[tex]f(x)=\frac{x+1}{e^{x}}[/tex]
a)
Vezi tabel de integrale in atasament
[tex]\int\limits^2_0{\frac{x+1}{\frac{x+1}{e^x} } } \, dx =\int\limits^2_0 {e^x} \, dx =e^x|_0^2=e^2-e^0=e^2-1[/tex]
b)
[tex]\int\limits^1_0 {e^{3x}\cdot \frac{(x+1)^2}{e^{2x}} } \, dx =\int\limits^1_0 e^x(x+1)^2\ dx=\int\limits^1_0x^2e^x\ dx+2\int\limits^1_0xe^x\ dx+\int\limits^1_0e^x\ dx=[/tex]
Calculam prima integrala prin parti
[tex]\int\limits^1_0x^2e^x\ dx=[/tex]
[tex]f=x^2\ \ \ \ \ f=2x\\\\g'=e^x\ \ \ \ g=e^x\\\\=x^2e^x|_0^1-2\int\limits^1_0xe^x\ dx[/tex]
[tex]\int\limits^1_0 {e^{3x}\cdot \frac{(x+1)^2}{e^{2x}} } \, dx =\int\limits^1_0 e^x(x+1)^2\ dx=\int\limits^1_0x^2e^x\ dx+2\int\limits^1_0xe^x\ dx+\int\limits^1_0e^x\ dx=\\\\=x^2e^x|_0^1-2\int\limits^1_0xe^x\ dx+2\int\limits^1_0xe^x\ dx+e^x|_0^1=e+e-e^0=2e-1[/tex]
c)
[tex]1-\int\limits^a_0 {\frac{\frac{x+1}{e^x} }{x+1} } \, dx =1-\int\limits^a_0e^{-x}\ dx[/tex]
[tex]1-\int\limits^a_0e^{-x}\ dx,1-\int\limits^b_0e^{-x}\ dx, 1-\int\limits^c_0e^{-x}\ dx[/tex]
sunt termeni consecutivi ai unei progresii geometrice
[tex]1-\int\limits^a_0e^{-x}\ dx=1+e^{-x}|_0^a=1+e^{-a}-1=e^{-a}[/tex]
[tex]e^{-a},\ e^{-b},\ e^{-c}[/tex] -sunt termeni consecutivi ai unei progresii geometrice
[tex]e^{-2b}=e^{-a}\cdot e^{-c}\\\\e^{-2b}=e^{-a-c}[/tex]
-2b=-(a+c)
2b=a+c⇒ a, b si c sunt termeni consecutivi ai unei progresii aritmetice
Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9835836
#BAC2022
#SPJ4
