 In this problem we need to find what is the maximum deflection of the following beam with flexural rigidity equal to EI and we need to use the method of direct integration. As you can see here this beam is simply supported, we have two pin supports here and we have a uniform load with load intensity equal to W and the reaction forces these supports are given. They are equal to WL divided by 2. Then since we have all the reaction forces and the applied forces we can start calculating what is the distribution of moments as a function of X. We have here our beam, this is the reaction force WL divided by 2 and this is the distributed load, so at a point X we have that first well first we can remember that we define this internal distribution of moments as positive so if we have compression in the upper part of the beam so this force here, this vertical reaction force is created in a moment here like this and therefore the internal reaction goes in the opposite direction then this is positive, so the contribution of this moment created by this force is positive and is equal to WL divided by 2 times the distance times X and now the contribution of the load in this case this is creating a moment here like this therefore the internal reaction goes like this and this is negative so this is negative and the magnitude is equal to the, well this is a rectangular load so first the area of this load which is equal to X times W and the distance is the distance from the centroid of this area to the point X so this is X divided by 2 times X divided by 2 then this is equal to WL divided by 2 times X minus WX squared divided by 2 then now we need to calculate what is the curvature of this beam so we can use moment curvature relationships we have that the moment is equal to minus EI second derivative of V VX so we can integrate here, therefore we have that M is equal to, as we said before then if we integrate here once we have that EI dV dx is equal to where A is the constant of integration and of course this is equal to the slope theta we can integrate once more then we have one equation for the slope and one equation for the deflection and now we need to determine what are the constants of integration so we can use boundary conditions we have that the deformed shape of this beam is something like this so it is symmetric then from this symmetry we have that at this point at the mid span of the beam we have that the slope here is equal to 0 as I said before because of symmetry and we have as well that at this support at this one for instance at X equal to 0 the deflection is equal to 0 so these are our two boundary conditions and we can apply them to solve for the constants of integration so from equation 1 we know that theta is equal to 0 at L halves so we have that then if we solve this equation we find that the constant of integration A is equal to WL to the power of positive 3 divided by 24 and if we apply the second boundary condition at 0 the deflection is 0 we have that B is directly equal to 0 therefore we can already calculate what is the final expression for the deflection we have that deflection from equation 2 we have that this is equal to now the problem is asking what is the maximum deflection of the beam so without solving we can see of course that the maximum deflection of the beam occurs at L halves but if you don't see that of course we can calculate it we can calculate dv dx equal to 0 and from this equation if we take derivatives here and we say that dv dx is equal to 0 we can calculate what is the point X for which the deflection is maximum and of course you will find that X is equal to L divided by 2 and then if we substitute this expression this point X here in this formula we find that the maximum deflection of the beam is equal to