Find the area between f ( x ) = x 2 − 4 f\left(x\right)=x^2-4 & g ( x ) = − x 2 + 4 g\left(x\right)=-x^2+4 .

this problem, we're asked to find the area between the function f of x equals x squared minus four and g of x equals negative x squared plus four. Now we're not actually given a graph here, so that means we need to go ahead and sketch these functions ourselves so that we really have a good idea of what exactly is going on and how exactly we should set up this integral in order to ultimately find the area between our two functions. Now the other thing that we can notice here is that we are not given a specific interval in order to bound our integral. So we need to figure out what this interval is ourself by determining the intersection points of our two functions. Now because we have to go ahead and graph these functions as well, let's go ahead and start there. So starting with that function f of x equals x squared minus four, I know that this is an upward facing parabola that's been translated down four units. So it has a vertex point down here where y is equal to negative four. Now thinking about the other points of this function, I can go ahead and plot some points by plugging in things like x equals one, x equals two. If I plug in x equals one here, so looking at this as a table x and f of x, if I plug in x equals zero, we know that that's negative four. That's our vertex point. Then if I plug in positive one here, one squared is one, one minus four is negative three. So I have one and negative three and because parabolas are symmetric, I know that I'm gonna have a point here at negative one negative three. Then if I plug in two, so two squared is two is four minus four is zero. So I'm gonna have another point right here. And then again, since we're symmetric again there on the other side. So with that in mind, I can go ahead and connect all of these points and plot that first function, connecting them as best I can. And that's my function f of x. And we also want to plot our function at g of x. G of x is negative x squared minus four. This is a downward facing parabola with a vertex that it's been translated up four points. So up here. So again, plotting a couple of points here, x and g of x. If we plug in zero, we know that that's positive four. Then if we plug in a one, one squared is one, but that's negative. So negative one plus four is three. So I have that point at one three symmetric. So I also have that on the other side, negative one three. Then if I plug in two here, two squared is four, but it's negative two. So negative four plus four is zero. So I can see that these are actually at the exact same points as my other one, two, zero. And I can go ahead and connect all my points here from either side of my parabola. And we can see that the area that we're looking for is right in between these two functions contained here in the middle with this sort of oval shape. Now we actually already found our intersection points because we see that they have points at two zero and also at negative two zero because they're symmetric. Now if we hadn't figured it out by graphing, we could go ahead and just set these two functions equal to each other. So I want to go ahead and do that method as well. If If we didn't already know our intersection points, we would need to take x squared minus four, set that equal to negative x squared plus four and solve for x. So if I add x squared to both sides here, that cancels over here, gives me two X squared minus four equals four. And then if I go ahead and move this four over by adding it, I get that two X squared is equal to eight. Then dividing both sides by two here that will cancel. I get that x squared is equal to four and then finally taking the square root on both sides, this gives me both of my intersection points that x is equal to plus or minus two. So we can see that that corresponds with what's going on with our graph here. We have our intersection point at negative two and at positive two. So these represent the bounds of our integral. So let's go ahead and set up that area integral here. So our area integral, our area is going to be from negative two to positive two. That's what we just found of our top function minus our bottom function. Now in this case, we can see that that function that's on top is our function g of x. Remember that in our area formula, these are just arbitrary functions. So if in a problem you're given functions that are called f of x and g of x, that doesn't mean that f of x is always your top function. That's why we need to sketch our graph and know exactly what's going on there. So this time we have our function g of x on top. So that's negative x squared plus four. And then we have our function f of x on the bottom. That's x squared minus four. So minus x squared minus four and then dx. So we wanna go ahead and evaluate this integral here. I'm gonna do some cleaning up by distributing this negative. So this is from negative two to positive two. Negative x squared minus x squared gives me negative two x squared and then plus four plus four again is plus eight. So this is negative two x squared plus eight dx and we want to go ahead and evaluate this integral. So finding my antiderivative here negative two x squared using the power rule, if I take that negative two and then multiply it by one third x cubed, that's that antiderivative, and then plus eight x from negative two to positive two. So if I go ahead and combine that negative two and that positive one third, this is just negative two thirds. And then I wanna go ahead and plug in my bounds. So starting with that upper bound, so this is negative two thirds times two cubed plus eight times two, and then we're subtracting off what's going on in that lower bound. I'm going to go to my second line here just because I don't have enough space, but this is negative two thirds times a negative two cubed plus eight times negative two, plugging in that lower bound. And I'm just gonna get some more space here continuing on this line down here. So plugging in all of this, we wanna look at what's going on. I'm gonna go ahead and do all of the algebra out loud. But if you wanna go ahead at this point and just type this right into your calculator to get an answer, that's totally fine as well. I'm just gonna talk through all of the algebra that I would be doing by hand if I could not access a calculator. So I have negative two thirds times two cubed, that's that first term there. Two cubed is eight, so this would be negative 16 over three plus 16 because that's eight times two. Then with that lower bound there, negative two cubed is going to be negative eight. So this becomes a positive 16 over three and then plus eight times negative two that's minus 16. Now if I distribute this negative in here this ends up giving me negative 32 over three and then plus 32 because I have a positive 16 and then a negative negative 16 which is another positive 16 and if we want to make these have a common denominator here this is negative 32 over three and then if I multiply 32 times three over three that's gonna give me 96. So this is plus 96 over three. Now combining those fractions into one, this is 64 over three if we wanna report that final answer as a fraction Or if you wanna report this as a decimal, we can plug that into our calculator or do some long division, and we should get about 21.33. Now feel free to let us know if you have any questions, and I'll see you in the next one.