Integration is used to find the area between a curve and the #x#- or #y#-axis, and the shaded region here is exactly that area (between the curve and the #x#-axis, specifically). So all we have to do is integrate #4x-x^2#.

We also need to figure out the bounds of integration. From your diagram, I see that the bounds are the zeros of the function #4x-x^2#; however, we have to find out numerical values for these zeros, which we can accomplish by factoring #4x-x^2# and setting it equal to zero:

#4x-x^2=0#

#x(4-x)=0#

#x=0##color(white)(XX)andcolor(white)(XX)##x=4#

We will therefore integrate #4x-x^2# from #0# to #4#:

#int_0^4 4x-x^2dx#

#=[2x^2-x^3/3]_0^4-># using reverse power rule (#intx^ndx=(x^(n+1))/(n+1)#)

#=((2(4)^2-(4)^3/3)-(2(0)^2-(0)^3/3))#

#=((32-64/3)-(0))#

#=32/3~~10.67#