I have the following integration problem. I am dealing the following integral:
$$\int_0^{10} max\left[ \frac{w}{0.1}, V \right]d F(w) = \int_0 ^{10} max\left[ \frac{w}{0.1}, V \right] f(w)dw $$ $$ with \quad F(w) \sim unif[0,10] \quad and \quad f(w) =0.1 $$
Where $V$ is a constant. Now, I found on other sources (like here how to solve double integral of a min function) that in these cases, one needs to divide the integration space to incorporate the maximization problem. Bearing that in mind, I came up with this:
$$ \int_0 ^{10} max\left[ \frac{w}{0.1}, V \right] f(w)dw = \int_{0.1V}^{10} \frac{w}{0.1}f(w)dw + \int_0^{0.1V} Vf(w)dw $$ So that, $$ \int_{0.1V}^{10} \frac{w}{0.1}f(w)dw + \int_0^{0.1V} Vf(w)dw = \frac{1}{0.1}\left[ \frac{w^2}{2} \right]^{10} _{0.1V} *0.1 + \left[V*w \right]_0^{0.1V} * 0.1 $$ $$= \frac{10^2}{2} - \frac{(0.1V)^2}{2} + (0.1)^2 V$$ $$= 50- \frac{0.01 V^2}{2} + 0.01 V$$
Still, I am not sure if this is the right way to go. 1) Is changing the integration limits the appropriate way to go and 2) If that were the case, is this result correct?
I really appreciate your comments.
