Find the shaded area between f ( x ) = x 3 + 2 x 2 \textcolor{orange}{f\left(x\right)=x^3+2x^2} & g ( x ) = x + 2 \textcolor{blue}{g\left(x\right)=x+2} .

In this problem, we're asked to find the shaded area between the two functions f of x equals x cubed plus two x squared and g of x equals x plus two. Now we can see based on the sketch of these functions on our graph here that we're going to have to set up two different integrals and add them together because our function f of x starts out on the top but then becomes the bottom function as it crosses that other function g of x. Now we aren't given the bounds of what these integrals are in the interval that we're finding this shaded area over, so that means that the first thing that we need to do here is actually determine what these intersection points of our functions are so that we can properly bound our integrals. So let's start there. We need to set these two functions equal to each other, so x cubed plus two x squared equals x plus two, and we need to solve this for x. So let's pull everything over to the same side by subtracting x and two on both sides. That then cancels everything on that right hand side and leaves us with x cubed plus two x squared minus x minus two equals zero. Now looking at this, this may be a bit tricky to figure out how we're gonna solve for x here. But looking at this, I'm gonna go ahead and try to factor this by grouping. If I group these first two terms together, then I can pull an x squared out of both of them, leaving me with that x plus two inside of that factor. Then for these next two terms, I can factor a negative one out of both of these. So this negative one factors out, leaving me with X plus two. So since these two factors are the same, I can then factor that X plus two out. So this is x plus two times x squared minus one, and this is then equal to zero. So now with these individual factors, I can set them equal to zero as well. So if I take x plus two, set that equal to zero. If I subtract two on both sides here, I'll get that x is equal to negative two. Then doing the same thing for x squared minus one, setting that equal to zero. Adding one to both sides here gives me that x squared is equal to positive one. Then taking the square root on both sides, I get that x is equal to plus or minus one. So I have my three intersection points here negative two, negative one, and positive one. Now I'm going to go ahead and label these points on my graph here just so that it's clear how exactly we're going to bound our integral. So with this point, we need our lowest value that's at negative two. Then this point here would be negative one, and then our final intersection point is here at positive one. So we know that we need to set up our our two integrals. Our first one bounded from negative two to negative one and our second bounded from negative one to positive one. So let's go ahead and actually set up our area integral in order to find this shaded area. So our area here is going to be equal to the integral from negative two to negative one of that top function minus our bottom function. In this case, we can see that the function on top is that function f of x, which is x cubed plus two x squared. So taking that x cubed plus two x squared and subtracting off that bottom function, which here is that x plus two. Now that is that first chunk of area right there. And then as our functions cross each other and now my function g of x is on top, I then need to add a second integral here from negative one to positive one of what is now my top function, x plus two, minus what is now my bottom function x cubed plus two x squared. Now I'm going to go ahead and do some simplification here. We have our area integrals properly set up, and now we just need to evaluate them using the fundamental theorem of calculus. So first thing I'm going to do here is actually clean this up and write these polynomials in standard form in the integrand, just to make it a bit easier to deal with. So this is the integral from negative two to negative one of X cubed plus two X squared, and then distributing that negative minus X minus two DX. And then for my second integral here, negative one to positive one of negative X cubed minus two X squared plus X plus two. And I created still with respect to X. Now I want to go ahead and find our anti derivatives here for that first integral. We have one fourth X to the power of four plus two thirds X cubed minus one half X squared minus two X, then bounded from negative two to negative one. Then for our second integral here, we have the same exact stuff, but now the sign is different. So this first term is going to be a negative one fourth x to the power four minus two thirds x cubed plus one half x squared plus two x bounded from negative one to positive one. Now I want to go ahead and plug in those upper and lower bounds. I'm going to continue this line over here just because I'm going to need some extra space. So this area is ultimately equal to one fourth times negative one to the power of four plus two thirds times negative one cubed minus one half times negative one squared minus two times negative one. So that's that upper bound of my first integral plugged in here. I'm going to put it in parentheses and then subtract off that lower bound plugged in. So this is one fourth times negative two to the power of four plus two thirds times negative two cubed minus one half times negative two squared minus two times negative two. So that's that upper bound. And this all just represents that first integral. So that's that first chunk of area that we're dealing with here. And we are adding this together with our other integral with those balance plugged in. So now working on the second half of that area here this is going to be negative one fourth times one to the power of four minus two thirds times one cubed plus one half times one squared plus two times one. So that's my upper bound. And then subtracting off that lower bound plugged in, my lower bound for this integral is negative one. So this is going to be negative one fourth times negative one to the power of four minus two thirds times negative one cubed plus one half times negative one squared plus two times negative one. So that is that lower bound and this is that second integral. So this all represents this integral right here. Now we have a lot of algebra to do if we're doing this by hand. You should go ahead and try to work through this algebra on your own and I'm gonna show you what simplified answers we get to. Since we're dealing with a lot of fractions you're going to want to put these with common denominators so that we can actually add these fractions and combine all of our like terms. Now what this first yellow or this first section that I have highlighted in yellow here will work out to eventually giving everything a common denominator doing all of our multiplication, this will eventually just give us five twelfths. So all of that works out to be five over 12. Now for this green section right here, all of this works out to be 32 over 12. So that's that green section right there. So all of that messy algebra, once we do all that multiplication, get all of our common denominators, we are just left with five over 12 plus 32 over 12 giving us a final answer here of 37 over 12. Now if you want to report your answer as a fraction this is your final answer. But if you want to report your answer as a decimal and you just plug everything into your calculator your final answer here should be about 3.083. Now again, this final answer here represents the shaded area that we see in our graph here. If you have any questions, feel free to let us know in the comments. I'll see you in the next one.

9. Graphical Applications of Integrals

Area Between Curves

Calculus

9. Graphical Applications of Integrals

Area Between Curves

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Multiple Choice

Find the shaded area between $f (x) = x^{3} + 2 x^{2}$ & $g (x) = x + 2$ .

2.25

3.08

0.42

2.67

Verified step by step guidance

Identify the points of intersection between the functions f(x) = x^3 + 2x^2 and g(x) = x + 2 by setting f(x) = g(x) and solving for x.

Solve the equation x^3 + 2x^2 = x + 2 to find the x-values where the curves intersect. This will involve rearranging the equation to x^3 + 2x^2 - x - 2 = 0 and finding the roots.

Once the points of intersection are found, determine the intervals over which f(x) is above g(x) and vice versa.

Use the formula for the area between curves: A = ∫[a to b] (f(x) - g(x)) dx + ∫[b to c] (g(x) - f(x)) dx, where a, b, and c are the x-values of intersection.

Evaluate the integrals separately over the determined intervals and sum the absolute values to find the total shaded area between the curves.

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Struggling with Calculus?

Find the shaded area between f(x)=x3+2x2\(\textcolor{orange}{f\left(x\right)=x^3+2x^2}\) & g(x)=x+2\(\textcolor{blue}{g\left(x\right)=x+2}\).

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Area Between Curves practice set

Find the shaded area between $f (x) = x^{3} + 2 x^{2}$ & $g (x) = x + 2$ .