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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 6

Lesson 6: Applying properties of definite integrals- Negative definite integrals
- Finding definite integrals using area formulas
- Finding definite integrals using area formulas
- Definite integral over a single point
- Integrating scaled version of function
- Switching bounds of definite integral
- Integrating sums of functions
- Worked examples: Finding definite integrals using algebraic properties
- Finding definite integrals using algebraic properties
- Definite integrals on adjacent intervals
- Worked example: Breaking up the integral's interval
- Worked example: Merging definite integrals over adjacent intervals
- Definite integrals over adjacent intervals
- Functions defined by integrals: switched interval
- Finding derivative with fundamental theorem of calculus: x is on lower bound
- Finding derivative with fundamental theorem of calculus: x is on both bounds
- Definite integrals properties review

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# Worked example: Merging definite integrals over adjacent intervals

Thinking about how to evaluate the sum of definite integrals over adjacent intervals.

## Want to join the conversation?

- Since the area below the graph is negative, why are the 7 and 6 positive on the graph?(3 votes)
- It's an assumption you can make. Integrals give the signed area, while it looks like the graph is giving the unsigned area.(5 votes)

- Why aren't we adding square units to the answers since we are calculating areas?(1 vote)
- It's because the area under a graph can mean different things. For example, the area under a speed-time graph gives the distance (which definitely isn't measured in square units). Hence, we leave integrals unitless until they have some physical significance.(2 votes)

- for the original function of the first question, -2 appears as the upper bound of the first part of the function and then appears as the lower bound of the second part of the function, so how will the fact that -2 appeared twice affect (or not affect) the result?(1 vote)
- This will not affect the result. If the upper bound of one definite integral is the same as the lower bound of another, we can simply consolidate them into one integral like Sal did. If we eyeball the graph, it looks like the area from -4 to -2 is about -3.5, and it looks the same for the area from -2 to 0. We can add these (-3.5 + (-3.5)), to get -7. Or we could have initially realized that we can express this as the definite integral from -4 to 0, which is still -7.(1 vote)

## Video transcript

- [Instructor] What we have here is a graph of y is equal to f of x, and these numbers are the
areas of these shaded regions, these regions between
our curve and the x-axis. What we're going to do in
this video is do some examples of evaluating definite
integrals using this information and some knowledge of
definite integral properties. So let's start with an example. Let's say we want to evaluate
the definite integral going from negative four
to negative two of f of x d of x plus the definite
integral going from negative two to zero of f of x dx. Pause this video, and see if you can evaluate this entire expression. So this first part of our expression, the definite integral from negative four to negative two of f of x dx, we're going from x equals negative four to x equals negative two. And so this would evaluate as this area between our curve and our x-axis, but it would be the negative of that area because our curve is below the x-axis. And we could try to estimate it based on the information they've given us, but they haven't given
us exactly that value. But we also need to figure
out this right over here. And here we're going from x
equals negative two to zero of f of x d of x, so that's
going to be this area. So if you're looking at the sum of these two definite integrals, and
notice the upper bound here is the lower bound here,
you're really thinking about, this is really going to
be the same thing as, this is equal to the definite integral going from x equals negative four all the way to x equals zero of f of x dx. And this is indeed one of
our integration properties. If our upper bound here is the
same as our lower bound here and we are integrating the same thing, well, then you can merge these two definite integrals in this way. And this is just going
to be this entire area. But because we are below the x-axis and above our curve here, it would be the negative of that area. So this is going to be
equal to negative seven. Let's do another example. Let's say someone were to ask you, walk up you on the street and say, quick, here's a graph, what is
the value of the expression that I'm about to write down, the definite integral going
from zero to four of f of x dx plus the definite integral
going from four to six of f of x dx? Pause this video, and see
if you can figure that out. Well, once again, this
first part right over here, going from zero to four, so what would be is it would be this area. It would be this five right over here, but then we would need
to subtract this area because this area is below our
x-axis and above our curve. We don't know exactly what this is. But luckily we also need, we need to take the sum of
everything I just showed, but plus, plus this right over here. And this, we're going from four to six, so it's going to be this area. Well, once again, when
you look at it this way, you can see that this expression is going to be equivalent to
taking the definite integral all the way from zero to six, zero to six of f of x dx. And once again, even if
you didn't see the graph, you would know that
because, in both cases, you're getting the definite
integral of f of x dx. And our upper bound here
is our lower bound here. So once again, we're able
to merge the integrals. And what is this going to be equal to? Well, we have this area
here, which is five. And then we have this area, which is six. That was given to us. But since it's below the
x-axis and above our curve, when we evaluate it as
a definite integral, it would evaluate as a negative six. So this is going to be
five plus negative six, which is equal to negative one. And we're done.