Integration. The Concept of Area (Part II)

Integration. The Concept of Area (Part II)

If we go further, we can find the area of a region described by the formula y equals b times the square root of x minus a squared.

When you take the positive sign, it describes the portion of this ellipse above the x-axis. When you take the negative sign, it corresponds to the lower half of this ellipse. We would like to find the area defined by a curve, which is shown in blue here, where this curve has y = f(x).

Thus, to find the area of this region between x = a and x = b, let us look at an example of this shape.

Now imagine we are looking at the curve shown in red, which has the equation y = x2. And suppose we want to find the area of the region under this function from x = 0 to x = 1, also above the x-axis.

How to find the area of this region here?

In this case, let us proceed as follows.

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We will cut the x-axis into thin rectangles of width 1/n and height 1/n so that we have n such rectangles. The first rectangle lies between 0 and 1/n, the second lies between 1/n and 2/n, and so on. The last rectangle contains fractions from n-1/n to 1/1 or in other words from 0 to 1.

As before, we tried to estimate the area under the curves by adding up rectangles. This time, however, we divide each rectangle into n equal parts (where n is the number of rectangles).

So the first piece is here, the width is 1/n and the height is zero, so this gives us the first term here. The width of the rectangle is 1/n and the height is 0/n square.

Now, consider the second piece. The second piece also has the same width of 1/n, and its height is given by (1/n)^2. This square comes from the equation of the function.

The third piece, again the width is 1/n and the height is (2/n)^2. This is the equation here: so this is the area of the last piece here, the width, 1/n, and the height is (n-1 /n)^2. We call this lower sum because it is below the actual area defined by this red curve.

We can simplify the above sum by noting that it is a summation of terms j, where j goes from zero to n squared minus one. The first term is j = 0, the second term is j = 1, and the last term is j = n - 1.

We can simplify the expression (1+x)n to one over n cube, take out this n and we get (1+x)n. Then sigma j square, where j goes from zero to n minus one. It's obvious that the summation of j square from zero to n minus one is equal to one over six times n times two n minus one.

So, this is for the lower sum. As we did with the circular disc, we also want to estimate the exact area by upper sums--circumscribe the region by rectangles and in this case, the first piece of the rectangle is this one, its width is again 1/n and its height is 1/n2.

The second piece is this one, and the height is two over n times the square root of three. This is the third piece, and this is the last piece.

In summary, similar to the above, we find that the total area of this region circumscribing the region we want to find this area is equal to this.

We once again make use of the well-known formula for summation of j squared. This time from j equals one to j equals n, so this is the value of the summation.

Therefore, the upper sum is called the lower sum and the exact area of the region we want to find lies between these two sums.

So, as in the case of the circular disc, we want to look at the lower sum and upper sum when n tends to infinity because we can see that these two sums will come closer and closer to the exact area as n gets larger. Therefore, we look at the lower sum and find that this expression tends to 1/3 as n tends to infinity.

In the same way that we consider the upper sum here, as n tends to infinity, it also tends to 1/3.

Therefore, the area of the region we are trying to find must lie between 1/3 and 1/3 of the upper sum. Thus, the area of this region is now defined to be 1/3.