Differenziation eines Integrals

Sei eine Funktion $\mathit{z}=\mathit{f}(\mathit{x},\mathit{y})$ innerhalb eines abgeschlossenen Rechtecks $\mathit{a}\le \mathit{x}\le \mathit{b}$ und $\mathit{c}\le \mathit{y}\le \mathit{d}$ für $\mathit{x}$ und $\mathit{y}$ stetig. Dann ist

$$\mathit{h}(\mathit{y})=\underset{\mathit{a}}{\overset{\mathit{b}}{\int }}\mathit{f}(\mathit{x},\mathit{y})\text{d}\mathit{x}$$

stetig in $[\mathit{c},\mathit{d}]$. Die Reihenfolge von Differenziation und Integration kann vertauscht werden, wenn $\mathit{h}(\mathit{y})$ in dem Intervall stetig ist

$$\begin{array}{rcl}\frac{\text{d}\mathit{h}}{\text{d}\mathit{y}}& =& \frac{\text{d}}{\text{d}\mathit{y}}\left(\underset{\mathit{a}}{\overset{\mathit{b}}{\int }}\mathit{f}(\mathit{x},\mathit{y})\text{d}\mathit{x}\right)\\ \frac{\text{d}\mathit{h}}{\text{d}\mathit{y}}& =& \underset{\mathit{a}}{\overset{\mathit{b}}{\int }}\frac{\partial \mathit{f}(\mathit{x},\mathit{y})}{\partial \mathit{y}}\text{d}\mathit{x}.\end{array}$$

Genauso ist

$$\mathit{g}(\mathit{x})=\underset{\mathit{c}}{\overset{\mathit{d}}{\int }}\mathit{f}(\mathit{x},\mathit{y})\text{d}\mathit{y}$$

stetig in $[\mathit{a},\mathit{b}]$ mit

$$\begin{array}{rcl}\frac{\text{d}\mathit{g}}{\text{d}\mathit{x}}& =& \frac{\text{d}}{\text{d}\mathit{x}}\left(\underset{\mathit{c}}{\overset{\mathit{d}}{\int }}\mathit{f}(\mathit{x},\mathit{y})\text{d}\mathit{y}\right)\\ \frac{\text{d}\mathit{g}}{\text{d}\mathit{x}}& =& \underset{\mathit{c}}{\overset{\mathit{d}}{\int }}\frac{\partial \mathit{f}(\mathit{x},\mathit{y})}{\partial \mathit{x}}\text{d}\mathit{y}.\end{array}$$

Beispiel

Sei $\mathit{z}=\mathit{f}(\mathit{x},\mathit{y})=6\mathit{x}{\mathit{y}}^2$ stetig in $\mathit{a}\le \mathit{x}\le \mathit{b}$, $\mathit{c}\le \mathit{y}\le \mathit{d}$. Dann ist

$$\begin{array}{rcl}\frac{\text{d}\mathit{g}}{\text{d}\mathit{x}}& =& \frac{\text{d}}{\text{d}\mathit{x}}\left(\underset{\mathit{c}}{\overset{\mathit{d}}{\int }}\mathit{f}(\mathit{x},\mathit{y})\text{d}\mathit{y}\right)\\ & =& \frac{\text{d}}{\text{d}\mathit{x}}\left({\left[2\mathit{x}{\mathit{y}}^3\right]}_{\mathit{c}}^{\mathit{d}}\right)\\ & =& 2({\mathit{d}}^3-{\mathit{c}}^3)\\ \frac{\text{d}\mathit{g}}{\text{d}\mathit{x}}& =& \underset{\mathit{c}}{\overset{\mathit{d}}{\int }}\frac{\partial \mathit{f}(\mathit{x},\mathit{y})}{\partial \mathit{x}}\text{d}\mathit{y}\\ & =& \underset{\mathit{c}}{\overset{\mathit{d}}{\int }}6{\mathit{y}}^2\text{d}\mathit{y}\\ & =& {\left[2{\mathit{y}}^3\right]}_{\mathit{c}}^{\mathit{d}}\\ & =& 2({\mathit{d}}^3-{\mathit{c}}^3).\end{array}$$

Für variable Integrationsgrenzen (nicht rechteckiges Integrationsbereich (Abb. 2) ) ist

$$\mathit{g}(\mathit{x})=\underset{{\phi }_1(\mathit{x})}{\overset{{\phi }_2(\mathit{x})}{\int }}\mathit{f}(\mathit{x},\mathit{y})\text{d}\mathit{y}=\mathit{I}(\mathit{x},{\phi }_1(\mathit{x}),{\phi }_2(\mathit{x})).$$

Die Ableitung ist (siehe Link) nach der Kettenregel

$$\begin{array}{rcl}\frac{\text{d}\mathit{g}}{\text{d}\mathit{x}}& =& \frac{\partial \mathit{I}}{\partial \mathit{x}}+\frac{\partial \mathit{I}}{\partial {\phi }_2}\frac{\text{d}{\phi }_2}{\text{d}\mathit{x}}+\frac{\partial \mathit{I}}{\partial {\phi }_1}\frac{\text{d}{\phi }_1}{\text{d}\mathit{x}}\\ & =& \underset{{\phi }_1(\mathit{x})}{\overset{{\phi }_2(\mathit{x})}{\int }}\frac{\partial \mathit{f}(\mathit{x},\mathit{y})}{\partial \mathit{x}}\text{d}\mathit{y}+\mathit{f}(\mathit{x},{\phi }_2(\mathit{x}))\frac{\text{d}{\phi }_2}{\text{d}\mathit{x}}-\mathit{f}(\mathit{x},{\phi }_1(\mathit{x}))\frac{\text{d}{\phi }_1}{\text{d}\mathit{x}}.\end{array}$$