can someone explain integrals?
Sort:
RealTactics960
7 days ago
0
#1
Idk how they work or what they are
Bro why u asking this on july
7 days ago
0
#6
Imagine you have a bunch of tiny blocks.
When you put all the blocks together, you can see how big something is.
An integral is just a way of adding up lots of tiny little pieces to find the whole thing.
Example:
• Imagine you have a wiggly line on a graph (like a hill).
• Under that line, there is space.
• An integral tells you how much space is under the line.
So think of an integral like:
🧱🧱🧱🧱🧱
Adding up lots and lots of tiny blocks to get the total area!
💡 In short:
Integral = Adding up all the little bits to find the whole thing.
If you want, I can give you a picture example or an even simpler story!
When you put all the blocks together, you can see how big something is.
An integral is just a way of adding up lots of tiny little pieces to find the whole thing.
Example:
• Imagine you have a wiggly line on a graph (like a hill).
• Under that line, there is space.
• An integral tells you how much space is under the line.
So think of an integral like:
🧱🧱🧱🧱🧱
Adding up lots and lots of tiny blocks to get the total area!
💡 In short:
Integral = Adding up all the little bits to find the whole thing.
If you want, I can give you a picture example or an even simpler story!
7 days ago
0
#7
Integrals are a concept in Calculus. They are represented by the symbol ∫. They are the inverse of derivatives, so if you don’t know derivatives, you might want to learn those first.
Essentially, integrals find the area under a curve. You can set upper and lower bounds if you want, creating what’s called a Definite Integral. But you don’t have to set bounds if you don’t want to. If you don’t set any bounds, it is an Indefinite Integral.
These integrals, as I stated before, find the net area under functions, but first, the basic concept of an integral is that it is able to reverse a derivative. This is why it might help to learn derivatives.
For example, let’s look at an indefinite integral on the basic function f(x) = 2, and try to find ∫f(x) dx. (We put the dx to the end of the integral.) The function f(x) = 2 isn’t a particularly complex function, so we’ll use it, just to get a baseline of what to expect.
To find the integral of f(x) when f(x) = 2, you must work backwards and figure out which function has a derivative of 2. We can see that the derivative of 2x is 2, meaning that the integral of 2 starts with 2x, since an integral and derivative cancel each other out, like addition and subtraction do. But you aren’t done yet, because you need to add +C to the end of the solution, making our final answer ∫f(x) dx = 2x + C. We add the +C because there are actually infinitely many solutions to the integral. All we need to do is to add a constant of our choice, and we have infinitely many functions whose derivatives are 2. So, we can easily cover them all by adding +C to the end, defining C as any real number. In a math class, you will be marked wrong if you forget the +C, but you likely won’t have the explain what the +C actually means, as the instructor will likely know what you mean.
Indefinite Integrals are trickier, but it’s the same thing with extra steps added to the end. Indefinite integrals are useful when trying to find the net area under the graph, by taking the area above y = 0 and subtracting the area under y = 0. Doing so will give the indefinite integral of that function, but in order to solve definite integrals, you’ll need to set some bounds.
To start, let’s find the integral of j(x) = 2, from the bounds of 0 to 1. ∫₀¹ j(2x) dx. This is a simple linear function, and by looking at the function and restricting it to be between x = 0 and x = 1, you are able to easily see that it is just a triangle, the area of which is 1, which is our answer. But let’s pretend we don’t know that for a second, because for most problems, we won’t.
First, we will find the indefinite integral of the function using the method above. The indefinite integral of 2x is x², so take that function, and we’ll give it a temporary name h(x) = x²· Plug in the values of the bounds, and subtract h(upper bound) - h(lower bound). If we remember, our bounds were 0 and 1. 1 is the upper bound, and 0 is the lower bound. So, subtract h(1) - h(0). Which is just 1² - 0², which is just 1 - 0, which is 1.
Check this! The area of a triangle is 1/2 × b × h, so setting our height to 2 and base to 1, we get 1/2 × 2 × 1, which is 1. It checks out!
I’ll admit that I don’t know much about Calculus, but I hope this helps!
For a differently worded answer, check out this website! (I just realized that I accidentally used the same examples, haha.)
https://www.mathsisfun.com/calculus/integration-introduction.html
