Rozwiązanie:
Zadanie [tex]1.[/tex]
[tex]$\int\limits^{1}_{0}(2x^{2}+4x+5)e^{x} \ dx[/tex]
Całka nieoznaczona:
[tex]$\int(2x^{2}+4x+5)e^{x} \ dx=\left|\begin{array}{ccc}f=2x^{2}+4x+5&dg=e^{x} \ dx\\df=4x+4 \ dx&g=e^{x}\end{array}\right|=[/tex]
[tex]$=(2x^{2}+4x+5)e^{x}-\int (4x+4)e^{x} \ dx=\left|\begin{array}{ccc}f=4x+4&dg=e^{x} \ dx\\df=4 \ dx&g=e^{x}\end{array}\right|=[/tex]
[tex]$=(2x^{2}+4x+5)e^{x}-(4x+4)e^{x}+ \int 4e^{x} \ dx=[/tex]
[tex]$=(2x^{2}+4x+5)e^{x}-(4x+4)e^{x}+4e^{x} +C=e^{x}(2x^{2}+5)+C[/tex]
Całka oznaczona:
[tex]$\int\limits^{1}_{0}(2x^{2}+4x+5)e^{x} \ dx=e^{x}(2x^{2}+5)\Big|^{1}_{0}=7e-5[/tex]
Zadanie [tex]2.[/tex]
[tex]$\int\limits^{1}_{0} \frac{2x^{4}+4x^{3}+5x^{2}}{x^{2}+1} \ dx[/tex]
Całka nieoznaczona:
[tex]$\int\frac{2x^{4}+4x^{3}+5x^{2}}{x^{2}+1} \ dx=\int 2x^{2}+4x+3-\frac{4x+3}{x^{2}+1} \ dx =[/tex]
[tex]$=\frac{2}{3}x^{3}+2x^{2}+3x-\int 2 \cdot \frac{2x}{x^{2}+1} +\frac{3}{x^{2}+1} \ dx =[/tex]
[tex]$=\frac{2}{3}x^{3}+2x^{2}+3x-2 \ln (x^{2}+1)-3\arctan x +C[/tex]
Całka oznaczona:
[tex]$\int\limits^{1}_{0} \frac{2x^{4}+4x^{3}+5x^{2}}{x^{2}+1} \ dx=\frac{2}{3}x^{3}+2x^{2}+3x-2 \ln (x^{2}+1)-3\arctan x\Big|^{1}_{0}=[/tex]
[tex]$=\frac{2}{3} +2+3-2\ln 2-\frac{3\pi}{4}=\frac{17}{3} -\ln 4-\frac{3\pi}{4}[/tex]
Zadanie [tex]3.[/tex]
[tex]$\int\limits^1_0 {2e^{4x+5}} \, dx[/tex]
Całka nieoznaczona:
[tex]$\int 2e^{4x+5} \ dx =2\int e^{4x+5} \ dx=\left|\begin{array}{ccc}u=4x+5\\du=4 \ dx\end{array}\right|=\frac{1}{2} \int e^{u} \ du =\frac{1}{2} e^{u}+C=[/tex]
[tex]$=\frac{1}{2} e^{4x+5}+C[/tex]
Całka oznaczona:
[tex]$\int\limits^1_0 {2e^{4x+5}} \, dx=\frac{1}{2}e^{4x+5} \Big|^{1}_{0}=\frac{e^{9}}{2} -\frac{e^{5}}{2} =\frac{e^{9}-e^{5}}{2}[/tex]
