. This is an integrationproblem, to calculate the center of mass (center ofgravity) for a continuous distribution. Specifically,we need the horizontal component of the center ofmass.The height varies from h to zero according to this2function: y(x) = h ( - 1).The constants hand e replace 1.00 m and 3.00 m. There is also athickness t and a density p. You need two integrals,the total mass and the center of mass. Possiblysurprisingly, you don't actually need the numbers t,h, and p.Axy(x)The column at x has a mass Am = (density * volume)= y(x) p t Ax. You add all the Am values to get thetotal mass M. The sum becomes an integral:М — pt|y(x) dæFor the center of mass, you add each column's x Am,and divide by M:ptx y(x) dxXcMCalculate xc. The only quantity you'll need is e = 3%3Dm.

Given: y(x)=hxl-12h=1.00 ml=3.00 m Also, the mass of the distribution can be calculated as:…

Answered: This is an integration problem, to…

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