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Integrate initial answer

we are now going to talk about integration using initial conditions and particular solutions if you take a look at the graph here you'll notice these are all the same shape what we have here are the general solutions of an antiderivative the antiderivative is X cubed minus X minus 2 and if you take a look at these graphs they all obviously are cubic graphs notice if you look closely though all of these curves although they have the same shape pass through different points on the y-axis that's because the constant of integration the C value would be different for each one of these curves the variable part is identical for all of them but what differs is the constant which of course we know affects the height of the curve in the coordinate plane you'll notice one of these curves running right here through the point 0 negative 2 is in boldface that is the case in which we were given an initial condition specifically that the point 2 comma 4 had to lie on the original curve you can see it up here on the upper right the point 2 comma 4 right there any time you're given an initial condition a condition that has to be true of the original function we are able to find a particular solution to the antiderivative meaning we are actually able to find what that plus C value is and we're going to take a look at examples of how you do this in this first one we are asked to find the general solution of f prime of x equals e to the X and then find the particular solution given that capital f of 0 is equal to 3 notice we are first asked to find a general solution that implies we want an answer that has plus C in it in order to find the original function we need to take the antiderivative of e to the X so the original function f of X is going to be found by doing the antiderivative of e to the X DX notice the notation I am using to set this up so think of the rules we know antiderivative of e to the X is simply e to the X but don't forget your constant of integration this is where we can now use the condition we're given the initial condition that F of 0 had to equal 3 well 0 is your X we can simply put that in for X and e to the X plus C f of X has to equal 3 the function needs to equal 3 you're essentially just setting up a little equation to allow you to solve for C each of the 0 of course is 1 so C is equal to 2 so what that means in the end is that our function f of X is e to the X plus 2 this is what we refer...