linerare-substitutions-aufgaben

6.42)

1. unbestimmtes Integral lösen (Grenzen ignoreiren)
2. Grezen einsetzen a) ${\int }_0^4(\frac{x}{3}-3{)}^{2}dx$ c) ${\int }_1^{2}2\ast \frac{1}{3-x}dx$ 6.43) a) ${\int }_0^1{e}^{4x-2}dx=$ $u=4x-2$ $\frac{du}{dx}=4$ $dx=\frac{1}{4}du$ $\int e^u\ast \frac{1}{4}du$ $\frac{1}{4}\int e^{u}du=\frac{e^u}{4}+c$ grenzen einsetzen $=\frac{e^{4\ast 1\ast 2}}{4}-\frac{e^{4\ast 0\ast 2}}{4}\approx 1.81$ c) ${\int }_0^{\pi /4}\sin 2xdx$ $u=2x$ $\frac{du}{dx}=2\Rightarrow \frac{1}{2}\ast du=dx$ $\int \sin u\ast \frac{1}{2}du=\frac{1}{2}\ast (-\cos u)+c$ $\frac{1}{2}\ast (-\cos 2x)+c$ $\frac{-\cos 2x}{2}{|}_0^{\pi /4}=\frac{-\cos \frac{\pi }{2}}{2}-\frac{-\cos 0}{2}=0.5$