# Properties of Definite Integrals Part I

#### Hckyplayer8

##### Full Member

Just going to take this one step at a time. The way you interpret this is on the interval [0,1] the integral of f(x) is 3 and if on the interval [0,3] the integral is 2. Correct?

So for (a) since the interval [1,3] encompasses two functions, the property I would use is the difference of two integrals?

yep

#### Jomo

##### Elite Member
Yes. I would prefer that you speak in area. If you have the area from 0 to 3 (and you do) and you take away the area from 0 to 1, then you do get the area from 1 to 3. So what is your answer?

#### Hckyplayer8

##### Full Member
If the area from 0 to 3 is 2 and the area from 0 to 1 is 3

2-3 = -1 for the differences of the two areas.

If that is correct, how does one interpret that? You can't have negative area, can you?

#### Jomo

##### Elite Member
Yes you can! Area under the x-axis is considered as negative area. You answer is correct! I am glad that I pressed you for an answer because I did not notice it was going to be negative.

#### pka

##### Elite Member
You must understand that if $$\displaystyle a<b<c$$ then
$$\displaystyle \int_a^c f = \int_a^b f + \int_b^c f$$
$$\displaystyle \int_a^c f = - \int_b^a f$$
$$\displaystyle \int_a^c \alpha f = \alpha \int_a^c f$$ and
$$\displaystyle \int_a^a f = 0$$

#### Hckyplayer8

##### Full Member
You must understand that if $$\displaystyle a<b<c$$ then
$$\displaystyle \int_a^c f = \int_a^b f + \int_b^c f$$
$$\displaystyle \int_a^c f = - \int_b^a f$$
$$\displaystyle \int_a^c \alpha f = \alpha \int_a^c f$$ and
$$\displaystyle \int_a^a f = 0$$
Yup.

So reversing the interval will reverse the outcome.

So in this case, the -1 becomes 1 for (b).

I understand that if there is no change in x, the answer is zero. If one has to think of these problems in terms of area, how is the area zero? Just by having a x coord, wouldn't there be something?

#### Dr.Peterson

##### Elite Member
I understand that if there is no change in x, the answer is zero. If one has to think of these problems in terms of area, how is the area zero? Just by having a x coord, wouldn't there be something?
The region over which you are integrating in $$\displaystyle \int_a^a f(x) dx$$ is a vertical line segment.

What is the area of a line segment?

#### Jomo

##### Elite Member
If the base is 0 (from a to a) then what would you think the area would be?

#### Hckyplayer8

##### Full Member
I'm just trying to think of it visually. If you draw a line segment, you can still measure its dimensions whether it be down to mm or some sort. When I think of zero, I think of seeing nothing at all.

#### Subhotosh Khan

##### Super Moderator
Staff member
I'm just trying to think of it visually. If you draw a line segment, you can still measure its dimensions whether it be down to mm or some sort. When I think of zero, I think of seeing nothing at all.
By definition:

line segments have "zero" "width and depth".

#### Dr.Peterson

##### Elite Member
I'm just trying to think of it visually. If you draw a line segment, you can still measure its dimensions whether it be down to mm or some sort. When I think of zero, I think of seeing nothing at all.
When we draw a line, it is really meant just as an illustration of the concept. A line in theory has no thickness, so it would be impossible to see a real line. The graphite or ink you see just marks where that invisibly thin line is; it is not the line itself.

#### Hckyplayer8

##### Full Member
Finally for (d), the integral of a constant times a function is the constant times the integral of the function.

In this case, 2*2 = 4.

#### pka

##### Elite Member
Finally for (d), the integral of a constant times a function is the constant times the integral of the function.
In this case, 2*2 = 4.
Yes indeed, that is correct.
[math}\begin{gathered}
\int_0^3 f = 2 \hfill \\
\int_0^3 {2f} = 2\int_0^3 f = 2 \cdot 2 = 4 \hfill \\
\end{gathered}[/math]

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#### Hckyplayer8

##### Full Member
Thank you all for posting.