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

we're asked to calculate the area of the shaded region between f of x and g of x contained specifically between x equals negative four and x equals negative two. Now we can see this shaded region on our graph here in between our two functions and we have our balance from x equals four x equals negative four to x equals negative two that's containing that area. Now looking at this, we can see that this is underneath the x axis, so you may be worried about how exactly this area integral will work out, but we don't even have to worry about the fact that this is below the x axis because we can set up our integral the same exact way. Bounded from a to b here from negative four to negative two of our top function minus our bottom function. So here we can see that the function on top is this f of x function, which happens to be negative two over x squared. Then our bottom function containing our area is this g of x, which is equal to negative one fourth x squared plus two x plus four. So let's go ahead and set up our area integral here. This is gonna be the area or the integral from a negative four to negative two. Those are the bounds given to us in our problem of our top function, which here is that negative two over x squared minus our bottom function, which is a longer polynomial one fourth negative one fourth x squared plus two x plus four integrated with respect to x. So with this area integral set up, all we have to do is evaluate it as we already know using all of our integral rules in order to find this shaded area. So let's go ahead and distribute this negative into our parentheses here and clean this up a little bit. So this is the definite integral from negative four to negative two. I'm gonna go ahead and write this first term as negative two times x to the power of negative two, just to make it more clear how we're going to use the power rule. And then distributing that negative, this is plus one fourth x squared minus two x minus four d x. So let's go ahead and use our power rule here and get our antiderivative. So for this first term negative two x to the power of negative two, if I put that constant negative two out front, just focus on that x to the power of negative two. If I add one to that power, that's gonna give me negative one dividing by negative one here. So that is the same thing as multiplying by negative one. So negative one times x to the power of negative one and then plus one fourth times one third x cubed, applying our power rule there. Then here two times a one half x squared minus four x and then bounded from negative four to negative two. Now before we plug in our bounds and use our fundamental theorem of calculus here, I'm going to go ahead and clean up all of the fractions and constants that are happening. So here a negative two times a negative one is gonna be positive two. And then x to the power of negative one is the same thing as one over x. So this is two over x plus one over 12 x cubed multiplying that one fourth times one third. And then here my two and one half are going to cancel. So this is minus x squared minus four x bounded from negative four to negative two. Now we can go ahead and apply those bounds since we've done some simplification here and we just have some algebra. So plugging in that upper bound here, this is two over negative two and then plus one twelfth times negative two cubed minus negative two squared minus four times negative two. So that is my upper bound plugged in. We have we have to subtract off that lower bound plugged in. So this is minus two over negative four plus one twelfth, negative four cubed minus negative four squared minus four times negative four. So that is everything all plugged in there. We just have some algebra to go through for now. You can go ahead and just plug all of this into your calculator at once to get your final answer. And if you're doing that, you can feel free to skip this video to the next couple of steps until we get that final answer, but if you want to see how I'm thinking through this algebraically and dealing with all of these fractions without using a calculator, you can keep watching from here. So looking at that first term, I'm just going to go term by term here. Two over negative two is negative one, so this is negative one. Negative two cubed is going to be negative eight. Negative eight over 12, if I reduce that fraction, is going to be negative two thirds. So this is minus two thirds, then negative two squared, that's four, and then I still have that minus up front, so this is minus four, and then negative four times negative two, that's positive eight. So we are adding eight here. So that's that upper bound and then subtracting off here, simplifying this fraction, two over negative four is going to be negative one half. So I have negative one half plus negative four cubed over 12. This fraction is going to reduce to actually negative 16 over three. And then negative four squared, that is 16 and I'm subtracting that off, so minus 16. And then negative four times negative four that is positive 16, so that's that term there. Now here I can already see that this negative 16 and this positive 16 are going to cancel, so those are getting canceled out. And then over here on this side I have a negative onetwo minus twothree minus four plus eight. I'm going to go ahead and focus on combining all those whole numbers first. Negative one and negative four is negative five plus eight is positive three. So this is three minuteus two thirds, and then here minus, I have these two fractions here that's negative one half and negative 16 over three. I'm going to go ahead and make that into a into fractions with like denominators to make it easier to combine them here. So this is a negative threesix and negative 32 over six, six giving those like denominators here. Now I want to go ahead and do the same thing to this three and this two thirds. If I give three a like denominator of three, that's the same thing as nine over three. So with these fractions simplified, nine thirds minus two thirds is seven thirds and then minus here negative three and negative 32, that is negative 35 over six. If I give these like denominators here, I can go ahead and multiply seven thirds by two to make this 14 plus 35 over six. And then this gives me a final fractional answer of 49 over six. So if you wanna report your final answer as a fraction, that is your final answer. But if you wanna go ahead and divide that out into a decimal, 49 divided by six gives us a final answer here of 8.17, rounding that off to two decimal places.

9. Graphical Applications of Integrals

Area Between Curves

Calculus

9. Graphical Applications of Integrals

Area Between Curves

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Multiple Choice

Calculate the area of the shaded region between $f (x)$ & $g (x)$ contained between $x = - 4$ & $x = - 2$ .

8.17

4.17

-12.17

0.50

Verified step by step guidance

Step 1: Identify the functions f(x) and g(x) from the graph. From the image, f(x) = -2/x^2 and g(x) = (1/4)x^2 + 2x + 4.

Step 2: Determine the interval of integration. The shaded region is between x = -4 and x = -2.

Step 3: Set up the integral to calculate the area between the curves. The area is given by the integral of [g(x) - f(x)] over the interval [-4, -2]. This can be expressed as: ∫[-4 to -2] [(1/4)x^2 + 2x + 4 - (-2/x^2)] dx.

Step 4: Simplify the integrand. Combine the terms inside the integral: (1/4)x^2 + 2x + 4 + 2/x^2.

Step 5: Evaluate the integral. Break it into separate integrals for each term: ∫[-4 to -2] (1/4)x^2 dx + ∫[-4 to -2] 2x dx + ∫[-4 to -2] 4 dx + ∫[-4 to -2] 2/x^2 dx. Compute each integral step by step to find the total area.