The Fundamental Theorem of Calculus (Part III)

The Fundamental Theorem of Calculus (Part III)

Let's examine another example. We want to find the area of a certain region. The function is y equals cosine x. The region from zero to pi.

When we integrate the function from zero to pi, we get a total signed area. That means areas are given a positive or negative sign.

According to Part Two of the Fundamental Theorem of Calculus, the integral of cosine X from zero to pi is equal to an anti-derivative of the function.

For example, here the sine of X equals zero, and you evaluate at two limits: pi and zero. Subtracting the sine of zero from zero yields zero; thus, the answer is zero.

The fact that the total area of the region bounded by the function y = cos x is zero can be seen from the vector diagram, because this part is given the positive sine and this part is given the negative sine.

Thus, these two functions cancel each other out.

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If we want to find the actual area of the region, we have to take this sine into account.

So we split the region into two pieces: one from 0 to π/2 and another from π/2 to π. The first gives us the area of this region but it is negative.

To find the real area, we need to compensate for the negative sign by adding 1 here and therefore the first integral equals this second one.

And because you have a minus sign here, you actually compute this to be two, which is exactly the area of these two regions together.

In this example, we find the area of a region bounded by a graph. We sketch the graph and identify the regions where it is positive piece or negative piece, then compensate for the sign in each of these cases as in part two of this example.

In another example, the region is now defined by two functions instead of a function and the x-axis.

Suppose we have two continuous functions, f(x) and g(x), defined over a closed interval [a, b]. Suppose that f(x) is always greater than or equal to g(x) over the region A, B. They therefore enclose a certain region here.

The area of a region enclosed between two curves can be found by evaluating the definite integral of the function f(x) - g(x) from A to B. This is defined as the area of the region between f(x) and g(x).

Example.

We have two functions, y = x and y = x2 - 2. The first function is a straight blue line; the second function is a parabola, red curve. We want to find the area of the region in between these two functions.

Firstly, we need to determine the points of intersection of these two curves.

Solving simultaneously the two equations above will yield x to be equal to x squared minus two. This is a very simple quadratic equation and its answer is x equals two and minus one.

Thus, the two intersection points are (-1, -1) and (2, 2). Therefore, the area of the region is given by this definite integral.

Integrate from this point minus one, to this point two here. The integrand is x minus x squared minus two.

This is a very simple integral, and we can find its anti-derivative by subtracting x cubed over three from here, plus two x from here, and then subtracting the value of the function at the upper limit minus the value at the lower limit. The final answer is 9/2.