$P(X<Y)$ where X and Y are exponential with means $2$ and $1$

Yes, your working is fine, you are integrating the relevant region over the joint pdf.

\begin{align} P(X < Y) &=\int_0^\infty P(X<Y|Y=y)f_Y(y) \, dy \\ &= \int_0^\infty \int_0^y f_X(x)f_Y(y) \,\, dx dy \\ &= \int_0^\infty \exp(-y)\left(1-\exp\left(-\frac{y}2 \right)\right) \,dy \\ &=\int_0^\infty \exp(-y) \, dy - \left( \frac{2}{3} \right)\int_0^\infty \left(\frac32\right)\exp\left( - \frac{3y}{2}\right) \, dy \\ &= 1- \frac23 \\ &= \frac13 \end{align}

In general if $X \sim \operatorname{Exp}(\lambda_X)$ and $Y \sim \operatorname{Exp}(\lambda_Y)$,

\begin{align} P(X < Y) &=\int_0^\infty P(X<Y|Y=y)f_Y(y) \, dy \\ &= \int_0^\infty \int_0^y f_X(x)f_Y(y) \,\, dx dy \\ &= \int_0^\infty \lambda_Y\exp(-\lambda_Yy)\left(1-\exp\left(-\lambda_X y \right)\right) \,dy \\ &=\int_0^\infty \lambda_Y\exp(-\lambda_Y y) \, dy - \left( \frac{\lambda_Y}{\lambda_X + \lambda_Y} \right)\int_0^\infty \left(\lambda_X+\lambda_Y\right)\exp\left( - (\lambda_X+\lambda_Y)y\right) \, dy \\ &= 1- \frac{\lambda_Y}{\lambda_X+\lambda_Y} \\ &= \frac{\lambda_X}{\lambda_X+\lambda_Y} \end{align}