AP®︎/College Calculus AB
- 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
By subdividing the stretch of numbers where you are integrating, you can break up an integral.
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- 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(20 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?(7 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?(3 votes)
- The set of points that lie exactly above c form a line segment, which has area 0. So the outcome is not affected.(7 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)
- Does this property require all values of f(x) on range [a,b] are above or below the x-axis?(1 vote)
- There's no rule like that. If you get your bounds right, the integral takes care of the positive and negative areas.
And to add on, incase you want the total area (irrespective of sign), you can integrate over the modulus of the function.(1 vote)
- 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.