1b)

${\int }_{-1}^1(2x+1)dx=x^2+x+c$
$1^2+1-((-1{)}^2-1)=2$

2)

${\int }_{-2}^{0}f(x)dx=0,8-0,3=0,5$
${\int }_0^{3}f(x)dx=2,9-1,1=1,8$
${\int }_{-1}^{2}f(x)dx=0,8+2,9=3,7$
${\int }_{-2}^{3}f(x)dx=0,8+2,6-(0,3+1,1)=3,4-1,4=2,4$

3a)

${\int }_0^2{x}^{2}dx=\frac{8}{3}$

b)

${\int }_0^2(x^2-1)dx=\frac{2}{3}$

c)

${\int }_{-1}^1{x}^{3}dx=4$

d)

${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos xdx=2$

e)

${\int }_0^1(e^x-2)dx$

4)

${\int }_1^4\frac{1}{x}dx=\log (4)$

${\int }_{-\sqrt{5}}^{\sqrt{5}}-\frac{1}{2}{x}^2+2,5dx\approx 7,4536$

${\int }_{-4}^{-1}\frac{1}{x^2}-1dx=-2.25$

S. 49 Nr. 10

10a)

Die Fläche ist das Gasvolumen, solange der Tank am Anfang leer war

b)

${\int }_0^{6}f(x)dx=3,5\ast 1,2+\frac{2,5\ast 1,2}{2}=4,2$
${\int }_6^{16}f(x)dx=\frac{(5+10)\ast 1,2}{2}=9$
${\int }_0^{18}f(x)dx=4,2+9+\frac{1,5\ast 0,8}{2}+\frac{0,8}{2}=13,7$