Find the area of the shaded region between f ( x ) = x 4 − x 2 f\left(x\right)=x^4-x^2 f ( x ) = x 4 − x 2 & g ( x ) = 3 x 2 g\left(x\right)=3x^2 g ( x ) = 3 x 2 .

this problem, we're asked to find the area of the shaded region between the functions f of x equals x to the power of four minus x squared and g of x equals three x squared. Now we can already see our functions sketched on our graph here so we can tell which function is on the top and which function is on the bottom. But we what we don't know is where exactly these functions intersect each other so that we can properly bound our integral and find this shaded area. So let's go ahead and start there. Now we need to figure out where exactly these functions intersect each other. So that means we need to set these functions equal to each other. So we want to set x to the power of four minus x squared equal to three x squared. Now I'm gonna go ahead and get everything over to the same side here. If I go ahead and subtract three x squared, that will cancel over here and I have x to the power of four and then negative x squared and negative three x squared is negative four x squared and that's all equal to zero. Now looking at this, I can go ahead and do some factoring here if I factor out an x squared. So factoring out an x squared here, this is x squared times x squared minus four, which is equal to zero. Now factoring this even further here, x squared minus four, is going to factor to x plus two times x minus two. So this is x squared times x plus two times x minus two, and that's all equal to zero. Now because I have these three separate factors and them being multiplied is equal to zero, I can set each of these individual factors equal to zero. So if I set x squared equal to zero, x plus two, and x minus two, and then solve for x in each of these little equations, I will have my intersection points. So for x squared being equal to zero, that would mean that x has to be equal to zero. Then for x plus two, if I subtract two from both sides, that tells me that x is equal to negative two. Then finally, if I add two to both sides for this equation, I get that x should be equal to positive two. So the question here is if I have three intersection points, which two of them am I supposed to use to bound my integral? But let's take a look at what's going on with our graph here. We can see that this point is x equals negative two. And then, of course, I have my zero point and my origin, and then I have x equal to positive two. Now at this point x equals zero, my function f of x kind of goes up and touches that function g of x, but then comes right back down. So that function g of x is on top the whole time. It's not that it intersects and then changes, so we don't have to worry about that. So that means that we could effectively split this up into two separate integrals from negative two to zero and then from zero to two because they do intersect at zero, but we don't have to do that because our function g of x is on top the whole time. So that means that we can just proceed in setting up our area integral with our intersection points of negative two and positive two because zero is just kind of an intermediate point where, yes, my functions do intersect each other, but they're not fully crossing and changing which function is on the top or on the bottom. So we can go ahead and set up our integral bounded from negative two to positive two of our top function, which in this case is g of x, that three x squared, minus our bottom function, which is x to the power of four minus x squared. Now we wanna go ahead and do some simplification of this integral. Before we get that antiderivative, I wanna go ahead and combine my like terms Three x squared, if I go ahead and distribute this negative, three x squared, and then I'm adding another x squared, that's gonna be four x squared, but I'm gonna put this in standard form. So I'm gonna put that x to the power of four upfront. So this is negative x to the power of four and then plus four x squared DX. So we can go ahead and find our antiderivative here, negative x to the power of four. That's gonna be negative one fifth X to the power of five, and then plus four over three X cubed bounded from negative two to positive two. Now we want to plug in those bounds here. So negative one fifth, two to the fifth, plus four thirds, two cubed, and then minus I'm gonna move this line here just so I have enough space. Minus negative one fifth times negative two to the power of five plus four thirds times negative two cubed. So at this point, if you have access to a calculator, you can just go ahead and plug this directly into your calculator, but I'm going to go ahead and talk through what algebra I would be doing if I didn't have access to a calculator and I was just working through this step by step by hand. Now starting with this first term, two to the power of five is 32, so this is negative 32 over five and then two cubed is eight. Eight times four is also 32, so this is plus 32 over three. So first two terms and then minus here again negative two to the power of five. This is going to be negative 32 negative 32. But then I have this negative one fifth. So that is positive 32 over five And then four thirds negative two cube, that's going to be negative eight times negative four, that's negative 32. So minus 32 over three. Now between these two sets of terms, we can see that we have some that we can go ahead and combine their fractions because they have common denominators. I have this negative 32 over five and then another negative 32 over five if I consider distributing this negative. So that's negative 64 over five. And then I have positive 32 over three and then another positive 32 over three because that's negative negative. That makes a positive. That gives us positive 64 over three. So plus 64 over three. Now because these two fractions do not have common denominators, I cannot just add them across, that's not how fractions work, but I can give them a common denominator by multiplying this negative 64 over five by by three over three and this 64 over three by five over five to give them that common denominator of 15. Now negative 64 times three, that's going to give us negative 192 and then plus 64 times five, that's going to be 320 and we gave them that common denominator. So going ahead and combining those fractions, now we can add those numerators across negative one ninety two and positive three twenty. That's going to give us positive one twenty eight with that common denominator of 15. So if we wanted to report our answer as a fraction this would be our final answer here, but if you want to report as a decimal here this would end up being 8.53 which is likely what you would get if you just at this step plugged everything into your calculator. So this gives us this shaded yellow area between our two functions. If you have any questions, feel free to let us know in the comments, and I'll see you in the next one.