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.

has critical points where the derivative is 0:

The second derivative is

and , which indicates a local minimum at with a value of .

At the endpoints of [-2, 2], we have and , so that has an absolute minimum of and an absolute maximum of on [-2, 2].

So we have

a negative exponent is translated as follows:

so, we have

18

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring).   so find the distance between (1,2) and (4,2).   that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3.   it is a regular polygon which means all sides are congruent.   you have 6 sides (hexa) so multiply 6 by 3 to receive 18.