8. Hauptsatz der Integral- und Differentialrechnung

8.1 Hauptsatz der Integral- und Differentialrechnung

8.1.1 Der Hauptsatz (The fundamental Theorem of Calculus, FTC)

7.1 Primitive

Let $I\subseteq \mathbb{R}$ be an interval and $f:I\to \mathbb{R}$. A differentiable function $F:I\to \mathbb{R}$ with $F^{\prime }=f$ is called a primitive (or antiderivative, Stammfunktion) of $f$.

Note: A primitive need not exist: by Darboux’s theorem, no differentiable $F:\mathbb{R}\to \mathbb{R}$ satisfies $F^{\prime }=\mathrm{sgn}$ for example.

7.2 Fundamental Theorem of Calculus

Let $f:[a,b]\to \mathbb{R}$ be continuous. Then:

1. For every $C\in \mathbb{R}$, the function

$$F(x)={\int }_a^{x}f(t)dt+C$$

is a primitive of $f$.
2. Every primitive $F$ of $f$ has this form for some constant $C$.

7.3 Unterteilung des Integrationsbereichs

Es seien $a<b<c\in \mathbb{R}$. Sei $f$ eine auf $[a,c]$ integrierbare Funktion. Dann gilt

$${\int }_a^{c}f(x)dx={\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx$$

Proof: (FTC) $f$ continuous thus integrable.

1. Fix $x_0\in [a,b]$ and we want to show $F^{\prime }(x_0)=f(x_0)$.

• Fix $\epsilon >0$; by continuity there is $\delta >0$ with

$$z\in [a,b],|z-x_0|<\delta ⟹|f(z)-f(x_0)|<\epsilon$$

• For $x\in (x_0,x_0+\delta )\cap [a,b]$, using $f(x_0)=\frac{1}{x-x_0}{\int }_{x_0}^{x}f(x_0)dt$

$$\begin{array}{rl}\left|\frac{F(x)-F(x_0)}{x-x_0}-f(x_0)\right|& =\left|\frac{1}{x-x_0}{\int }_{x_0}^x(f(t)-f(x_0))dt\right|\\ & \le \frac{1}{x-x_0}{\int }_{x_0}^x|f(t)-f(x_0)|dt\\ & <\frac{1}{x-x_0}{\int }_{x_0}^x\epsilon dt=\epsilon ,\end{array}$$

	- since $t \in [x_0, x] \subset [x_0, x_0 + \delta)$ forces $|f(t) - f(x_0)| < \varepsilon$ by continuity
- The case $x \in (x_0 - \delta, x_0)$ is symmetric. Hence the difference quotient converges to $f(x_0)$, i.e. $F'(x_0) = f(x_0)$.

2. Let $F$ be any primitive. Since ${\left({\int }_a^{x}f(t)dt\right)}^{\prime }=f(x)$ by (i)

$${\left(F(x)-{\int }_a^{x}f(t)dt\right)}^{\prime }=F^{\prime }(x)-f(x)=f(x)-f(x)=0\forall x\in (a,b),$$

so $F(x)-{\int }_a^{x}f(t)dt$ is constant on $[a,b]$.

7.4 Piecewise continuous derivative

Es sei $f:[a,b]\to \mathbb{R}$ bis auf endlich viele Ausnahmestellen $x_1,x_2,\dots ,x_n$ auf $[a,b]$ stetig. Dann ist

$$F(x):={\int }_a^{x}f(y)dy+C$$

bis auf die Ausnahmestellen $x_1,x_2,\dots ,x_n$ auf $[a,b]$ stetig differenzierbar mit $F^{\prime }(x)=f(x)$, $x\notin \{x_1,x_2,\dots ,x_n\}$

7.5 Integral vs. Derivative

If $F:[a,b]\to \mathbb{R}$ is continuously differentiable, then for all $x\in [a,b]$

$$F(x)=F(a)+{\int }_a^x{F}^{\prime }(t)dt.$$

Proof: Since $F$ is a primitive of $F^{\prime }$, the FTC yields $F(x)={\int }_a^x{F}^{\prime }(t)dt+C$. Evaluating at $x=a$ yields $C=F(a)$.

Riemann Integral and Primitives

If $f:[a,b]\to \mathbb{R}$ is continuous and $F$ is a primitive of $f$, then

$${\int }_a^{b}f(t)dt=F(b)-F(a).$$

Proof: follows trivially from 7.5 with $F^{\prime }=f$ and $x=b$.

8.2 Integrationstechniken

We write $[h(x){]}_a^b:=h(b)-h(a)$, with the convention ${\int }_b^{a}hdx=-{\int }_a^{b}hdx$.

8.2.1 Partielle Integration

Integration by Parts

If $f,g:[a,b]\to \mathbb{R}$ are continuously differentiable, then

$${\int }_a^{b}f(x)g^{\prime }(x)dx=[f(x)g(x){]}_a^b-{\int }_a^b{f}^{\prime }(x)g(x)dx.$$

Proof: From $(fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime }$, rearranging and integrating via the previous corollary

$${\int }_a^{b}f{g}^{\prime }dx={\int }_a^b(fg{)}^{\prime }dx-{\int }_a^b{f}^{\prime }gdx=[fg{]}_a^b-{\int }_a^b{f}^{\prime }gdx.$$

8.2.2 Substitution

Integration by Substitution (1st Form)

Let $I,J\subseteq \mathbb{R}$ be intervals, $f:I\to J$ continuously differentiable, $g:J\to \mathbb{R}$ continuous. For $[a,b]\subseteq I$,

$${\int }_a^{b}g(f(x))f^{\prime }(x)dx={\int }_{f(a)}^{f(b)}g(y)dy.$$

Proof: Fix $y_0\in J$ and set $G(y)={\int }_{y_0}^{y}g(t)dt$, so $G^{\prime }=g$. By the chain rule $(G\circ f{)}^{\prime }=g(f)f^{\prime }$. Integrating and applying the corollary

$${\int }_a^{b}g(f(x))f^{\prime }(x)dx=G(f(b))-G(f(a))={\int }_{f(a)}^{f(b)}g(t)dt.$$

Integration by Substitution (2nd Form)

Let $I,J\subseteq \mathbb{R}$ be intervals, $f:I\to J$ be $C^1$, $g:J\to \mathbb{R}$ continuous. Let $[a,b]\subseteq I$ with $f^{\prime }(x)\ne 0$ on $[a,b]$, and let $f^{-1}:[f(a),f(b)]\to \mathbb{R}$ be the inverse of $f{|}_{[a,b]}$. Then

$${\int }_a^{b}g(f(x))dx={\int }_{f(a)}^{f(b)}g(y)(f^{-1}{)}^{\prime }(y)dy.$$

Note: if $h$ is $C^1$ with $\forall x{h}^{\prime }(x)\ne 0$, then $h^{\prime }$ has constant sign.

• So $h$ is strictly monotone (via MVT) and invertible;
• $h^{-1}$ is continuous and differentiable
• $(h^{-1}{)}^{\prime }=\frac{1}{h^{\prime }\circ h^{-1}}$, which is again continuous.
  • Differentiate $h(h^{-1}(y))=y$
  • $h^{\prime }(h^{-1}(y))\cdot (h^{-1}{)}^{\prime }(y)=1$.

Proof: Write $g(f(x))=\frac{g(f(x))}{f^{\prime }(x)}{f}^{\prime }(x)=\frac{g(f(x))}{f^{\prime }\circ f^{-1}(f(x))}{f}^{\prime }(x)$

• apply the 1st form with $\frac{g}{f^{\prime }\circ f^{-1}}$ in place of $g$:

$${\int }_a^{b}g(f(x))dx={\int }_{f(a)}^{f(b)}\frac{g(y)}{f^{\prime }(f^{-1}(y))}dy,$$

• and $\frac{1}{f^{\prime }(f^{-1}(y))}=(f^{-1}{)}^{\prime }(y)$.

Example: ${\int }_0^{{\ln }_2}\frac{1}{1+e^x}dx$ with $f(x)=e^x$ and $g(x)=\frac{1}{1+x}$. Because there’s no $f^{\prime }(x)$ factor to be seen, we use substitution part 2.

• $f^{-1}(x)=ln(x)$ and $(f^{-1}{)}^{\prime }(x)=1/x$.

Tips & Tricks

Use Euler instead of Trig identities

Instead of knowing that ${\cos }^2(x)=1/2+\cos (2x)$, we can also expand ${\cos }^2(x)=(\frac{e^{ix}+e^{-ix}}{2}{)}^2$ and integrate this, which is much easier!

Substitution

For substitution we use the fact that

$$\int g^{\prime }(x)f(g(x))dx=F(g(x))+C$$

you massage the function into looking like that and replace all occurrences of the $g(x)$ with $u$.
Then you get $\int f(u)du$ and solve to get $F(g(x))$.

Here we clearly see:

• rewrite $y=g(x)$. Then $g^{\prime }(x)=\frac{dy}{dx}$.
• $f^{\prime }(g(x))=\frac{df}{dy}$ as we derive $f^{\prime }$ with regards to $y$, our new $g(x)$!

We know $du=g^{\prime }(x)dx$ (intuitively: small change in u = slope of g at x + small change in x → we replace $g(x)=u$ inside $f(u)$).

Notice we can also do it the other way round: We have $du=g^{\prime }(x)dx$ but also $dx=\frac{1}{g^{\prime }(x)}du$.
Instead of replacing $\int \dots g^{\prime }(x)\dots dx$ by $\int \dots du$ we can also insert the derivative:

$$\int f(g(x))dx=\int \frac{f(u)}{g^{\prime }(x)}du$$

Ex: substituting $u=\sqrt{x}$. Then $du=\frac{1}{2\sqrt{x}}dx$ and $dx=2\sqrt{x}du=2udu$:

$$\int \frac{1}{1+\sqrt{x}}=\int \frac{1}{1+\sqrt{u^2}}\cdot 2udu=\int \frac{2u}{1+u}$$

then using polynomial division $\frac{2u}{1+u}=2-\frac{2}{1+u}$ which gives $\int 2-2\int \frac{1}{1+u}=2u-2\ln |1+u|$.

With bounds

$${\int }_a^b{g}^{\prime }(x)f(g(x))dx={\int }_{g(a)}^{g(b)}f(u)+C$$

→ Watch out for bounding, as often $|x|<1$ is required. Then we have $\ln |...|$ not $\ln (...)$.

Trigonometric

Here we replace $x$ by $h(\theta )$ to use trig identities. Then we get $dx=h^{\prime }(\theta )d\theta$.

Partialbruchzerlegung