thingthingidknamethinganfangmathewiderholungsthigniawfjasifjasifjsifj

![](https://i.imgur.com/JmbaNlA.png) i) änderungsrate geht runter also ändert es sich weniger, dh nimmt ja immer noch zu ii) volumen des becks was im intervall [1;2] angefüllt wird iii) im intervall [0;4] das volumen im tank in m

$$\int (1-3x{)}^{3}dx$$

Substitution: $u=1-3x$ $u^{\prime }=\frac{du}{dx}=-3$ $dx=\frac{du}{-3}=-\frac{1}{3}du$

$-\frac{1}{3}\int u^{3}du=-\frac{1}{3}\ast \frac{u^4}{4}+C=\frac{-u^4}{12}+C=-\frac{(1-3x{)}^4}{12}+C$ c) $\int \sqrt{1-x}dx$ Substitution: $u=1-x$ $u^{\prime }=\frac{du}{dx}=-1$ $dx=\frac{du}{-1}=-du$ $-\int u^{\frac{1}{2}}du=-\frac{2}{3}\ast u^{\frac{3}{2}}+C$ Rücksubstitution: $-\frac{2}{3}\ast (1-x{)}^{\frac{3}{2}}=\frac{-2\ast \sqrt{(1-x)}}{3}=-\frac{2}{3}\ast (1-x)\ast \sqrt{1-x}$

f) $\int (2t-x{)}^{2}dx$ Substitution: $u=2t-x$ $u^{\prime }=\frac{du}{dx}-1$ $dx=\frac{du}{-1}=-du$

$-\int u^{2}ux=-\frac{u^3}{3}+C$ Rücksubstitution: $-\frac{1}{3}\ast (2t-x{)}^3+C$

6.78a) $\int \frac{2}{2+x}dx$ Substitution: $u=2+x$ $dx=du$ $2\ast \int \frac{1}{u}du=2\ast \ln |u|+C$ Rücksubstitution: $2\ast \ln |2+x|+C$

${\int }_0^4(\frac{x}{3}-3{)}^{2}dx$ Substitution: $u=\frac{x}{3}-3$ $u^{\prime }=\frac{du}{dx}=\frac{1}{3}dx=3du$ $3{\int }_0^{4}du=3\ast \frac{4}{3}={\left(\frac{x}{3}-3\right)}^3=22.32$

Partielle integrtion wiederholungslop

$\int u\ast v^{\prime }dx=u\ast v-\int u^{\prime }\ast vdx$

6.50 a) $\int x\ast \cos xdx=x\ast (-\sin x)-\int 1\ast (-\sin x)dx$ x = u cos x = v’ = $x\ast \sin x+\cos x+c$

b) $\int t\ast \sin (2t)d=t\ast \left[-\frac{1}{2}\cos (2t)\right]-\int 1\ast \left[-\frac{1}{2}\cos (2t)\right]dt$ c) $\int t\ast \cos \frac{t}{5}dt=t\ast 5\ast \sin \frac{t}{5}-\int 5\ast \sin \frac{t}{5}dt$ d)

$\int x\ast c^x\ast dx=x\ast e^x-\int e^{x}dx=x\ast e^x-e^x+c=e^x(x-1)+c$ 6.51 a) ${\int }_{-1}^{1}dx$