Main content

### Course: AP®︎ Calculus AB (2017 edition) > Unit 8

Lesson 7: Definite integrals properties- Definite integrals on adjacent intervals
- Definite integral over a single point
- Switching bounds of definite integral
- Worked examples: Finding definite integrals using algebraic properties
- Using multiple properties of definite integrals
- Integrating sums of functions
- Integrating scaled version of function
- Worked examples: Definite integral properties 2
- Definite integrals over adjacent intervals
- Definite integral properties (no graph): function combination
- Definite integral properties (no graph): breaking interval
- Warmup: Definite integral properties (no graph)
- Finding definite integrals using algebraic properties
- Definite integrals properties review

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Integrating sums of functions

If you know the integrals of two functions, what is the integral of their sum?

## Want to join the conversation?

- Im still a bit confused about the "Integrals" notation, ¿ what do we mean with dx? Im thinking it refers to (DeltaX) from the sum, but im not sure.(14 votes)
- When Δx becomes infinitesimally small, we call it dx, or the differential.(27 votes)

- Would it be correct to say that the definite integral of a sum is the sum of the definite integrals is a result of the limit property lim x->a f(x) + g(x) is the same as lim x->a f(x) + lim x->a g(x)? I'm not completely sure if dealing with the limit of the areas of geometric shapes with infinitely small widths follows the same rules as taking regular limits and if my intuition regarding this issue is correct.(8 votes)
- It is a combination of the limit addition property AND the sum (sigma notation) addition property (which is basically just the commutative property for addition). Just think of the definite integral as the Riemman sums. You add f(x_1) + g(x_1) + f(x_2) + g(x_2) + .... But you can use the commutative property and rewrite it as f(x_1) + (f_x2) + ... + g(x_1) + g(x_2) + ... So you see how the sum of f(x) + g(x) is exactly the sum of f(x) + the sum of g(x). Also the delta_x can be factored out of the original sum, and redistributed to the two resulting sums, so it does not affect out commutative property. Finally you end up with a limit of two sums, which can be split into two sums of limits.(13 votes)

- So is he taking the graph of y=g(x) and putting it on top of y=f(x)?

I'm visualizing him taking the graph of y=g(x), moving it, and when it's on top of the y=f(x), having the x-axis sort of shift from a straight line to a curve accommodating into the shape of y=f(x), and then the top of y=g(x) would change accordingly.

Is this an accurate description?(6 votes)- The main take-away of this video, though it is not explicitly stated, is that the integral of the sum of two functions is equal to the sum of the integrals of each function, that is:

∫(f(x) + g(x))dx = ∫f(x)dx + ∫g(x)dx.

Now since addition is commutative, the order of f(x) and g(x) does not matter.

Now, what Sal was doing was graphically showing this. The procedure he was using to illustrate goes something like this:

At x=a, add the heights f(a) and g(a) to get the combined height and graph it.

Now pick another point, x_1, such that a<x_1<b and add f(x_1) and g(x_1) to get the combined height and graph that.

Continue, for as many additional points as you care to, until you reach b, now add f(b) and g(b) to get the combined height and graph that.

Now connect the points you just graphed and that is the graph of f(x) + g(x) and the integral of that area ∫(f(x) + g(x))dx is equal to the sum of the areas of each of the functions f and g, that is:

∫(f(x) + g(x))dx = ∫f(x)dx + ∫g(x)dx.

Hope that helped.(14 votes)

- Isn't this the similar idea as the sum rule for differentiaition?(7 votes)
- Correct - they are both linear operators. That is, integration and differentiation are linear operations.(7 votes)

- What's the difference between indefinite and definite integrals?(3 votes)
- An
*indefinite integral*of a function, also called an*antiderivative*of the function, is another function whose derivative is the original function. For example, suppose an antiderivative of 𝑓 is 𝐹. Then, the following equation is satisfied:

𝐹' = 𝑓

So indefinite integration is the reverse process of differentiation. Instead of going from the original function to its derivative, you're going from the derivative to an original function!

A*definite integral*WRT (with respect to) 𝑥 of a function of 𝑥 is the*signed area*bounded by the curve and the 𝑥-axis over some interval. For example, the definite integral WRT 𝑥 from 0 to 1 of the function 𝑓(𝑥) = 𝑥 is the area of the region bounded by the line from 𝑥 = 0 to 𝑥 = 1 and the 𝑥-axis. This region is a right triangle, and its area can be computed easily without any calculus (1/2). On the other hand, if we had a function like 𝑔(𝑥) = -𝑥, and we integrate WRT 𝑥 from 0 to 1, we are looking at an area*underneath*the 𝑥-axis. We define area underneath the 𝑥-axis to be*negative area*. This is what is meant by the term*signed area*: area above the 𝑥-axis is defined as positive and area under the 𝑥-axis is defined as negative.

So this may make you wonder why these two seemingly unrelated concepts both fall under the same name of "integration". In fact, there is a very useful and beautiful connection between these two concepts! This connection is called the**Fundamental Theorem of Calculus**. It states that the definite integral of the function 𝑓 WRT 𝑥 from 𝑥 = 𝑎 to 𝑥 = 𝑏 is equal to:

𝐹(𝑏) – 𝐹(𝑎)

Where 𝐹 is an antiderivative of 𝑓. I suggest you look into all of this on KA! Comment if you have questions!(4 votes)

- Good video. I have a question. What happens when we integrate (x+2). We can either do \int x+2 dx, or \int x dx + \int 2 dx. When x =2 first gives 18 and second gives 16. I've had something similar in an exercise i was doing about differential equations. I used the second method to solve it and the book used the first. We came up with different answers (after also substituting the original conditions), so i'm assuming that i'm wrong. I'm confused here. If someone can help with this, i'd appreciate it. Thanks in advance.(4 votes)
- Are we basically adding two definite integrals?(3 votes)
- would the graph of f(x) + g(x) be the same as the graph of g(x) + f(x)?(2 votes)
- Rules of addition/subtraction of function are the same as the addition of numbers: the Commutative Property holds.(5 votes)

- Does it mean that the graph of g(x)+f(x) would be the other way around that is the shifted version of the growing exponential function g(x)?(3 votes)
- Addition is commutative, so f + g = g + f. You could construct g + f in the opposite way that f + g was constructed in the video, but you will get the same area.(2 votes)

- What is the definiton of definite integral of a function?(2 votes)
- If F(x) = the indefinite integral of f(x) dx

then, the definte integral of f(x) over the interval [a, b] is F(b) - F(a).

A detailed explanation can be found at:

http://mathworld.wolfram.com/DefiniteIntegral.html(3 votes)

## Video transcript

- So we have a couple of functions here. This is the graph of y is equal to f of x. This is the graph of y is equal to g of x. And we already know something. Or we know ways to represent the area under the curve y equals f of x. Between these two points x is equal to a, and x is equal to b. So this area, right over here, between the curve and the x axis, between x equals a and x equals b, we know we can write that
as the definite integral, from a to b, of f of x, dx. And we can do the same thing over here. We could call this area, let me pick a color that I have not used. Well, this is slightly different green. I could call this area right over here, the area right under the
curve y is equal to g of x. And above the positive x axis, between x equals a and x equals b, we could call that the definite integral. From a to b of g of x, dx. Now given these two things, let's actually think about
the area under the curve of the function created by the
sum of these two functions. So what do I mean by that? So let me, this is
actually a fun thing to do. Let me start again. That's exactly what we have over here. This is the graph of y is equal to f of x. But what I want to do,
is I want to approximate the graph of y is equal
to, so my goal is to graph y is equal to f of x, plus g of x. So for any given x,
it's going to be f of x. So that's that the f of x. And then I'm gonna add the g of x to it. So what will that look like? So that's gonna look like, let's see. When x is zero, g of x looks like it's about that length right there. I'm obviously approximating it. So I'm gonna have to add
that length right over here. So it'll probably be right around there. At x equals a, it's a little bit more. And so but now my f of x curve
is gone, or has increased. But I take that same distance above it. If I add the g of x there, it gets me right about there. Once again I'm just eyeballing it, trying to get an approximation, giving me an intuition actually
for f of x, plus g of x is. I'm just trying to add g of x for given x. Now let's see if I'm a little bit, let's say that I'm between a and b, g of x is about that
distance right over there. So, if I wanted to put that
same distance right over here, it gets me right about there. And then when x is equal to
b, g of x is about that big, so I have to add that
length, which is about, looks something like that. That actually looks a little bit too much. Maybe something like that. So, if I were to add
the two, I'd get a curve that looks something like this. And maybe it just keeps on
going higher and higher. So, this is the curve. Or it's a pretty good
approximation of the curve of f of x plus g of x. Now an interesting question
is, is what would be well, you know how we
can represent this area. So the area under the
curve f of x plus g of x, above the positive x axis, between x equals a and x equals b, we know we can represent that as, let me see, I have not used pink, yet. So, this area right over here, we know that that could be represented as the definite integral, from a to b of f of x, plus g of x, dx. Now the question is, how does this thing relate to d, or how does this area relate to these areas right over here? Well the important thing to realize is this area that we have in yellow, that's going to be this
area, right over here. That one's pretty clear. But how does this area in green
relate to this area there? And to think about that, we just have to think about
well what does an integral mean? What does it represent? We've already thought about these like these really small rectangles. And we're taking the
sum of the limit of an infinite number of these
infinitely thin rectangles. But when we're thinking
about Riemann sums, we're thinking about okay
we have some change in x, and then you multiply it
times essentially the height, which is going to be the value of the function at that point. Well over here, you could
have the same change in x. You can have the exact same change in x. And what is the height right over here? Well, that's going to be this
exact height right over here. You saw that when we constructed it. This is going to be the
g of x at that value. So even though the
rectangles look like they're kinda shifted around a little bit, and they're actually all
shifted up by the f of x, the heights of these rectangles that I'm drawing right over here, are exactly the same thing as the heights of the rectangles that
I'm drawing over here. They, once again, they're
all just shifted up and down by this f of x function. But these are the exact same rectangles. And they have the exact same heights. And the limit as you get
more and more of these by making them thinner and thinner, is gonna be the same
as the limit as you get more and more of these as
you get thinner and thinner. And so this area right over here, and I'm obviously not
doing a rigorous proof, I'm giving you the intuition for it, is the exact same thing as
this area right over here. So the area under this curve, the definite integral from a
to b of f of x plus g of x, dx, is just going to be the sum of
these two definite integrals. And you might say, "Oh this is obvious". Or maybe it's not so obvious. But when is this actually useful? Well, as you later actually learn to evaluate these integrals, you'll see that one of
the most powerful ideas is being able to decompose
them in this way. To say, okay, if I'm taking
the definite integral from zero to one of x squared, plus sine of x, which you may or may not
have learned to do so far, you can at least start to break this down. You can say okay, well this
is going to be the same thing as the integral from
zero to one of x squared, dx, plus the integral from zero
to one of sine of x, dx. And you see this is one
of the most powerful principles of definite
integrals, to make them, when you start to try to compute them, or even sometimes conceptualize
what they're representing.