When f(x)=0, x=0 or 2. Because k is a type of scaling factor, we can plot y=f(x)=x(2-x)^2 to see what the graph looks like. What we see is that there is a hump between x=0 and x=2. The effect of k is merely to change the height of the hump (a maximum). The area under the hump can be broken down into very thin rectangles of height y and width dx. The area is the sum of the areas of these rectangles: ydx. As dx becomes infinitesimal, the sum is the integral, which I'll write as kS[0,2](x(2-x)^2dx), where [0,2] are the limits.
kS[0,2](x(4-4x+x^2)dx)=kS[0,2]((4x-4x^2+x^3)dx)=k[0,2](2x^2-4x^3/3+x^4/4)=k(8-32/3+4)=4k/3. To make this area 1, k=3/4.
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