Second Fundamental Theorem of Calculus

Second Fundamental Theorem of Calculus#

Theorem (Fundamental Theorem of Calculus II)

Let \(f\) be continuous on \([a,b]\), and suppose \(F\) is an antiderivative of \(f\) there;

that is, \(F'(x)=f(x)\) for all \(x\in[a,b]\).

Then

\[ \int_{a}^{b} f(x)\,\mathrm{d}x \;=\; F(b)\;-\;F(a). \]

We provide astandard proof of the second part of the Fundamental Theorem of Calculus II (FTC II):

Proof. Fundamental Theorem of Calculus II.

Let \(f\) be continuous on \([a,b]\), and suppose \(F\) is an antiderivative of \(f\) there; that is, \(F'(x)=f(x)\) for all \(x\in[a,b]\).

We will show that

\[ \int_a^b f(x)\,\mathrm{d}x \;=\; F(b) - F(a). \]

This is equivalent to showing that the difference \(F(b)-F(a)\) is equal to the Riemann integral \(\int_a^b f(x)\,\mathrm{d}x\).

To do this, we will use the following steps:

  1. Partition the interval.
    Choose any partition

    \[ a = x_0 < x_1 < \dots < x_{n-1} < x_n = b \]

    of \([a,b]\), and in each subinterval \([x_{i-1},x_i]\) pick an arbitrary sample point \(\xi_i \in [x_{i-1},x_i]\).

  2. Riemann sum for \(\int_a^b f\).
    Because \(f\) is continuous, the Riemann sums

    \[ \sum_{i=1}^n f(\xi_i)\,\bigl(x_i - x_{i-1}\bigr) \]

    converge (as the mesh \(\max_i(x_i - x_{i-1})\to0\)) to \(\int_a^b f(x)\,\mathrm{d}x\).

  3. Mean value on each subinterval.
    On each \([x_{i-1},x_i]\), apply the mean value theorem to \(F\):

    since \(F\) is differentiable, there exists \(c_i\in(x_{i-1},x_i)\) with

    \[ F(x_i)-F(x_{i-1}) \;=\; F'(c_i)\,\bigl(x_i - x_{i-1}\bigr) \;=\; f(c_i)\,\bigl(x_i - x_{i-1}\bigr). \]

  4. Compare sums.
    Thus the telescoping sum

    \[ F(b)-F(a) \;=\; \sum_{i=1}^n \bigl[F(x_i)-F(x_{i-1})\bigr] \;=\; \sum_{i=1}^n f(c_i)\,(x_i - x_{i-1}). \]

    Notice each \(c_i\) lies in the corresponding subinterval, just like the Riemann sample points \(\xi_i\).

  5. Refine the partition.
    As we let the mesh of the partition go to zero, continuity of \(f\) implies that both sums

    \[ \sum_{i=1}^n f(c_i)\,(x_i - x_{i-1}) \quad\text{and}\quad \sum_{i=1}^n f(\xi_i)\,(x_i - x_{i-1}) \]

    converge to the same limit.

    The first sum is exactly \(F(b)-F(a)\) for every partition and the second sum converges to \(\int_a^b f(x)\,\mathrm{d}x\).

  6. Conclude.
    Therefore

    \[ F(b)-F(a) \;=\; \lim_{\text{mesh}\to0}\sum_{i=1}^n f(c_i)\,(x_i - x_{i-1}) \;=\; \int_a^b f(x)\,\mathrm{d}x, \]

    which completes the proof. ◻

Key idea: by slicing \([a,b]\) into tiny pieces, on each little piece the average rate of change of \(F\) equals \(f\) at some interior point; summing those up exactly telescopes to \(F(b)-F(a)\), and the same sums approximate the integral \(\int_a^b f\).