Fundamental Theorem of Calculus - Proof of The Second Part

Proof of The Second Part

This is a limit proof by Riemann sums. Let f be (Riemann) integrable on the interval, and let f admit an antiderivative F on . Begin with the quantity F(b) − F(a). Let there be numbers x₁, ..., x_n such that

It follows that

Now, we add each F(x_i) along with its additive inverse, so that the resulting quantity is equal:

The above quantity can be written as the following sum:

Next we will employ the mean value theorem. Stated briefly,

Let F be continuous on the closed interval and differentiable on the open interval (a, b). Then there exists some c in (a, b) such that

It follows that

The function F is differentiable on the interval ; therefore, it is also differentiable and continuous on each interval . According to the mean value theorem (above),

Substituting the above into (1), we get

The assumption implies Also, can be expressed as of partition .

Notice that we are describing the area of a rectangle, with the width times the height, and we are adding the areas together. Each rectangle, by virtue of the Mean Value Theorem, describes an approximation of the curve section it is drawn over. Also notice that need not be the same for all values of i, or in other words that the width of the rectangles can differ. What we have to do is approximate the curve with n rectangles. Now, as the size of the partitions get smaller and n increases, resulting in more partitions to cover the space, we will get closer and closer to the actual area of the curve.

By taking the limit of the expression as the norm of the partitions approaches zero, we arrive at the Riemann integral. We know that this limit exists because f was assumed to be integrable. That is, we take the limit as the largest of the partitions approaches zero in size, so that all other partitions are smaller and the number of partitions approaches infinity.

So, we take the limit on both sides of (2). This gives us

Neither F(b) nor F(a) is dependent on, so the limit on the left side remains F(b) − F(a).

The expression on the right side of the equation defines the integral over f from a to b. Therefore, we obtain

which completes the proof.

It almost looks like the first part of the theorem follows directly from the second. That is, suppose G is an antiderivative of f. Then by the second theorem, . Now, suppose . Then F has the same derivative as G, and therefore F′ = f. This argument only works, however, if we already know that f has an antiderivative, and the only way we know that all continuous functions have antiderivatives is by the first part of the Fundamental Theorem. For example if f(x) = e−x2, then f has an antiderivative, namely

and there is no simpler expression for this function. It is therefore important not to interpret the second part of the theorem as the definition of the integral. Indeed, there are many functions that are integrable but lack antiderivatives that can be written as an elementary function. Conversely, many functions that have antiderivatives are not Riemann integrable (see Volterra's function).