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## AP®︎/College Calculus AB

### Unit 6: Lesson 1

Exploring accumulations of change

# Definite integrals intro

AP.CALC:
CHA‑4 (EU)
,
CHA‑4.A (LO)
,
CHA‑4.A.1 (EK)
,
CHA‑4.A.2 (EK)
,
CHA‑4.A.3 (EK)
,
CHA‑4.A.4 (EK)
Definite integrals represent the area under the curve of a function and above the 𝘹-axis. Learn about the notation we use to write them and see some introductory examples.

## Want to join the conversation?

• what is dx suppose to represent in the notation?
• Integration basically means summing up the area of infinitesimally thin rectangles under a function in a given interval [a, b]. The infinitesimally small width of the rectangles is denoted dx. Hope this helps!
• Is there a way can do this process without graph?
• At in the video, Sal describes the mathematical notation for the definite integral of f(x) evaluated from a to b. There is no need for a graph at all, simply find the antiderivative of f(x), then evaluate that antiderivative at b, and subtract the antiderivative evaluated at a.
• is this video supposed to be longer than ? it feels like he got cut off. I didnt really learn anything from this because it feels like he didnt finish. I also see other people in the comments saying this but with no response.
• Nope! This is how long this specific video is supposed to be. It continues later, but that's when you're supposed to learn more.
• Hey hey! So there's a second coordinate plane in this video, but no graph or function. I might realize there's a continuation of this video somewhere, but wanted to catch it early in case I forget. Is there supposed to be more to this?

Thanks a bunch!
• Hi! I am a bit confused with what dx means in the video ... does anyone know?

Thanks!
• dx,is always "a small change in x" it amounts to the limit of x as x goes to 0. or whatever is next tot he d
• what is the difference between exact area and definite integral?
• The definite integral gives you a SIGNED area, meaning that areas above the x-axis are positive and areas below the x-axis are negative. That is why if you integrate y=sin(x) from 0 to 2Pi, the answer is 0. The area from 0 to Pi is positive and the area from Pi to 2Pi is negative -- they cancel each other out.

If you want to use integration to find a geometric (or exact, as you called it) area, you have to make all of the negative areas positive. One way to do this is with absolute values or just ignoring negative areas. For example, integrating y=|sin(x)| is not fun, so you normally break that into two integrals (0 to Pi and Pi to 2Pi) and ignore negative signs in the answers.
• Where is the video for finding the exact area using Riemann Sums?
• (One little clarification: This is written assuming you already know what Riemann Sums are. If you don't, you can go check out the videos on that.)

It's nowhere, because there's no real way to do that. Riemann Sums are more of a way to get a sense of how much area is under the curve. We can talk about approximating with Riemann Sums, but actually evaluating them when the lines are infinitesimally small is a whole different story. There's a more precise (and most of the time easier) method called the Fundamental Theorem of Calculus that can tell you the exact answer. You can learn about that a bit later in this course. If using that doesn't work, we have computers that can approximate the area with Riemann Sums a lot more precisely than a person can, however, the answer will never be exact.

I'm a student, just like you, so I hope I know what I'm talking about, and that this response made any sense.
• Can you combine 25pi/2 -6?
• Using the least common denominator, 25π/2 - 12/2 = (25π-12)/2
• Hi! So in the notation for the definite interval, you wrote 'a' towards the bottom of the symbol and 'b' at the top. Is it convention to keep the smaller value at the bottom, or can it work the other way round as well i.e. keeping the 'a' at the top and 'b' at the bottom? Thanks. Your videos are really helpful :)
(1 vote)
• The value at the bottom is the number the integral starts at, and the value at the top is the integral ends at. Typically, the ending value will be larger than the starting value, so the value on top will be larger. However, you can have the integral start at a large value and end at a smaller value, so the small value would be on top. (Using the second part of the Fundamental Theorem of Calculus, we can show that reversing the direction in which an integral is evaluated multiplies the integral by -1.)