Analysis III TT23, Fundamental theorems of calculus

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@State the first fundamental theorem of calculus (differentiation undoes integration).

ftc-first-state

Suppose that $f$ is integrable on $(a, b)$ and $F : [a, b] \to \mathbb R$ is defined by

\[F(x) := \int^x _ a f\]

Then $F$ is continuous and if $f$ is continuous at $c \in (a, b)$ then $F$ is differentiable at $c$ and $F’(c) = f(c)$.

@State the second fundamental theorem of calculus (integration undoes differentiation).

ftc-second-state

Suppose $F : [a, b] \to \mathbb R$ is continuous on $[a, b]$ and differentiable on $(a, b)$. Suppose furthermore that its derivative $F’$ is integrable on $(a, b)$. Then

\[\int^b _ a F' = F(b) - F(a)\]

Can you give an example of an integrable function $F$ that is differentiable everywhere but whose derivative $F’$ is not integrable everywhere?

differentiable-with-non-integrable-derivative-example
\[F(x) = \begin{cases} x^2\sin\frac1{x^2} &\text{if } x\ne 0 \\ 0 &\text{if } x = 0 \end{cases}\]

When proving the latter part of the first fundamental theorem of calculus, i.e.

Suppose that $f$ is integrable on $(a, b)$ and $F : [a, b] \to \mathbb R$ is defined by

\[F(x) := \int^x _ a f\]

Then $F$ is continuous and if $f$ is continuous at $c \in (a, b)$ then $F$ is differentiable at $c$ and $F’(c) = f(c)$.

You know $f$ is continuous, so $\exists h$ such that $\forall x \in [c, c+h]$, $ \vert f(x) - f(c) \vert < \varepsilon$. Then what expression do you consider that the fact $F$ is differentiable at $c$ follows from?

ftc-first-proof-key-expression
\[ \vert F(c + h) - F(c) - hf(c) \vert = \left \vert \int^{c+h} _ c (f(x) - f(c)) \text d x\right \vert \]

Proofs



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