Suppose under the null hypothesis, the density of a test statistic T is $f_{H_0}(t)=2t\space I(t∈[0,1])$, and under the alternate, the test statistic T is $f_{H_a}(t)=2(1−t) \space I(t∈[0,1])$.
Suppose $H_0$ is true (so T has density $f_{H_0}(t)$.) What is the probability that $f_{H_0}(T)/f_{H_a}(T)≤0.5$?
Is this the correct way to solve this question: $\int_0^{0.5}\frac{2t}{2(1-t)}dt$ ?
By definition, $$\require{cancel}\begin{align} {f_{H_0}(T)\over f_{H_a}(T)}\le 0.5&\iff{\cancel2T \over \cancel2(1-T)}\le0.5\\ &\iff2T \le 1-T \\ &\iff T\leq 1/3 \end{align}$$ Now, under the null hypothesis, the probability that $T\le 1/3$ is given by $$\mathbb P(T\le 1/3\mid H_0) = \int_0^{1/3}f_{H_0}(t)\ dt =\int_0^{1/3} 2t\ dt $$