Answer
$$2$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\int_x^1 {6ydydx} } \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^1 {\left[ {3{y^2}} \right]_x^1dx} \cr
& = \int_0^1 {\left[ {3{{\left( 1 \right)}^2} - 3{{\left( x \right)}^2}} \right]dx} \cr
& = \int_0^1 {\left( {3 - 3{x^2}} \right)dx} \cr
& {\text{Integrate with respect to }}x \cr
& = \left[ {3x - {x^3}} \right]_0^1 \cr
& {\text{Evaluating}} \cr
& = \left[ {3\left( 1 \right) - {{\left( 1 \right)}^3}} \right] - \left[ {3\left( 0 \right) - {{\left( 0 \right)}^3}} \right] \cr
& = 2 \cr} $$
