Plot `y=e^(x),y=e^(x)+1 and y=e^(x)-1.`

Area of the region bounded by y=e^(x),y=e^(-x) and the line x=1 is

The area of the region bonded by y=e^(x),y=e^(-x),x=0 and x = 1 is (a) e+(1)/(e) (b) log(4/e) (c) 4log(4/e) (d) e+(1)/(e)-2

The area bounded by y=e^(x),y=e^(-x) and x = 2 is

Find the area (in sq. unit) bounded by the curves : y = e^(x), y = e^(-x) and the straight line x =1.

If y=f(x) is the solution of equation ydx+dy=-e^(x)y^(2) dy, f(0)=1 and area bounded by the curve y=f(x), y=e^(x) and x=1 is A, then

Suppose y=e^(a x)+e^(b x) , find y''

The area bounded by the curve y y=e^|x|,y=e^(-|x|),x >= 0 and x <= 5 is

(ii) e^(y)((dy)/(dx)+1)=e^(x)

The solution of the differential equation (dy)/(dx)+1=e^(x+y), is a. (x+y)e^(x+y)=0 b. (x+C)e^(x+y)=0 c. (x-C)e^(x+y)=1 d. (x+C)e^(x+y)+1=0

If y = (e^(x)+1)/(e^(x)-1), " then" (y^(2))/2 + (dy)/(dx) is equal to