I know it’s the “area under the curve”, but what does that mean exactly? Is there a physical or tangible way to explain it?
I understand that a derivative is rate of change at a specific point, and something like acceleration is rate of change of speed. But how can I visualize that speed is the “integral” of acceleration? What does that mean, and how does it relate to the area underneath?
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If area under the curve doesn’t explain it to you then just think of it as the total amount. For instance if you have a hose spraying water into a bucket then the integral of the amount of water being sprayed is the total amount in the bucket.
Think about what speed does to position. After a t amount of time with a constant velocity v, your position will have changed by v*t. If you picture them on a graph, where time is the horizontal axis and velocity is the vertical axis, there’s a horizontal line. The area under that line, a rectangle, is given by one side times the other, so v*t. As you can see, that area is equivalent to your change in position, or total distance travelled. It does not say anything about your current position, though, unless you know what it was originally, which is why the general solution for an integral always has that +c. In this case, c is your original position.
You can then also use this definition when the velocity isn’t constant and is instead some line whose integral, the area under it, represents the change in position for the given interval. You can extrapolate this to the other derivatives of position, too. The integral of acceleration will be the total change in speed.
Just flip your phrasing around. A derivative is the rate of change if a function. An integral of that derivative is just the current value of that function. Speed is the derivative of distance traveled. If you have a function of speed, the integral of the speed function is your total distance.
If you plot the function, it’s the area between the line and the x axis. If you plot a straight line and find the integral, you will get the area of the triangle formed by the line, the x axis and a vertical line at the end. Look up a reimman sum, it might help you visualize it.
If that doesn’t help, it’s how much you’re accumulating. If your function is velocity, he integral is how far you’ve gone.
It’s an accumulation.
The more acceleration you add, and the longer you add it, the more speed builds up. Similarly the more speed you add for more time, the more distance builds up.
The “area under the curve” is really missing the point. Yes, the operations will give the same result but it’s not what you’re working out.
Think of it kinda like taxes. Throughout the year you make money. the derivative would be your paycheck; 500/2 weeks. The integral would be the annual gross income. 12,000. The sum of every paycheck throughout the year.
I think it’s easier to start with position/velocity.
Say you have a car driving straight down a road at a constant velocity, let’s say 10m/s.
The derivative of our position function tells us how fast our position is changing, aka our speed. And in this case where our speed is constant that would just be a straight line at 10.
If our velocity function is a straight line, the integral is the area under that line and would tell us the distance we’ve traveled at any point in time. And that’s pretty easy to visualize since it’s just a rectangle.
Our width is 10, and at 1 second our length is 1. So the area under our line at 1 second is 10. In other words, we’ve traveled 10 meters.
And then again, at 2 seconds the area under the line is 10×2=20 meters. At 3 seconds it’s 10×3=30, etc. And that’s just a position function: Pos = 10*t
And that should make some sense intuitively too as your functions/lines get more complicated. If you draw a line showing the rate something is changing, when that line goes down the area under that line increases slower. If that line goes up, the area increases faster.
I’ve got an analogy that covers both derivatives and integrals!
Your function is the speed of your car at any given time. Plugging the time in the functions gives you the speed.
Derivative is your acceleration. Integral is the distance driven up until that point in time.
I find velocity to be an easy example. The derivative of a velocity function is the acceleration function, the change in velocity. The integral of velocity is the position function, your current position over time.
One way to look at an integral is the definite integral. As opposed to the indefinite integral, which is just a function, a definite integral includes start and end points on the x axis (it can get more complicated but screw that). Solving a definite integral is what people refer to as “the area under the curve”, that is the area between your original function line and the x axis. For velocity, a definite integral of velocity is the total position change, or distance covered, between two times. For example, driving for an hour at 50mph constantly, you could take the definite integral of 50mph over the hour to get how much distance you covered, which happens to be 50 miles. Finding the “area under the curve” is easy for linear functions, as setting a start and end x value makes a polygon on the graph that you can just use geometry on to get the area, but you need the calculus magic to do this for more curvy functions like x squared.
Another way to look at a velocity integral is the current position at the given time. If you pick an x value on the velocity function, you can plug that x into the integral function to get the position at that time. Remember that constants are lost when differentiating, so the integral of a function is always some function “plus C”, the unknown constant. You can never know for sure what the exact position of something is if you only have the velocity function, but it will always differ from the true value by a constant “C” value. For velocity, the C value in the integral could be seen as your starting point. Maybe you’re driving across the country, but you start 10 miles in from one of the borders, so C may be “+10”.
Here’s one for you: I had to estimate the amount of gravel to fill a ditch. The ditch varied in depth from one end to the other (increasing, though not perfectly regularly). I measured the depth of the ditch at regular intervals and interpolated a function from those points. The integral of that function was the amount of gravel I would need.
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