when y=1+2e^-x. what is the area enclosed by the curve, the lines x=1 and x=0 and the line y=1?

Answer 1

1+2e^-x > 1

the area between this line and y = 1 is

integral 2e^-x = -2e^-x (because it is definite so constant is dropped)

at x =1 it is -2/e
at x =0 it is -2
so area = (2-2/e) that is positive

Answer 2

Since e^-x > 0, 1+2e^-x > 1, so y=1+2e^-x is the top curve, and y=1 is the bottom.

To get the area, integrate (top-bottom) between the given x-values. You get

integral from 0 to 1 of (1+2e^-x)-1 dx =
integral from 0 to 1 of 2e^-x = -2e^-x|0,1
plug in the limits and you're done