Applying Properties of Definite Integrals
Introduced| English | Meaning |
|---|---|
| linearity/lɪˈnɪərɪti/ | linearity |
Rules that make integrals easier
- Definite integrals obey clean algebra — the same friendly rules limits and derivatives have.
- These properties let you split, combine, and rearrange integrals without recomputing areas.
- Most follow directly from "integral = signed area."
- Learn the handful here and hard-looking integrals often simplify in one step.
$\int fg\,dx$ equals $\left(\int f\,dx\right)\left(\int g\,dx\right)$.
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There is no product rule for integrals.
Linearity: split sums and pull out constants
- Constant multiple: $\displaystyle\int_a^b k\,f(x)\,dx = k\int_a^b f(x)\,dx$.
- Sum / difference: $\displaystyle\int_a^b \big(f\pm g\big)\,dx = \int_a^b f\,dx \pm \int_a^b g\,dx$.
- Together these are linearity 线性性质 — integrate term by term, coefficients out front.
- Just like differentiation, you handle a combination one piece at a time.
Integrals over adjacent intervals add; reversing the limits changes the sign
Given $\int_0^4 f\,dx=10$ and $\int_0^4 g\,dx=3$, find $\int_0^4 (2f-g)\,dx$.
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$2(10)-3=17$.
Zero width and swapping limits
- Zero width: $\displaystyle\int_a^a f(x)\,dx = 0$ — no interval, no area.
- Swap the limits: $\displaystyle\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$ — reversing direction negates it.
- So the order of the limits carries a sign.
- These two keep bookkeeping consistent when limits move around.
What is $\int_5^5 f(x)\,dx$ for any $f$?
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Zero width interval → $0$.
If $\int_1^4 f\,dx=7$, then $\int_4^1 f\,dx=$
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Swapping limits negates: $-7$.
Select all valid definite-integral properties.
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The first three hold; the product "rule" is false.
Splitting an interval
- Adjacent intervals: $\displaystyle\int_a^c f\,dx = \int_a^b f\,dx + \int_b^c f\,dx$.
- You can break one integral into pieces at any interior point $b$ and add them.
- This is essential for piecewise integrands and for using given sub-areas.
- It also runs backward: combine two adjacent integrals into one.
Split the area at an interior point
y = x²
The area from $a$ to $c$ equals the area from $a$ to $b$ plus $b$ to $c$ — integrals split over adjacent intervals.
If $\int_0^2 f\,dx=5$ and $\int_2^6 f\,dx=8$, find $\int_0^6 f\,dx$.
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Adjacent intervals add: $5+8=13$.
There is no "product rule" for integrals: $\int fg\,dx \neq \left(\int f\,dx\right)\left(\int g\,dx\right)$. Linearity only splits across sums and constant multiples, never products or quotients. And remember swapping the limits flips the sign — a very easy point to drop.
Given $\displaystyle\int_0^4 f\,dx=10$ and $\displaystyle\int_0^4 g\,dx=3$, find $\displaystyle\int_0^4\big(2f-g\big)\,dx$.
- Linearity: $\displaystyle\int_0^4 2f\,dx - \int_0^4 g\,dx = 2\int_0^4 f\,dx - \int_0^4 g\,dx$.
- $=2(10)-3=17$.
- No need to know $f$ or $g$ themselves — the properties do it.
Definite-integral properties: linearity ($\int kf = k\int f$; $\int (f\pm g)=\int f\pm\int g$), zero width ($\int_a^a f=0$), swapping limits negates ($\int_b^a f=-\int_a^b f$), and splitting ($\int_a^c=\int_a^b+\int_b^c$). There is no product rule for integrals.
Write from swapped limits
- The integral of f from 1 to 4 equals 10.
- The same integral is wanted from 4 to 1.
What is the swapped integral, and what is the integral from 2 to 2?
Direction of the limits carries the sign.
The integral from 1 to 4 is 10. The integral from 4 to 1 is what?
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Swapping the limits negates the value.