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

`\frac1e\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}`