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

[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

 

[MarkFL]

Define the boundaries:

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

$$g(x)=0$$

Calculate the places where the boundaries intersect:

$$f(x)=g(x)$$

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

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

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

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

Now, we see that:

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

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

$$f(x)>g(x)$$

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

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

Let's utilize some symmetry here, and use:

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

And we may state:

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

Using the even-function rule, we now have:

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

 

[Noah1]

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

 

[HallsofIvy]

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