Question

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.

Answer 1

$f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}$

$f$ has critical points where the derivative is 0:

$2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1$

The second derivative is

$f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}$

and $f''(-1)=\frac2e>0$, which indicates a local minimum at $x=-1$ with a value of $f(-1)=\frac1e$.

At the endpoints of [-2, 2], we have $f(-2)=1$ and $f(2)=e^8$, so that $f$ has an absolute minimum of $\frac1e$ and an absolute maximum of $e^8$ on [-2, 2].

So we have

$\dfrac1e\le f(x)\le e^8$

$\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx$

$\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}$