The Fundamental Theorem of Calculus (Part II)
Before we prove the Fundamental Theorem of Calculus, we need to define the Mean Value Theorem for Definite Integrals. This theorem states that given a continuous function f(x) defined over a closed interval [a,b] there exists a point c in the interval [a,b] such that f(c) equals one over b minus a times the definite interval.
The right-hand side is the average value of the function f. The Mean Value Theorem tells us how to see this.
Now, the average value of the function must lie between the maximum value of f and the minimum value of f. This follows from the max-min inequality we saw earlier. Since f is continuous on [a,b], this inequality applies to every value in that interval.
Thus, by the Intermediate Value Theorem, the average value lying between the minimum and maximum must equal a certain point c in the interval [a,b]. This gives us the Mean Value Theorem for a definite integral.
We will now prove part one of the Fundamental Theorem, which concerns the derivative of the capital F(x) function.
According to the definition of the derivative, the derivative of capital F(x) is equal to the differentiation d of the definite integral from a to x with respect to x. So again, according to definition, the derivative is equal to the ratio of F(x+h)-F(x), divided by h and consider the limit h tends to zero.
Now, according to the definition, capital F(x+h) is this area. Capital F(x) is this area. We subtract them and get the definite integral here. And we divide by h, considering that h tends to zero.
Now the Mean Value Theorem for Definite Integral, which we saw earlier, implies that this expression is equal to the value of the function at a certain point (c_x,h),
where (c_x,h) lies between x and x+h.
As h tends to zero, this function must tend to x, because it lies between x and x+h. Therefore, when h tends to zero, this value tends to f(x) because f is continuous. Now suppose both G(x) and F(x) are antiderivatives of the function f. We know previously that the difference of F(x) and G(x) is a constant, say C.
So in particular, the constant C has to be equal F(a) minus G(a).
If F(a) is equal to zero, then C equals -G(a). Therefore, F(x) = G(x) – G(a).
Saying differently, the definite integral of f from a to b is equal to F(b) minus G(a).
This is part two of the Fundamental Theorem of Calculus. We will see some examples of applications of the Fundamental Theorem of Calculus. We want to find dy/dx for each of the following functions of y in terms of x.
According to part one of the Fundamental Theorem, dy/dx is equal to f(x). The integrand f is t squared cosine t, so you replace t with x. The answer is x squared times cosine x.
In part b here, the variable x is in the lower limit, so we first of all switch it to the upper limit.
If we switch the limits x and 1, we get a negative sign here.
Now we can integrate this directly as in part a. This gives us the value of the integrand evaluated at x. So the derivative of this one is equal to 1 over x square, and then the minus is carried over here.
For this part, the answer is that when the variable x is in the lower limit, we get a negative sign.
In part c, the example here, the variable is not just x; it’s a certain function of x. In this case, we first let F(u) be this definite integral with the variable u.
Thus the one we are going to look at equals F(2x).
Now let’s use the Chain Rule to find dF/dx. We start with an expression for F(2x), which means we want to differentiate F(2x) with respect to x.
By using the Chain Rule, we find that F'(2x) times d(2x)/dx equals 2xd/dx. Thus, 2xd/dx is the correct answer. The Fundamental Theorem of Calculus tells us that F'(u) = ud/du for all functions F, where u represents any constant not equal to 0.
Therefore, one must differentiate 2x with respect to itself and then divide by two. The answer is F(2x) where u = 2x.
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