Condition (1):

Since, f(x)=e^-x sinx is a combination of exponential and trigonometric function which is continuous.

⇒ f(x)= e^-x sinx is continuous on [0,π].

Condition (2):

Here, f’(x)= e^-x (cosx – sinx) which exist in [0,π].

So, f(x)= e^-x sinx is differentiable on (0,π)

Condition (3):

Here, f(0)= e^-0 sin0=0

And f(π)= e^-πsinπ =0

i.e. f(0)=f(π)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(0,π) such that f’(c)=0

i.e. e^-c (cos c – sin c) =0

i.e. cos c-sin c = 0

i.e.

Value of

Thus, Rolle’s theorem is satisfied.