# What is the area bounded by y = 8 – 2x - x^2 and the x-axis?

• MHB
• Noah1
In summary, the region bounded by the graph of the function y = 8 – 2x - x^2 and the x-axis has an area of 9 units.
Noah1
Calculate the area of the region bounded by the graph of the function y = 8 – 2x - x^2 and the x-axis
Y = 8 - 2- x^2
0 = 8 – 2 – x^2
(-x – 4)(x – 2)
- x – 4 = 0 and x – 2 = 0
-x = 4 x = 2
X = - 4

Do I do this?

Y = 8 -2x -x^2
= 8x - (2x^2)/2 - x^3/3
= 8 - x^2 - x^3/3
= [8x - x^2 - x^3/3] – [8x - x^2 - x^3/3]
= [(8x 2) - 2^2 - 2^3/3] – [(8 x -4) - 〖-4〗^2 - 〖-4〗^3/3]
= [16 – 4 + 8/3] – [-32 + 16 - 64/3]
= 141/3 – [-16 - 311/3]
= 141/3 – [-16 + 211/3]
= 141/3 - 5 1/3
Area = 9 units

Define the boundaries:

$$\displaystyle f(x)=8-2x-x^2$$

$$\displaystyle g(x)=0$$

Calculate the places where the boundaries intersect:

$$\displaystyle f(x)=g(x)$$

$$\displaystyle 8-2x-x^2=0$$

$$\displaystyle x^2+2x-8=0$$

$$\displaystyle (x+4)(x-2)=0$$

$$\displaystyle x\in\{-4,2\}$$

Now, we see that:

$$\displaystyle f(-1)=9>g(-1)$$

And so we conclude that on $(-4,2)$, we must have:

$$\displaystyle f(x)>g(x)$$

And so the area $A$ in question is given by:

$$\displaystyle A=\int_{-4}^{2} f(x)-g(x)\,dx=\int_{-4}^{2} 8-2x-x^2\,dx$$

Let's utilize some symmetry here, and use:

$$\displaystyle u=x+1\,\therefore\,du=dx$$

And we may state:

$$\displaystyle A=\int_{-3}^{3} 8-2(u-1)-(u-1)^2\,du=\int_{-3}^{3} 9-u^2\,du$$

Using the even-function rule, we now have:

$$\displaystyle A=2\int_{0}^{3} 9-u^2\,du=2\left[9u-\frac{1}{3}u^3\right]_0^3=\frac{2}{3}\left[27u-u^3\right]_0^3=\frac{2}{3}\left(3^4-3^3\right)=2\cdot3^2\left(3-1\right)=(2\cdot3)^2=6^2=36$$

Thank you I have my final exam this afternoon to finish high school.

Noah said:
Calculate the area of the region bounded by the graph of the function y = 8 – 2x - x^2 and the x-axis
Y = 8 - 2- x^2
0 = 8 – 2 – x^2
(-x – 4)(x – 2)
- x – 4 = 0 and x – 2 = 0
-x = 4 x = 2
X = - 4

Do I do this?

Y = 8 -2x -x^2
= 8x - (2x^2)/2 - x^3/3
No, if Y= 8- 2x- x^2 then Y is not also equal to 8x- (2x^2/2)- x^3/3!
Yes, I know you meant that this is anti-derivative of Y but writing
"Y= ****
= ..."
means that they are both equal to Y.

= 8 - x^2 - x^3/3
= [8x - x^2 - x^3/3] – [8x - x^2 - x^3/3]
and this is quite obviously equal to 0! Again, this is not what you mean.

= [(8x 2) - 2^2 - 2^3/3] – [(8 x -4) - 〖-4〗^2 - 〖-4〗^3/3]
This is what you should have written before- although it is not a good idea to use "x" as the variable in one expression and as the multiplication sign in the next! Use "(8)(2)" instead.

= [16 – 4 + 8/3] – [-32 + 16 - 64/3]
= 141/3 – [-16 - 311/3]
= 141/3 – [-16 + 211/3]
= 141/3 - 5 1/3
Area = 9 units

## What is the function used to calculate the area of the region bounded by the graph?

The function used to calculate the area of the region bounded by the graph is the definite integral.

## What are the limits of integration for this problem?

The limits of integration for this problem will depend on the points of intersection between the graph of the function and the x-axis. These points can be found by setting the function equal to 0 and solving for x.

## How do you set up the definite integral for this problem?

To set up the definite integral, we first need to determine the function that represents the area between the graph and the x-axis. This can be found by taking the absolute value of the function y = 8 - 2x - x^2. Then, we use the limits of integration to create the integral expression: ∫[limits of integration] |8 - 2x - x^2| dx.

## How do you solve the definite integral to find the area?

To solve the definite integral, we can use integration techniques such as substitution or the power rule. Once the integral is solved, the resulting expression will represent the area of the bounded region.

## What is the significance of calculating the area of this region?

Calculating the area of this region helps us understand the behavior of the function and the relationship between the graph and the x-axis. It also allows us to find the total amount of space within the boundaries of the function, which can be useful in various real-world applications.

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