 Welcome to the GVSU Calculus screencasts. In this episode, we'll see how initial conditions determine specific anti-derivatives of our function. And just as a quick reminder, recall that an anti-derivative of a function little f is a function big f, whose derivative is little f. Consider the example of the function little f, defined by 2-x squared if x is less than 1, and 1 over x if x is bigger than or equal to 1. Our goal is to find one anti-derivative big f of little f that satisfies the initial condition big f of 0 is negative 1. The graph of little f is shown in red, and we can use this graph as we've done in the past to determine general information of a big f, where big f is increasing, decreasing, concave up, concave down. However, the graph of little f isn't made up of linear pieces or circular pieces as we've worked within the past, so we're not going to be easily able to use the net signed area interpretation of the depth and integral to get any information about big f. However, we can find anti-derivatives of little f for each of the pieces of little f, and we'll use those anti-derivatives to piece together an anti-derivative big f. So we want to find anti-derivatives for each piece of little f. The piece that defines little f on x minus 1 is 2 minus x squared. Pause the video for a moment, find all the anti-derivatives of 2 minus x squared, and then resume the video when you're ready. One anti-derivative of 2 minus x squared is 2x minus 1 third x cubed, because the derivative of 2x minus 1 third x cubed is 2 minus x squared. We know that every anti-derivative of 2 minus x squared, then, is just some translation of this 2x minus 1 third x cubed. So all the anti-derivatives of 2 minus x squared have the form 2x minus 1 third x cubed plus c, where c is some constant. So now consider the part of f that's defined when x is greater than or equal to 1. Pause the video, find all the anti-derivatives of 1 over x, and resume when you're ready. We know one anti-derivative of 1 over x is the natural log of x, so every anti-derivative of 1 over x is the natural log of x plus some constant. We'll call that constant k, and note that this constant k is potentially different than the constant c we found in our earlier anti-derivative. So putting these pieces together gives us capital F, our anti-derivative, a little f, having the form 2x minus 1 third x cubed plus c if x is less than 1, and the natural log of x plus k if x is bigger than or equal to 1, where c and k are some probably different constants. Our job now is to find these constants c and k so that our particular anti-derivative of f satisfies the condition big f of 0 is negative 1. Pause the video for a moment and use the condition that f of 0 is negative 1 to find the value of the constant c. Go ahead and resume when you're ready. Now if we want f of 0 to have a value, notice that 0 is less than 1, and for any x less than 1, f of x is 2x minus 1 third x cubed plus c. This means that we need to have f of 0, big f of 0, equaling 2 times 0 minus 1 third times 0 cubed plus c. Now if f of 0 is going to be negative 1, then this equation tells us that c has to be negative 1. This tells us that big f of x is 2x minus 1 third x cubed minus 1 if x is less than 1, and the natural log of x plus k if x is bigger than or equal to 1, and on the interval x less than 1 the graph of f is this looks like this blue curve that we've drawn. To finish finding big f all we need to do now is find the value of this constant k. Now pause the video for a moment and consider the fact that big f is a differentiable function, and think about how that can help us find the value of k, and resume the video when you're ready. Recall the fact that a differentiable function is continuous and smooth, so in particular our big f is a continuous function and hence must be continuous at x equals 1. So the left hand limit of f at 1 and the right hand limit of f at 1 have to both exist and be the same. Now the right hand limit of f at 1 is the natural log of 1 plus k, and the left hand limit at 1 is 2 times 1 minus 1 third times 1 cubed minus 1, so those two quantities have to be the same. Solving for k gives us k equals 2 thirds. So we found our anti-derivative big f. It's 2x minus 1 third x cubed minus 1 if x is less than 1, and the natural log of x plus 2 thirds of x is bigger than or equal to 1, and the graph of f is shown in blue. And this example illustrates how an initial condition determines a specific anti-derivative of little f. So suppose we wanted to find a different anti-derivative, maybe one that we'll call g that satisfies the condition g of 2 equals 0. Pause the video for a moment and find a rule for an anti-derivative big g so that g of 2 is 0. Then draw the graph of g and big f on the same set of axes and see if you can figure out how they're related. And resume the video when you're ready. Now the same process that we used before shows that big g as an anti-derivative of little f has to have the form 2x minus 1 third x cubed plus c if x is less than 1 and the natural log of x plus k if x is bigger than or equal to 1. And we need to find the constants c and k. We want g of 2 to be 0, and notice that we use the rule ln x plus k when we evaluate g at numbers that are bigger than 1. So g of 2 is ln 2 plus k. We want that to be 0. That makes big k the opposite of the natural log of 2. Now to find the constant big c, we use the fact that big g has to be continuous at x equals 1 again. And then we evaluate the expression 2x minus 1 third x cubed plus c at 1, ln x minus ln 2 at 1, set those equal, solve for c, and we get big c to be negative 5 thirds minus the natural log of 2. So we found our formula for big g of x. It's 2x minus 1 third x cubed minus 5 thirds minus the natural log of 2 if x is less than 1 and ln x minus ln 2 if x is bigger than or equal to 1. And notice that g of x and f of x differ by a constant. If we look when x is less than 1, they differ by 2 thirds plus ln 2 and the same when x is bigger than or equal to 1. So g of x is equal to big f of x minus 2 thirds minus ln 2 and f and g just differ by a constant. And if we draw the graphs of big f and big g, big f in blue, big g in magenta, we can see that big g is just a vertical translation of the graph of big f, illustrating the fact that big f and big g just differ by a constant. That concludes this edition of this GDSU screencasts, and we look forward to seeing you again in the future.