Sketch the region bounded by f(x)=−(x−2)2+5\textcolor{blue}{f\left(x\right)=-\left(x-2\right)^2+5} & g(x)=4x\textcolor{orange}{g\left(x\right)=4x} on the interval [0,1]\left\lbrack0,1\right\rbrack. Then set up an integral to represent the region's area and evaluate.

Integral = ∫ 0 1 − x 2 + 1 d x \int_0^1-x^2+1dx ; Area = 0.67 0.67

Sketch the region bounded by f ( x ) = − ( x − 2 ) 2 + 5 \textcolor{blue}{f\left(x\right)=-\left(x-2\right)^2+5} & g ( x ) = 4 x \textcolor{orange}{g\left(x\right)=4x} on the interval [ 0 , 1 ] \left\lbrack0,1\right\rbrack . Then set up an integral to represent the region's area and evaluate.

this problem, we're asked to sketch the region bounded by the two functions f of x equals negative x minus two squared plus five and g of x equals four x on the interval from zero to one. And then we want to set up an integral to represent that region's area and evaluate. So let's go ahead and get started by graphing here. I'm gonna start with my easiest function because we have g of x equals four x, which we know is just going to be a straight sloped line. So if I go ahead and graph that here, I know that I have a point at zero zero and then also at one four. So if I go ahead and connect those with my line here, that's what g of x looks like. So that's my function g of x. Now we wanna graph f of x, which we know to be a parabola because we have this x minus two squared. And it's also written in vertex form, meaning that we can easily tell what the vertex of our parabola is. So if I have x minus h plus k, then this minus h, that makes h two. So I have a vertex located at two and then k here is is five. So that vertex is located at two, five. I can go ahead and just plot that point on my graph here, two, five. Now I can tell that this is a downward facing parabola because of this negative sign out front. And really the easiest thing to do here is just to go ahead and plug in some points so that we can see exactly where they fall. So I'm just gonna set up a chart here, x and f of x. So looking at my graph, I only need to plot a few points here, and it would be helpful to know what our function is when x is equal to zero, one, two, and three. Now we already know we have a point located at two, five. That is our vertex. So looking at our other points here, if we plug zero into f of x here, negative zero minus two squared plus five, that's going to give us a positive one. I'm gonna go ahead and graph that here. Zero one. Then if I plug a one in here, one minus two is negative one, negative one squared is positive one. But I have that negative out front. Negative one plus five is going to be positive four. So this is where my two functions actually intersect each other. And then since this vertex is right here, I know that this is going to be a symmetric parabola. So I'm going to have another point at three, four because it's starting to fall down on the other side past that vertex. So I know that it's symmetric there and this is what my parabola generally looks like on my graph here, just connecting all of those points. So now we want the area in between these two functions from zero to one. So we wanna go ahead and shade in that area here, specifically from zero to one. So I'm shading it in in yellow, and that's the area that we're looking for between f of x and g of x. So now that we have sketched our region with our graph here, we want to go ahead and set up an integral to represent this region's area and then actually evaluate it and find that area. So we want to set up a definite integral of our top function minus our bottom function. Looking at my graph here, I can see the function on top is my parabola f of x and my function on the bottom is g of x that line. So we already know that this area is going to be the region from zero to one. So those are the bounds of my integral that was given to me in my problem statement. And then that top function f of x, which is negative x minus two squared plus five, and then subtracting off g of x, so that is four x, and then integrating with respect to x. So with this integral set up, we just need to evaluate it. Now it's gonna be easier here if we go ahead and multiply everything out in order to get this as an actual polynomial written out in standard form as opposed to vertex form. So going ahead and doing that here, negative x minus two squared plus five. This is actually going to be negative x squared plus four x plus one, just multiplying all of that out. And then we're subtracting off that four x DX. Now looking at this, I have a positive four x and a negative four X. So that's our those are actually going to cancel each other out here. And that leaves me with just the integral from zero to one of negative X squared plus one DX. So let's go ahead and apply the power rule here, find this anti derivative. If I have negative x squared, the integral of this is negative one third x cubed and then plus x for the integral or anti derivative of that positive one. Then we're bounded here from zero to one. So all that's left to do is plug in our upper and lower bound, applying our fundamental theorem of calculus here. This is going to be negative one third times one cubed plus one. And then if I plugged zero into any of these terms, I would just get a zero. So we're subtracting off zero there. We really just have to worry about this negative one third. So subtracting off zero there. We really just have to worry about this negative one third times one cubed plus one. So one cubed is just one, so this is just negative one third plus one. We can think of one as being three over three. So if I take negative one third add three over three that's going to give me positive two thirds. So if you want to report your answer as a fraction you are all done here, But if you want this as a decimal, you should have gotten 0.67, just rounding that to two decimal places. Let us know if you have any questions here and let's keep going.

Sketch the region bounded by f ( x ) = − ( x − 2 ) 2 + 5 \textcolor{blue}{f\left(x\right)=-\left(x-2\right)^2+5} & g ( x ) = 4 x \textcolor{orange}{g\left(x\right)=4x} on the interval [ 0 , 1 ] \left\lbrack0,1\right\rbrack . Then set up an integral to represent the region's area and evaluate.

Integral = ∫ 0 1 − x 2 + 1 d x \int_0^1-x^2+1dx ; Area = 0.67 0.67

Integral = ∫ 0 1 ( − x 2 + 9 ) d x \int_0^1\left(-x^2+9\right)dx ; Area: 0.33 0.33

Integral = ; ∫ 0 4 ( − x 2 + 9 ) d x \int_0^4\left(-x^2+9\right)dx Area: 21.33 21.33

Integral = ∫ 0 1 − x 2 + 1 d x \int_0^1-x^2+1dx ∫ 0 1 − x 2 + 1 d x ; Area = 22.67 22.67

9. Graphical Applications of Integrals

Area Between Curves

Calculus

9. Graphical Applications of Integrals

Area Between Curves

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Multiple Choice

Sketch the region bounded by $f (x) = - {(x - 2)}^{2} + 5$ & $g of x equals 4 x$ on the interval $open bracket 0 comma 1 close bracket$ . Then set up an integral to represent the region's area and evaluate.

Graph showing the area between two curves, f(x) in blue and g(x) in orange, highlighted in yellow.

Integral = $\int_{0}^{1} (- x^{2} + 9) d x$ ; Area: $0.33$

Graph showing two curves, f(x) in blue and g(x) in orange, with the area between them shaded in yellow.

Integral = $\int_{0}^{1} - x^{2} + 1 d x$ ; Area = $0.67$

Graph showing the area between two curves, f(x) in blue and g(x) in orange, highlighted in yellow on the interval [0,1].

Integral = ; $\int_{0}^{4} (- x^{2} + 9) d x$ Area: $21.33$

Graph showing two curves, f(x) in blue and g(x) in orange, with the area between them shaded in yellow.

Integral = $\int_0^1-x^2+1dx$ ; Area = $22.67$

Verified step by step guidance

First, identify the functions f(x) = -(x-2)^2 + 5 and g(x) = 4x. These are a downward-opening parabola and a linear function, respectively.

Next, determine the points of intersection of f(x) and g(x) within the interval [0, 1]. Set f(x) equal to g(x) and solve for x: -(x-2)^2 + 5 = 4x.

Simplify the equation: -(x^2 - 4x + 4) + 5 = 4x, which simplifies to -x^2 + 4x - 4 + 5 = 4x. Further simplify to -x^2 + 1 = 0.

Solve for x: x^2 = 1, giving x = 1 (since we are only considering the interval [0, 1]).

Set up the integral to find the area between the curves from x = 0 to x = 1: A = ∫[0 to 1] [(f(x) - g(x))] dx, which becomes A = ∫[0 to 1] [-(x-2)^2 + 5 - 4x] dx.

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