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Worked examples: Definite integral properties 2

Sal evaluates definite integrals of functions given their graphs. He does so using various properties of integrals.

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• This is weird without any function equations, almost too easy. Is this just to get us used to the notation?
• Yes, there is a huge missunderstanding for integrals UNDER the x-axis, and I believe many teachers want to eliminate this by explaining it in the beginning of the topic.
• For anyone struggling with the notion of a "negative area", just think of this:

Imagine the graph in the video above is representing the velocity of a vehicle over a certain time. (velocity = y axis, time = x axis). Let's also consider only the right half of the graph (ie. only positive values of X).

By looking at the graph, you can see that at x = 0 (ie. when time is equal to 0), our velocity is still peaking, but soon it starts going down and finally reaches zero when time is equal to 3. Then we get a "negative" velocity, which essentially means our vehicle is moving in the other direction (it's "coming back", so to say). It's velocity keeps increasing in the opposite direction until roughly time = 5, then it starts decreasing again until it finally reaches zero again at time = 7.

The area under the function is supposed to give us the distance traveled. On the interval of time from 0 to 3 we crossed a positive distance, but then on the interval 3 to 7 we crossed a "negative" distance, ie. we came back around a certain distance. So to get the total distance traveled (ie. the "total area under the curve" as we call it), you have to treat the area under the X axis as if it were negative so that the distance covered during this period of time can be subtracted from the distance covered over the first (ie. where x is between 0 and 3) period of time.

So again, if we were to treat this as the graph of the velocity of a vehicle over the course of 7 hourse (x from 0 to 7), what the graph is essentially telling us is:

"From hour (0) to hour (3) the vehicle traveled a total of 7 miles (7 coming from the area under the curve over the interval (0,3)). Then the vehicle turned around and traveled back towards its initial starting point, covering 3 miles (3 coming from the area under the cuver over the interval (3, 7))"

So the car went forward 7 miles, then traveled back 3 miles. How many miles did it actually cross?

7 - 3 = 4.

(In the video we are of course scaling these values by a factor of 2 and 3, respectively). This is just an example to help you get a grasp on what is actually happening here in terms of "real world applications" and why it is structured in the way that it is.