Essential Knowledge 3.2C1

Essential Knowledge 3.2C1 Students will know that in some cases, a definite integral can be evaluated by using geometry and the connection between the definite integral and area.

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We're going straight to this Essential Knowledge because I believe it will make the formulas much easier to remember. Here we see what an integral represents graphically, which is a bunch of rectangles. This will make it easier to imagine the rectangles created by integrals when you use the formulas.

First, let's start with what we know.

Example

If I'm going 20 mph, how far will I drive after 3 hours?

You could answer this with or without calculus. Obviously if each hour makes 20 miles, 20 mi/hr x 3 hr = 60mi. Now let's see what your calculation would look like on a graph.

Graphically

First, the graph of our velocity. Our velocity stayed 25 mi/hr throughout our journey, so it's a pretty simple graph.

Now let's think. We basically used the formula distance = velocity x time, or

D =v x t

This looks similar to a fundamental math equation: the area of a rectangle

A = b x h

And since we have a graph, we can actually make this rectangle. Since time is measured in the x-direction, 3 hours will be our base. Since velocity is measured in the y-direction, 20 mi/hr will be our height.

So you can see, the answer to this problem is actually the same thing as finding the area of the rectangle under the curve.

How does this relate to calculus? Well, we were given velocity and asked to find position. Remember, velocity is the derivative of position. And since integrals and derivatives are inverses, if velocity is the derivative of position, then position must be the integral of velocity. Take a moment to make sure this all makes sense.

OK, putting it all together, what we actually wanted to do was find an integral, and we were able to find it by finding the area of the rectangle under the curve.

You could do this for a million different types of integrals with a million different graphs, and it will always work. As you do more integrals in more situations, this will make more sense, but for now, you'll have to trust me on this critical rule:

The value of an integral can be found by finding the area under the curve

Notice that we only care about a specific interval of the curve. In our example, we cared about the interval from t = 0 hr to t = 3 hr.

Algebraically

We could also solve this using integral rules. We know position is the integral of velocity with respect to time, so let's integrate. (Note: I'm going to leave out a lot of details here. Just focus on the concept for now, we'll worry about the details later.)

We'd like to determine C so we can get an answer. So let's think of something to substitute. Well, we know when the car initially begins moving, it hasn't traveled any miles yet. In other words, when t = 0, D = 0. Substituting:

0 = 20mi/hr(0) + C

0 = 0 + C

C = 0

So our equation becomes

D = (20mi/hr)t

Substituting t = 3hr because that was what the question asked for:

D = (20mi/hr)(3hr) = 60 mi

And you see we get the same answer using a graph or using integration rules.

Example

Mr. Rose dropped a ball from 1000 ft above ground. We know the velocity of the ball is v = (-32ft/sec²)t. How far has the ball dropped after 2 seconds?

How many times will we use this example? No one knows, it just keeps being useful? Anyways, we know from seeing this example a million times that the answer needs to be 936 ft. Let's see how we can get it

Graphically

Let's graph the curve of velocity.

We know that the value of the integral is equal to the area made by the curve. Now we need to be precise. It's equal the area between the curve and the x-axis. Let's go ahead and find that.

This time the area makes a triangle. The base is 2 sec, since that's what the question asked for. The height is -64ft/sec because we used the equation given to us when t = 2 sec

v = (-32ft/sec²)t = (-32ft/sec²)(2sec) = -64 ft/sec

Solving the for the area of a triangle, we get

A = 1/2bh

A = 1/2(-64ft/sec)(2sec) = -64ft

So if the ball traveled -64 ft (in other words, 64 ft in the negative y-direction, aka 64 ft down) and the ball started 1000 ft in the air, well then, the position of the ball at 2 seconds would be

y = 1000ft - 64ft = 936ft

Exactly as we expected.

Using integral rules

= (-16ft/sec²)t² + C

Now we gotta solve for C. Well we know that at the time the ball is released, the height of the ball is 1000 ft. Substituting t = 0 sec y = 1000ft

y = (-16ft/sec²)t² + C

1000 ft = (-16ft/sec²)0² + C

C = 1000 ft

The equation becomes

y = (-16ft/sec²)t² + 1000 ft

Wow

This is the equation we had when we did this example the first time! Anyways, from here we just plug in t = 2sec and see what happens

y = (-16ft/sec²)(2sec)² + 1000 ft

y = (-16ft/sec²)4sec² + 1000 ft

y = -64ft + 1000 ft = 936ft

As you see, we get the same answer either way