Exploring Areas
Now that we have a simple understanding of the second part of the FTC, it's time to see it in action!
We know that by using the second formula, we can find the specific area from some a to x where x is between a and b. This means that given a time interval that the ladybug travels, we can break up the distance covered and look at specific time intervals more closely.
Let's see some examples using our Nspire calculator. If you have a TI-83 or equivalent this will also work. Just graph the function or plug in the function when taking the integral and derivative.
To make our example more interesting, let's let f(x) = x^2. This means that the ladybug's velocity is now x^2.
Again, go to the home screen and add a graph page. Then let f1(x) = x^2
Then, find the area between x = 0 and x = 10.
What do you get as the area? If you calculated this by hand do you reach the same result? (Round down)
We know that by using the second formula, we can find the specific area from some a to x where x is between a and b. This means that given a time interval that the ladybug travels, we can break up the distance covered and look at specific time intervals more closely.
Let's see some examples using our Nspire calculator. If you have a TI-83 or equivalent this will also work. Just graph the function or plug in the function when taking the integral and derivative.
To make our example more interesting, let's let f(x) = x^2. This means that the ladybug's velocity is now x^2.
Again, go to the home screen and add a graph page. Then let f1(x) = x^2
Then, find the area between x = 0 and x = 10.
What do you get as the area? If you calculated this by hand do you reach the same result? (Round down)
Now, what would we do if we want to find the distance covered between a = 0 seconds and x = 2 seconds?
To do this, add a second function and type in the following integral into your calculator. Now, we want the integral from 0 and 2 of x^2 in terms of x. The following graph is below:
To do this, add a second function and type in the following integral into your calculator. Now, we want the integral from 0 and 2 of x^2 in terms of x. The following graph is below:
Let's stop here and analyze the split-screen table. When we look at the first column for f(x), we see the y-values for each x-value. This makes sense. Now, let's look at f2(x). When we look at when x=10 , we see 333.333. This is what we got from our calculations, so we know we're correct!
Now, calculate the integral from 0 to 2. What do you get? Is this similar to what's in the table?
Now, calculate the integral from 2 to 10. Again, what is your result and is this the same as what is in the table?
Now, what happens when you add the two areas from the two separate integrals we just calculated? Compare this sum to what the answer is when x = 10.
Are they different or similar? Why do you think this is?
Summarizing our discovery here, we can safely say that using the FTC part 1 tells us the total distance a ladybug has crawled given a time interval and the velocity of the ladybug's movement.
However, we can also break up the integral into two or more separate integrals as long as we choose our beginning and end points correctly. If we split up the distance covered with two integrals as we did above, we do this by taking the integral of the velocity function from a=0 seconds to some x seconds where a < x < b. Then we compute a second integral from x to b of the same function as long as the ladybug is still traveling at that velocity. When we add the two values together, this will equal the same value if we found the distance covered with one integral. Thus, we can split up the areas under a curve and add them together to reach the same value.
How does this help us? This allows us to look at different areas under the curve from different time intervals between [a, b]. Thus, we can see how far the ladybug has traveled between a = 0 seconds and x = 2 seconds for example. Or we can see the distance covered between 3 seconds and 7 seconds. Accumulating the area under the curve at any time intervals we please gives us a lot of freedom!
However, we can also break up the integral into two or more separate integrals as long as we choose our beginning and end points correctly. If we split up the distance covered with two integrals as we did above, we do this by taking the integral of the velocity function from a=0 seconds to some x seconds where a < x < b. Then we compute a second integral from x to b of the same function as long as the ladybug is still traveling at that velocity. When we add the two values together, this will equal the same value if we found the distance covered with one integral. Thus, we can split up the areas under a curve and add them together to reach the same value.
How does this help us? This allows us to look at different areas under the curve from different time intervals between [a, b]. Thus, we can see how far the ladybug has traveled between a = 0 seconds and x = 2 seconds for example. Or we can see the distance covered between 3 seconds and 7 seconds. Accumulating the area under the curve at any time intervals we please gives us a lot of freedom!
Now, let's see what happens when we take the derivative of the integral. Similarly in the activity we just did, we see that taking the derivative of the integral gives us what? Think about this again and relate this to finding the ladybug's position from the origin of a graph.
We know that the derivative and the integral are inverse operations. How does knowing this help us?
We know that the derivative and the integral are inverse operations. How does knowing this help us?
By taking the derivative of the integral, we are left with the f(t) function. From there, we can simply substitute in our x value: f(x). Seems like magic doesn't it? But if you don't remember what we talked about with inverses, visit that page again here!
The derivative and integral "undo" each other, and thus, we are left with just f(t) with x as our value to substitute in. So finding the distance covered by the ladybug was never that difficult to begin with!
What's cool about FTC part 2 is that we can use any x value that we want. From the exercises I've been providing, I've said a < x < b. However, b could be 1,000 seconds instead of 10 seconds. So if that was the case, we can choose any x value between 0 and 1,000. If I didn't put a limit on what b had to be, then x can be any x where the limit exists along f(t). Thus, we can find the distance traveled by the ladybug from any time period, and we can also split up time periods as we'd like.
You're almost finished! Click here to see the conclusion page.
The derivative and integral "undo" each other, and thus, we are left with just f(t) with x as our value to substitute in. So finding the distance covered by the ladybug was never that difficult to begin with!
What's cool about FTC part 2 is that we can use any x value that we want. From the exercises I've been providing, I've said a < x < b. However, b could be 1,000 seconds instead of 10 seconds. So if that was the case, we can choose any x value between 0 and 1,000. If I didn't put a limit on what b had to be, then x can be any x where the limit exists along f(t). Thus, we can find the distance traveled by the ladybug from any time period, and we can also split up time periods as we'd like.
You're almost finished! Click here to see the conclusion page.