. $\displaystyle\bar f = \frac{1}{b-a}\int_a^b f$ — the height of the rectangle with the same integral.
The integral is signed. Where $f < 0$ the contribution is negative. The integral is area above minus area below; total area is $\int |f|$, and it is a different number.
Another way: picture
A curve crossing the axis. The region above counts positive, the region below negative, and the integral is the difference. Reflect the part below the axis upwards and you are computing $\int|f|$ instead — a different picture answering a different question, and the notation for the two looks almost identical.
Another way: steps
To use the properties on integrals given by their values:
No integrand is ever needed.
Two consequences of additivity and orientation are worth having to hand, because they turn many integrals into no calculation at all.
Odd integrand, symmetric interval. If $f(-x) = -f(x)$ then $$\int_{-a}^{a} f = 0,$$ because the two halves cancel exactly: substituting $u = -x$ on $[-a, 0]$ turns it into the negative of the integral on $[0,a]$.
Even integrand, symmetric interval. If $f(-x) = f(x)$ then $$\int_{-a}^{a} f = 2\int_0^a f.$$
So $\int_{-1}^{1} x^3\,dx = 0$ without any work, and $\int_{-2}^{2}x^2dx = 2\int_0^2 x^2 dx$. Both are worth spotting before starting, and the first is worth spotting before spending five minutes on an antiderivative.
The same reasoning covers a periodic integrand over a whole number of periods: the pieces are identical, so the integral is that many times the integral over one period.
"The integral is the area under the curve." Only for a non-negative integrand. Otherwise it is a signed quantity, and a question asking for area wants $\int|f|$ — which means splitting at the zeros of $f$ and negating the negative pieces.
Splitting a product. $\int fg \ne \left(\int f\right)\left(\int g\right)$. Linearity is about sums and constant multiples, and there is no product rule for integrals of this kind at all.
Ignoring the order of the limits. $\int_4^1$ is the negative of $\int_1^4$, and a sign lost here propagates silently.
Assuming additivity needs $b$ between $a$ and $c$. With the orientation convention it holds regardless — that is precisely what the convention buys.
Reading the average value as the average of the endpoint values. $\bar f = \frac{1}{b-a}\int f$ is an average over the whole interval, and equals $\frac{f(a)+f(b)}{2}$ only for a linear $f$.
It is how the integral is introduced everywhere, and it is right exactly when $f \ge 0$. Once the integrand goes negative the integral and the area part company, and the two questions have different answers that the same notation is used for.
$\int_0^{2\pi}\sin x\,dx = 0$. The region between the curve and the axis plainly has area, and the area is $4$. Nothing has gone wrong: the integral counted the second hump as negative, because the rectangles there have negative height, and the definition never promised otherwise.
So when a problem asks for area, the work is: find where $f$ changes sign, split there, and integrate $|f|$ — which means integrating $f$ on each piece and negating the pieces where it was negative. And when a problem asks for a net change — displacement rather than distance, profit rather than turnover — the signed integral is the one that is wanted, and taking absolute values would be the error instead.
Given $\int_0^3 f = 7$, $\int_0^1 f = 2$ and $\int_3^5 f = -4$, find $\int_5^1 f$.
Nothing about $f$ is known or needed.
$\int_1^5 f = \int_1^3 f + \int_3^5 f$, and $\int_1^3 f = \int_0^3 f - \int_0^1 f = 7 - 2 = 5$. So $\int_1^5 f = 5 - 4 = 1$.
Additivity, twice.
The limits asked for are reversed, so $\int_5^1 f = -1$.
Orientation last.
$f(x) = x$ on $[-2, 3]$. The integral: the triangle below the axis on $[-2,0]$ has signed contribution $-2$, the one above on $[0,3]$ contributes $4.5$.
Split at the zero.
$\int_{-2}^{3} x\,dx = -2 + 4.5 = 2.5$.
Signed.
Total area: $2 + 4.5 = 6.5$, which is $\int_{-2}^{3}|x|\,dx$. Two different numbers from the same picture, and only the question decides which is wanted.
Both $x^3$ and $x$ are odd, so the integrand is odd.
Check the symmetry first.
The interval is symmetric about $0$, so the two halves cancel.
The integral is $0$ — with no antiderivative computed and no arithmetic done.
Match each integral fact to the valid conclusion.
| $6\le\int_1^4f\le15$ | The signed integral is nonnegative | The signed integral is nonpositive | Integrate $|f|$ or split at zeros | |
|---|---|---|---|---|
| $2\le f(x)\le5$ on $[1,4]$ | ||||
| $f(x)\ge0$ on $[a,b]$ | ||||
| $f(x)\le0$ on $[a,b]$ | ||||
| Total area with an axis crossing |
Match each integral property to the conclusion it permits.
| Rewrite as $3\int_a^bf+\int_a^bg$ | Rewrite as $\int_a^bf+\int_b^cf$ | It equals $-\int_a^bf$ | Use $\int_a^b|f(x)|\,dx$ for total area | |
|---|---|---|---|---|
| $\int_a^b(3f+g)$ | ||||
| $\int_a^c f$ with $a<b<c$ | ||||
| $\int_b^a f$ | ||||
| $f$ is below the axis on part of $[a,b]$ |
$\int_a^b f = 7$ and $\int_a^b g = 5$. Find $\int_a^b (4f + g)$.
Answer:
$\int_0^5 f = 8$ and $\int_0^2 f = 9$. Find $\int_2^5 f$.
Answer:
$\int_1^4 f = 4$. What is $\int_4^1 f$, and why?
| Signed accumulation from $a$ to $b$ | $-\int_a^b f$ | 0 | $\int_a^b f+\int_b^c f$ | |
|---|---|---|---|---|
| $\int_a^b f$ | ||||
| $\int_b^a f$ | ||||
| $\int_a^a f$ | ||||
| $\int_a^c f$ |
$f$ is continuous on $[\,0, 6\,]$ with $0 \le f(x) \le 8$ throughout. What is the largest $\int_0^{6} f$ can be?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$\int_0^{2\pi} \sin x \, dx = 0$. Does that mean the region between $\sin x$ and the axis has zero area?
| Contributes positively to the integral | Contributes negatively to the integral | Can be zero through cancellation | Use the integral of $|f|$ | |
|---|---|---|---|---|
| Graph lies above the axis | ||||
| Graph lies below the axis | ||||
| Definite integral across an axis crossing | ||||
| Total geometric area |
You can manipulate integrals by their properties and tell a signed integral from an area. Say in your own words why reversing the limits changes the sign, and what would break without that convention. Next: the fundamental theorem, where the two halves of the course meet.
8. Your turn: $\int_{-3}^{3}(x^3 + x)\,dx$, step 3