Question

f(x)= e^x+1, x<sau egal decat -1 si 2+x, x>-1.
Aratati ca functia f admite primitive pe R.

Ajutor!

Answer 1

$f(x)= \left \{ {{e^x+1; x \leq -1} \atop {2+x;x\ \textgreater \ -1}} \right.$

Fie F o primitiva a ei, deci

$\int\limitsa {(e^x+1)} \, dx=e^x+x+C_1$ pt x≤-1

si

$\int\limits {(2+x)} \, dx = 2x + x^2 +C_2$ pt x>-1

Impunem F continua deci

$\lim_{x \to \-1;x\ \textless \ -1} (e^x+x+C_1)= \lim_{x \to \-1;x\ \textgreater \ -1}(2x+x^2+C_2)$

Deci $\frac{1}{e} -1+C_1=-2+1+C_2$

adica $\frac{1}{e} +C_1=C_2$

Vom nota pe $C_2=k ; C_1=k- \frac{1}{e}$

Primitivele lui f sunt:

$F= \left \{ {{e^x+x+k- \frac{1}{e};x \leq -1 } \atop {2x+x^2+k;x\ \textgreater \ -1}} \right.$