Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the comparison property to find the lower and upper bounds for I = \displaystyle \int_{-2}^{2} f(x) \, dxI = ∫ − 2 2 f ( x ) d x.

[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]