A new car battery is sold for 100 with a 3-year limited warranty. If the battery fails at time t (0<t<3), the battery manufacture will refund 100(1-t/3). After analyzing the battery performance, the battery manufacturer uses the continuous uniform distribution on the interval (0,n) as the model for time until failure of the battery (n years). The battery manufacturer determines that the expected cost of the warranty is 10. Find n.
So the answer in the book is simply $\int_0^3100(1-t/3)\cdot\frac1n dt=10$ (solve for n)
However, I don't understand why they are integrating in terms of t since isn't $f(n)=\frac1n$ therefore, in my mind, it should be
$E(N)=\int_0^3100(1-t/3)\cdot\frac1n dn$ (range 0<$n$<3 since if $t>0$ no payment)
The failure time $T$ is a random variable. The support's upper bound $n$ is a constant.
You do not integrate with respect to constants; they have only one value.
The expectation is a weighted integral of the function over the support with respect to the random variable.
$$\begin{split}\mathsf E(N(T)) &=\int_0^n N(t)~ f_T(t)~\mathsf d t \\ 10 & = \int_0^{\min(3,n)} 100(1-t/3)\cdot~\tfrac 1n~\mathsf d t\\&= 15\end{split}$$