Răspuns:
[tex]\int\limits^1_ {-1} \,f(x) dx =\int\limits^0_{-1} \,sinx dx +\int\limits^1_{0} \, \frac{x}{x+2} dx[/tex]
I1=-cosx║₋₁⁰=-(cos0-cos(-1))=-(1-cos(-1)=
-1+cos(-1)
--------[tex]\frac{x}{x+2} =\frac{x+2-2}{x+2}[/tex]=
[tex]\frac{x+2}{x+2} -\frac{2}{x+2}[/tex]=
1-[tex]\frac{2}{x+2}[/tex]
I1=[tex]\int\limits^1_ {0} \,(1-\frac{2}{x+2} ) dx[/tex]=[tex]\int\limits^1_ {0} \, dx -2\int\limits^1_ {0\}{} } \,\frac{1}{x+2} dx[/tex]
x║₀¹-2ln(x+2)║₀¹=
1-0-2(ln(1+2)-ln(0+2))=
1-2(ln3-ln2)=
1-2ln3/2
I=I1+I2
I=-1+cos(-1)+1-ln3/2=
cos(-1)-ln3/2
Explicație pas cu pas:
