The Integral Notation

The integral notation was chosen very carefully to get an important idea across:

We are summing up lots of tiny rectangles. Imagine a tiny rectangle below the derivative to visualise this:

From this, we can see that the area of a single rectangle is just:

We know that , so:

So the area represents a tiny change in the original function . Now if we sum up lots of small areas, we get a big area (A):

Which means that a big area equals a big change in y as well:

We can be more precise about this using the ideas by Riemann.

Riemann sums

This section can be safely skipped, but it explains the ideas more completely.

The Riemann sum uses rectangles to approximate the area under the curve. It gives us a precise definition of the integral notation, but it is also a little intimidating (but we explain it all below):

It tells us that the integral (the left hand side) is:

• By definition the same thing as. We wrote this using: (
  

• The limit as we make take a huge number of intervals. This is written with the notation:
  

• Of the sum sum of each. Written as:
  

• Little rectangle with an area of
  

  • So the height is
    

  • The width is
    

• As , the number of rectangles will closely approximate the area under the curve.

This can be seen in the picture below. As we increase the number of rectangles, the area underneath the graph is much more closely approximated.

Finding the change in a function

If we differentiate , then. To find the change in , we would integrate the derivative between x = 1 and x = 3

The integral equals the area of the derivative between to as shown in the graph:

This is is simply the area of the trapezium:

Therefore, by definition, the change in the original function from to is . Now it’s easy to find the area below a trapezium, but if the shape was more complicated, we’d need a better technique.

Verifying the Result

We can then verify this by finding the values of directly from the graph of as well:

Now , which is 14 and which is 6:

Notice that the value we found with the integral perfectly matches the change in the original function.

Note on Constants

The original function was and the constant term was 5. Imagine instead, we used ;

Notice that the completely disappears. This is because the integral only tells us the change in y between two y values, not the y values themselves. This also means we can not find the constant value of the original function, unless we are provided with additional information.

Using the Theorem

is graphed below.

Evaluate .

Solution:

We are asked to find the area under the curve from to .

Visually, we can see that all we need to do is find the area of the semicircle and the area of the triangle, then add them together.

So using the theorem,

The semicircle has a radius of 2, and hence its area is:

The triangle has base 3 and height 6, hence:

Therefore;

Why this is true (visual)

The left hand side is the integral from a to b, and the right hand side is the sum of the integral from a to c and the integral from c to b.

To put it simply, the green area plus the yellow area is equal to the integral from x=a to x=b.

Evaluating the both sides by using the definition of an integral, we get:

Using the Theorem:

Supposing the theorem is true, let’s try integrating:

The theorem says that we can swap the bounds and pick up a negative sign:

This makes it much easier to find the area, since we’re integrating a positive area with the upper bound bigger than the lower bound.

Why this is true

We can easily see that if we compare :

Integrating the green area from 4 to 2 is the same as integrating the red area from 2 to 4.

Swapping the bounds of an integral

By definition:

Therefore:

Using the theorem

Evaluate

Since is odd, and the bounds are -12 and 12,

Why this is true

Examine the following picture of a randomly chosen odd function:

Since the function has rotational symmetry, and the areas labelled 1,2 and 3 from to cancel out the areas labelled 1’,2’ and 3’: we can see this is true because:

• Consider
  

from to , this is explai

Using the theorem

Suppose we wanted to evaluate , where is graphed below;

Step 1: It is clear that the function is symmetrical about the y axis, therefore the area in the first quadrant is equal to the area in the second quadrant. As a result, we can find the area in the first quadrant, then double it;

Step 2: The integral from x=0 to x=7 can be split up into two trapeziums and a rectangle, like so:

Step 3: Area of smaller trapezium (orange):

Step 4: Area of larger trapezium (red):

Step 5: Area of rectangle (purple):

Step 6: Adding up the three regions, we get:

Why this is true

It is clear that all the shaded regions where x>0 are exactly equal to the regions where x<0, given that the bounds are x=-a and x=a.

Therefore;

Using the theorem

**

Why this is true

This is because translating any 2d shape to the left or right does not change its area.

Furthermore, if we flip an integral about the y axis, the area of the integral doesn’t change (this is simply because flipping a 2d shape does not change its area).

We can see where this comes from using a simple geometric proof:

The two green areas are clearly exactly equal to each other, as long as we integrate from right to left.