\( \int \limits_{2}^{\sqrt{e+3}} \frac{x}{x^{2}-3} d x \)
Substitution:
\(u=x^2-3\) \( \frac{du}{dx} =2x\) \( dx =\frac{1}{2x}du\)
Grenzen des Integrals verändern:
untere Grenze: \(x^2=u+3\) \(x=±\sqrt{u+3}\) \(2=±\sqrt{u+3}\) \(4=u+3\) \(u=1\)
obere Grenze: \(\sqrt{e+3}=±\sqrt{u+3}\) \(e+3=u+3\) \(u=e\)
\( \int \limits_{1}^{e} \frac{x}{u} \cdot \frac{1}{2x}du=0,5\int \limits_{1}^{e} \frac{du}{u} =[0,5 ln( u)] _{1}^{e}=0,5\) mit \(ln (e)=1\) und \(ln (1)=0\)
\( f(x)=e^{0,25 x} \) \( f'(x)=e^{0,25 x} \cdot 0,25 \)
\( f'(4)=e^{0,25 \cdot4} \cdot 0,25 =0,25e\)
Tangente:
\( f(4)=e\)
\( \frac{y-e}{x-4} =0,25e\)
\(y-e =0,25e \cdot x-e\)
\(y=0,25e \cdot x\)
\(A_1=\int\limits_{0}^{4} e^{0,25x}dx=\int\limits_{0}^{4} e^{\frac{x}{4}}dx\)
Substitution: \(u= \frac{x}{4} \) \(du= \frac{1}{4}dx \) \(dx= 4du \)
Grenzen des Integrals verändern:
untere Grenze: \(x=4u \) \(0=4u \) \(u=0 \)
obere Grenze: \(4=4u \) \(u=1 \)
\(A_1=\int\limits_{0}^{1}e^{u} \cdot 4du=4\int\limits_{0}^{1}e^{u}du=[4e^{u}]_{0}^{1}=4e-4\)
\(A_2=\frac{4e}{2}=2e\)
\(A=A_1-A_2=4e-4-2e=2e-4\)
