Fundamental theorem of calculus · Selected prerequisite proofs

L5.3.1: Integral of an integrable derivative equals the endpoint difference; full stated generality.

Theorem 5.3.1.

Let

F \colon [a,b] \to \R

be a continuous function, differentiable on

(a,b)\text{.}

Let

f \in \sR\bigl([a,b]\bigr)

be such that

f(x) = F'(x)

for

x \in (a,b)\text{.}

Then

\begin{equation*} \int_a^b f = F(b)-F(a) . \end{equation*}

Proof.

Let

P = \{ x_0, x_1, \ldots, x_n \}

be a partition of

[a,b]\text{.}

For each interval

[x_{i-1},x_i]\text{,}

use the mean value theorem to find a

c_i \in (x_{i-1},x_i)

such that

\begin{equation*} f(c_i) \Delta x_i = F'(c_i) (x_i - x_{i-1}) = F(x_i) - F(x_{i-1}) . \end{equation*}

See Figure 5.5, and note that the area of the

i

th rectangle is

F(x_{i})-F(x_{i-1})\text{,}

and the total area of all three rectangles pictured is

F(x_{i+1})-F(x_{i-2})\text{.}

The idea is that taking smaller and smaller subintervals, the total area of all these rectangles converges to the integral of

f\text{.}

A diagram of a graph of a function in dark bold marked as y equals f of x equals F prime of x and three subintervals of the x coordinates. The middle subinterval is labeled as going from x sub quantity i minus 1 to x sub i and is of length Delta x sub i. A point c sub i is marked inside this subinterval and a dashed line goes up vertically until f of c sub i where it hits the graph of f. A shaded rectangle of this height and the subinterval as the base is drawn and labeled with ’area equals f of c sub i times Delta x sub i equals F of x sub i minus F of x sub quantity i minus 1’. The other two subintervals are similar except on the left we replace i with i minus 1 and on the right we replace i with i plus 1. — Figure 5.5. Mean value theorem on subintervals of a partition approximating the area under the curve.

Using the notation from the definition of the integral,

m_i \leq f(c_i) \leq M_i\text{,}

and multiplying by

\Delta x_i

gets

\begin{equation*} m_i \Delta x_i \leq F(x_i) - F(x_{i-1}) \leq M_i \Delta x_i . \end{equation*}

We sum over

i = 1,2, \ldots, n

to get

\begin{equation*} \sum_{i=1}^n m_i \Delta x_i \leq \sum_{i=1}^n \bigl(F(x_i) - F(x_{i-1}) \bigr) \leq \sum_{i=1}^n M_i \Delta x_i . \end{equation*}

In the middle sum, all the terms except the first and last cancel and we end up with

F(x_n)-F(x_0) = F(b)-F(a)\text{.}

The sums on the left and on the right are the lower and the upper sums, respectively. So

\begin{equation*} L(P,f) \leq F(b)-F(a) \leq U(P,f) . \end{equation*}

We take the supremum of

L(P,f)

over all partitions

P

and the left inequality yields

\begin{equation*} \underline{\int_a^b} f \leq F(b)-F(a) . \end{equation*}

Similarly, taking the infimum of

U(P,f)

over all partitions

P

yields

\begin{equation*} F(b)-F(a) \leq \overline{\int_a^b} f . \end{equation*}

As

f

is Riemann integrable, we have

\begin{equation*} \int_a^b f = \underline{\int_a^b} f \leq F(b)-F(a) \leq \overline{\int_a^b} f = \int_a^b f . \end{equation*}

The inequalities must be equalities and we are done.

L5.3.3: Primitive is Lipschitz; derivative equals f at every point of continuity, one-sided at endpoints.

Theorem 5.3.3.

Let

f \colon [a,b] \to \R

be a Riemann integrable function. Define

\begin{equation*} F(x) \coloneqq \int_a^x f . \end{equation*}

First,

F

is continuous on

[a,b]\text{.}

Second, if

f

is continuous at

c \in [a,b]\text{,}

then

F

is differentiable at

c

and

F'(c) = f(c)\text{.}

Proof.

As

f

is bounded, there is an

M > 0

such that

\babs{f(x)} \leq M

for all

x \in [a,b]\text{.}

Suppose

x,y \in [a,b]

with

x > y\text{.}

Then

\begin{equation*} \babs{F(x)-F(y)} = \abs{\int_a^x f - \int_a^y f} = \abs{\int_y^x f} \leq M\sabs{x-y} . \end{equation*}

By symmetry, the same also holds if

x < y\text{.}

So

F

is Lipschitz continuous and hence continuous.

Now suppose

f

is continuous at

c\text{.}

Let

\epsilon > 0

be given. Let

\delta > 0

be such that for

x \in [a,b]\text{,}

\sabs{x-c} < \delta

implies

\babs{f(x)-f(c)} < \epsilon\text{.}

In particular, for such

x\text{,}

we have

\begin{equation*} f(c)-\epsilon < f(x) < f(c) + \epsilon. \end{equation*}

Thus if

x > c\text{,}

then

\begin{equation*} \bigl(f(c)-\epsilon\bigr) (x-c) \leq \int_c^x f \leq \bigl(f(c) + \epsilon\bigr)(x-c). \end{equation*}

When

c > x\text{,}

then the inequalities are reversed. Therefore, assuming

x \neq c\text{,}

we get

\begin{equation*} f(c)-\epsilon \leq \frac{\int_c^{x} f}{x-c} \leq f(c)+\epsilon . \end{equation*}

As

\begin{equation*} \frac{F(x)-F(c)}{x-c} = \frac{\int_a^{x} f - \int_a^{c} f}{x-c} = \frac{\int_c^{x} f}{x-c} , \end{equation*}

we have

\begin{equation*} \abs{\frac{F(x)-F(c)}{x-c} - f(c)} \leq \epsilon . \end{equation*}

The result follows. It is left to the reader to see why is it OK that we just have a non-strict inequality.

L5.3.5: Oriented one-variable substitution, including noninjective g and endpoint range values.

Theorem 5.3.5. Change of variables.

Let

g \colon [a,b] \to \R

be a continuously differentiable function, let

f \colon [c,d] \to \R

be continuous, and suppose

g\bigl([a,b]\bigr) \subset [c,d]\text{.}

Then

\begin{equation*} \int_a^b f\bigl(g(x)\bigr)\, g'(x)\, dx = \int_{g(a)}^{g(b)} f(u)\, du . \end{equation*}

Proof.

As

g\text{,}

g'\text{,}

and

f

are continuous,

f\bigl(g(x)\bigr)\,g'(x)

is a continuous function of

[a,b]\text{,}

therefore it is Riemann integrable. Similarly,

f

is integrable on every subinterval of

[c,d]\text{.}

Define

F \colon [c,d] \to \R

by

\begin{equation*} F(y) \coloneqq \int_{g(a)}^{y} f(u)\,du . \end{equation*}

By the second form of the fundamental theorem of calculus (see Remark 5.3.4 and Exercise 5.3.4),

F

is a differentiable function and

F'(y) = f(y)\text{.}

Apply the chain rule,

\begin{equation*} \bigl( F \circ g \bigr)' (x) = F'\bigl(g(x)\bigr) g'(x) = f\bigl(g(x)\bigr) g'(x) . \end{equation*}

Note that

F\bigl(g(a)\bigr) = 0

and use the first form of the fundamental theorem to obtain

\begin{gathered} \int_{g(a)}^{g(b)} f(u)\,du = F\bigl(g(b)\bigr) = F\bigl(g(b)\bigr)-F\bigl(g(a)\bigr) \\ = \int_a^b \bigl( F \circ g \bigr)' (x) \,dx = \int_a^b f\bigl(g(x)\bigr) g'(x) \,dx . \qedhere \end{gathered}