Answer to Question #190517 in Calculus for Joshua

Question

The region bounded by y = 3-e^-x, the x-axis, x =2 and the y-axis

Answer 1

solution :-

$y = 3-e^{-x}$

limits

x=0

x=2

$\int _{x=0}^{x=2}3+e^{-x}dx$

=$[3x+e^{x}]_0^2$

=5.13 answer