[tex]f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}[/tex]
[tex]f[/tex] has critical points where the derivative is 0:
[tex]2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1[/tex]
The second derivative is
[tex]f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}[/tex]
and [tex]f''(-1)=\frac2e>0[/tex], which indicates a local minimum at [tex]x=-1[/tex] with a value of [tex]f(-1)=\frac1e[/tex].
At the endpoints of [-2, 2], we have [tex]f(-2)=1[/tex] and [tex]f(2)=e^8[/tex], so that [tex]f[/tex] has an absolute minimum of [tex]\frac1e[/tex] and an absolute maximum of [tex]e^8[/tex] on [-2, 2].
So we have
[tex]\dfrac1e\le f(x)\le e^8[/tex]
[tex]\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx[/tex]
[tex]\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}[/tex]