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## Class 12 math (India)

### Course: Class 12 math (India) > Unit 10

Lesson 5: Definite integral properties- Integrating scaled version of function
- Integrating sums of functions
- Definite integral over a single point
- Definite integrals on adjacent intervals
- Definite integral of shifted function
- Switching bounds of definite integral
- Worked examples: Finding definite integrals using algebraic properties
- Using multiple properties of definite integrals
- Definite integral properties (no graph): function combination
- Worked examples: Definite integral properties 2
- Definite integral properties (no graph): breaking interval
- Warmup: Definite integral properties (no graph)
- Finding definite integrals using algebraic properties
- Examples leveraging integration properties
- Definite integrals properties review

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# Definite integral properties (no graph): breaking interval

Given the definite integral of f over two intervals, Sal finds the definite integral of f over another, related, interval.

## Video transcript

- [Voiceover] We're
given the definitive role from one to four of f
of x dx is equal to six. And the definitive role for one to seven of f of x dx is equal to 11. And we wanna figure out
the definite integral from four to seven of f of x dx. So at least in my brain,
I'm visualizing these as areas between the
curve y equals f of x, and the x axis. So let's just draw that,
we don't know exactly what f of x is, but we can
draw an arbitrary f of x. Just to help us visualize things. So let me draw, so that is,
draw it in a bolder color. So that's our y axis. And this is our x axis. And let's see, all the action is happening between x, from x, from one to seven. So we can go one, we can go one, two, three, four, five, six, seven, and we can even go to eight if we like. But the important numbers,
let's see we're dealing with one, two, three, four, five, six, seven, and then we go to eight. And let me just draw the
graph y equals f of x, and I'm just gonna draw
something arbitrary here. So let's say the graph of y
equals f of x looks like that. Y is equal to f of x. And of course I put it on my axis. That's the x axis, that is the y axis. And let's think about what each of these integrals represent. So the integral, the definite
integral from one to four, well that's going to be,
we're going to be going from one to four, right over here. So this is the definite
integral from one to four. This area under the
curve, between the curve of the x axis, dx, which is equal to six. And now let's see, we also have the region that goes from four to seven. We have this region right over here. And that area is represented
by this definite integral, the one that we need to figure out. The definitive role from four to seven of f of x dx, we need to figure that out. And they also, what else do we have? So let me underline this. So that's that, this is
the area of the region between x equals four and x equals seven under y equals f of x above the x axis. And then they also gave us
this last piece of information. Which is, let me do this in another color, the definitive role from one to seven. Well that's going from one. One all the way to seven. So that's the sum of these
two regions right over here. So we could re-write this as the definite integral from one to four of f of x dx plus the definite integral from four to seven of f of x dx is equal to, the definite integral from one to seven. One to seven of f of x dx. And notice what's going on here. The first one just goes to the area from one to four, then we
go from four to seven, so if you add those together,
that's gonna be the area from one to seven, from one to seven. And so they gave us a
lot of this information, they tell us that this
right over here is six, let me do that same color. They tell us that this
is six, they tell us that this is 11, and so we have six plus this is equal to 11. Well six plus one is equal to 11. Well this thing right over here must be equal to this thing right,
the definite integral from four to seven must be equal to five. This must be equal to five. Another way to think about
it, if this region right over, having trouble switching colors, if this region right over here is six, so that has an area of
six, and the whole region if everything has an area of 11, so if that plus that has an area of 11, well then the stuff that we don't know, this orange region, is going to be 11 minus six. So this region right over here is going to have an area of five.