Suppose that the operating lifetime of a certain type of battery is an exponential random variable with $\theta$ $= 2$ (measured in years). Find the probability that a battery of this type will have an operating lifetime of between 1 and 3 years.
My attempt:
First:
$f(t)=(1/2)e^{-t/2}$ for $1<t<3$
$$P(T>1) = \int_1^{\infty} f(t) dt$$
$$P(T>1) = 1-\int_0^{1} (1/2)e^{-t/2} dt$$
$$P(T>1) =1+e^{-t/2}|1,0 $$ $$P(T>1) = 1+e^{-\frac{-1}{2}}-e^{0} =1+e^{-1/2} -1 = e^{-1/2} $$
Then
$f(t)=(1/2)e^{-t/2}$ for $1<t<3$
$$P(T>3) = \int_3^{\infty} f(t) dt$$
$$P(T>3) = 1-\int_0^{3} (1/2)e^{-t/2} dt$$
$$P(T>3) =1+e^{-t/2}|3,0 $$ $$P(T>3) = 1+e^{-\frac{-3}{2}}-e^{0} =1+e^{-3/2} -1 = e^{-3/2} $$
Now
$e^{-1/2}-e^{-3/2}=.38340$
I was thinking is this the correct way to solve this problem. Can someone please help correct this if the working is wrong? Can someone also please show me a simpler way if possible?