Area bounded by the curves y = ex, y = e- x and the straight line x = 1 is (in sq units)
C.
Given curves are y = ex and y = e- x
The point of intersection is
e- x = ex
x = 0
Then, y = 1
So, the point of intersection is (0, 1).
Area of bounded region ABC
The solution of the differential equation (kx - y2 ) dy = (x2 - ky) dx is
x3 - y3 = 3kxy + C
x3 + y3 = 3kxy + C
x2 - y2 = 2kxy + C
x2 + y2 = 2kxy + C
The solution of the differential equation is
y = ex + C
y = x + ex + C
y = xex + C
y = x(ex + 1) + C