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## Calculus 2

### Unit 1: Lesson 8

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
- Functions defined by integrals: challenge problem
- Definite integrals properties review

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# Definite integrals on adjacent intervals

AP.CALC:

FUN‑6 (EU)

, FUN‑6.A (LO)

, FUN‑6.A.1 (EK)

, FUN‑6.A.2 (EK)

By subdividing the stretch of numbers where you are integrating, you can break up an integral.

## Want to join the conversation?

- why does c have to be in between a and b

if c was greater than b, the integral from c to b would be negative, and thus the integral from a to b would be equal to the integral from a to c minus the integral from b to c(18 votes)- In later problems you will see that instead of "a" and "b" you might have "a" = to "x^2" and "b" = to the "square root of x". This makes it necessary to split it up with "c" between "a" and "b" so that the integral from "x^2"(a) to "root x" (b) is equal to the integral of "c" to "root x"(b) minus the integral "c" to "x^2"(a). This makes it possible to solve the problem.(7 votes)

- I don't understand he says to use this property to solve when there's a discontinuity in a function, did we not say first thing in fundamental theorem of calculus that f has to be continuous between a and b?(6 votes)
- You can separate the integral into two parts. So integrate from a to b, for let say for f(x). Then integrate from b to c for g(x)(0 votes)

- point c is calculated twice, so how will this affect (or not affect) the final outcome?(2 votes)
- The set of points that lie exactly above c form a line segment, which has area 0. So the outcome is not affected.(5 votes)

- Is there a more formal name for this property?(2 votes)
- This reminds me of countable set,which has infinite number,but can match to the nonnegative integers set.The integral of a,b is equal to the limit of its Riemann sum,which is constructed of infinite small rectangles.If divide the integral into parts,each part is also constructed of infinite small rectangles.Is there any connection between the number set and the integral?(1 vote)
- I can see two connections between the set {x | x €
**I**}(2 votes)

## Video transcript

- So we've depicted here the area under the curve F of X above the X-axis, between the points X
equals A and X equals B. And we've denote it as the definite integral from A to B of F of X, DX. Now what I wanna do with
this video is introduce a third value, C, that
is in between A and B. And it could be equal to A
or it could be equal to B. So let me just introduce
it, right, just like that. And I could write that A
is less than or equal to C, which is less than or equal to B. And what I wanna think
about is, how does this definite integral relate to
the definite integral from A to C and the definite
integral from C to B. So let's think through that. So we have the definite
integral from A to C of F of X. Actually I've already used
that purple color for the function itself, so we use green. So we have the integral
from A to C of F of X, DX, and that of course is
going to, that's going to represent this area right
over here, from A to C under the curve F of X, above
the X-axis, so that's that. And then we can have the integral from C to B of F of X, DX,
and that of course is going to represent this
area right over here. Well the one thing that
probably jumps out at you is that the entire area from
A to B, this entire area is just a sum of these two smaller areas. So this is just equal to
that plus that over there. And once again, you might say, "Why is this integration property useful?" That if I found a C that is in
this interval that's greater than or equal to A and it's
less than or equal to B, "Why "is it useful to be able to
break up the integral this way?" Well as you'll see this is
really useful, it can be very useful when you're
doing disc, when you're looking at functions that
have discontinuities, if they have step
functions you can break up the larger integral
into smaller integrals. You'll also see that this
is useful when we prove the fundamental theorem of Calculus. So in general, this is actually a very, very, very useful technique. Let me actually draw an
integral where it might be very useful to utilize that property. So if this is A, this is B,
and let's say the function, I'm just gonna make it
constant over an interval. So it's constant from there
to there, and then it's, and then it drops down
from there to there. Let's say the function looked like this. Well you could say that the
larger integral, which would be the area under the curve,
it would be all of this. Let's just say it's a gap right there or it jumps down there. So this entire area you
can break up into two, you can break up into two smaller areas. So you could break that up into that area right over there, and then this area right over here using this integral property.