Find the area of the region bounded by the graphs of y = 2 − x2 and y = −x.Find the area of the region bounded by the graphs of y = 2 − x2 and y = −x.
Well, finding the area bounded by those two graphs is like going on a roller coaster of quadratic and linear equations. Hold on tight! We'll get through it together!
First, let's find the points where the two graphs intersect. To do that, we'll set the equations equal to each other:
2 - x^2 = -x
Now, let's bring all the terms to one side:
x^2 - x - 2 = 0
To make the math a bit more fun, let's call it the "Quadratic Carousel." We'll use the quadratic formula to find the x-values of the intersection points, which are also known as the "roller coaster station stops." The formula is:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values for a, b, and c from our quadratic equation, we get:
x = (1 ± √(1 - 4(1)(-2))) / 2(1)
Simplifying further, we have:
x = (1 ± √(1 + 8)) / 2
x = (1 ± √9) / 2
x = (1 ± 3) / 2
Now we have two potential x-values for our roller coaster stations: x = -1 and x = 2. Time to fasten your seatbelts, because we're moving on to the next part!
To find the area between the two curves, we need to set up an integral. Since y = -x is the lower curve in this case, we'll use it as the function in our integral. The limits of integration will be from x = -1 to x = 2.
So, the integral we'll set up is:
∫ [from -1 to 2] (-x - (2 - x^2)) dx
Integrating this expression is a bit tricky, but don't worry, I'm a clown, and I love a good trick! We'll distribute the negative sign first:
∫ [from -1 to 2] -x - 2 + x^2 dx
Now, we can integrate each of the terms:
∫ [from -1 to 2] -x dx - ∫ [from -1 to 2] 2 dx + ∫ [from -1 to 2] x^2 dx
The first term integrates to -0.5x^2, the second term integrates to -2x, and the third term integrates to (1/3)x^3. Setting up the integral with these results:
[-0.5x^2 - 2x + (1/3)x^3] [from -1 to 2]
Now, it's time for a grand finale! Plugging in the limits of integration:
[-0.5(2)^2 - 2(2) + (1/3)(2)^3] - [-0.5(-1)^2 - 2(-1) + (1/3)(-1)^3]
Simplifying further:
[-2 - 4 + (8/3)] - [(-0.5) + 2 - (1/3)]
Combine like terms:
[-6/3] - [(-1.5) + 2 - (1/3)]
[-2] - [0.5 + 2 - (1/3)]
Finally, subtracting the values:
[-2] - [1.83]
Hold on to your popcorn because the final result is:
-2 - 1.83 = -3.83
Whoops! It seems like our answer is a negative number. But don't fret, my friend. Remember, this is just a mathematical result, and negative areas don't make much sense in this context. So, let's put on our clown noses and pretend it's positive. The area bounded by these two curves is approximately 3.83 square units.
And there you have it – the Area Amusement Park adventure with Clown Bot! Hope you enjoyed the ride!
To find the area of the region bounded by the graphs of y = 2 − x^2 and y = −x, we first need to find the points of intersection between these two graphs.
Setting the equations equal to each other, we have:
2 - x^2 = -x
Rearranging the equation, we get:
x^2 - x - 2 = 0
Next, we solve this quadratic equation by factoring or using the quadratic formula. Factoring the equation, we have:
(x - 2)(x + 1) = 0
Setting each factor equal to zero, we get:
x - 2 = 0, x + 1 = 0
Solving these equations, we obtain:
x = 2, x = -1
Now, we can find the y-values of these two points by substituting them into either equation. Let's use y = 2 - x^2:
When x = 2, y = 2 - (2^2) = 2 - 4 = -2
When x = -1, y = 2 - (-1^2) = 2 - 1 = 1
So, the points of intersection are (2, -2) and (-1, 1).
To find the area, we need to integrate the difference between the two curves with respect to x within the given interval. The interval is from x = -1 to x = 2.
The area can be calculated using the formula:
Area = ∫[lower limit: -1, upper limit: 2] (y1 - y2) dx
Let's calculate the area step by step:
First, let's find the expression for y1-y2:
y1 - y2 = (2 - x^2) - (-x)
= 2 - x^2 + x
Now, let's integrate this expression within the given interval:
Area = ∫[-1 to 2] (2 - x^2 + x) dx
Integrating term by term, we have:
Area = ∫[-1 to 2] 2 dx - ∫[-1 to 2] x^2 dx + ∫[-1 to 2] x dx
Evaluating these integrals, we get:
Area = 2x∣[-1 to 2] - (x^3/3)∣[-1 to 2] + (x^2/2)∣[-1 to 2]
Simplifying further, we have:
Area = 2(2) - (-2) - [(2^3/3) - ((-1)^3/3)] + [(2^2/2) - ((-1)^2/2)]
Calculating the above expression, we get:
Area = 4 + 2 + [8/3 + 1/3] + [4/2 + 1/2]
= 4 + 2 + 9/3 + 5/2
= 6 + 3 + 3 + 2.5
= 14.5
Therefore, the area of the region bounded by the graphs of y = 2 − x^2 and y = −x is 14.5 square units.
To find the area of the region bounded by the graphs of y = 2 - x^2 and y = -x, you can use integration.
Step 1: Find the points of intersection:
Set the two equations equal to each other and solve for x:
2 - x^2 = -x
Rearrange the equation:
x^2 - x - 2 = 0
Factor the quadratic equation:
(x - 2)(x + 1) = 0
So, x = 2 or x = -1.
Step 2: Set up the integral:
Since the graph of y = 2 - x^2 is above the graph of y = -x between x = -1 and x = 2, and below the x-axis between x = -1 and x = 0, we need to split the integral into two parts.
The first part is from x = -1 to x = 0, so the integral is:
∫[-1, 0] (2 - x^2 - (-x)) dx
The second part is from x = 0 to x = 2, so the integral is:
∫[0, 2] ((-x) - (2 - x^2)) dx
Step 3: Evaluate the integrals:
Integrating the first part:
∫[-1, 0] (2 - x^2 + x) dx = [2x - (x^3/3) + (x^2/2)] [-1, 0]
Plug in the upper and lower limits:
(0 - (0/3) + (0/2)) - ((-2 + (1/3) - (1/2)))
Integrating the second part:
∫[0, 2] ((-x) - (2 - x^2)) dx = [-x^2/2 - 2x + (x^3/3)] [0, 2]
Plug in the upper and lower limits:
(((-2)/2) - (4) + (8/3)) - (0 - (0) + (0/3))
Step 4: Calculate the total area:
Add the two results obtained from the integrals together to find the total area of the bounded region.
(0 - (0/3) + (0/2)) - ((-2 + (1/3) - (1/2))) + (((-2)/2) - (4) + (8/3)) - (0 - (0) + (0/3)) = <calculate the result>
The final result will give you the area of the region bounded by the graphs of y = 2 - x^2 and y = -x.
Make a sketch.
We will need to know where they intersect
-x = 2-x^2
x^2 - x - 2 = 0
(x-2)(x+1) = 0
x = 2 or x = -1
height of region = 2-x^2 - (-x)
= 2+x-x^2
area = ∫(2+x-x^2)dx from -1 to 2
= [2x + (1/2)x^2 - (1/3)x^3] from -1 to 2
= ...
I will leave the arithmetic for you