find the area of the region bounded by y=1-2x^2 and y=|x|
the region is symmetric, and the curves intersect at (1,1)
a = 2∫[0,1] (2-x^2)-x dx = 7/3
or, using horizontal strips of width dy, you need to switch curves at (1,1)
a = 2(∫[0,1] y dy + ∫[1,2] √(2-y) dy) = 2(1/2 + 2/3) = 7/3
To find the area of the region bounded by the given equations, we need to determine the points where these two equations intersect.
Let's start by setting the two equations equal to each other:
1 - 2x^2 = |x|
Now, we can consider two cases, one where x is positive and one where x is negative.
Case 1: x is positive
1 - 2x^2 = x
Rearranging the equation, we get:
2x^2 + x - 1 = 0
Solving this quadratic equation using the quadratic formula, we get:
x = (-1 + √9) / 4 or x = (-1 - √9) / 4
Since we are looking for the positive value of x, x = (-1 + √9) / 4
Case 2: x is negative
1 - 2x^2 = -x
Rearranging the equation:
2x^2 - x + 1 = 0
Solving this again using the quadratic formula, we get:
x = (-(-1) ± √((-1)^2 - 4*2*1)) / (2*2)
x = (1 ± √(-7)) / 4
Since it is not possible to take the square root of a negative number, there are no real solutions for this case.
Therefore, we only consider the positive value of x, x = (-1 + √9) / 4.
Now, we need to find the y-coordinate values of the intersection points by substituting the x values into one of the equations.
For y = 1 - 2x^2:
When x = (-1 + √9) / 4, y = 1 - 2((-1 + √9) / 4)^2
Simplifying, we get:
y = 1 - 2(1/16)(-1 + √9)^2
Expanding, we have:
y = 1 - 2(1/16)(-1 + 3)
y = 1 - 2(1/16)(2)
y = 1 - 2/8
y = 1 - 1/4
y = 3/4
So, the intersection point is ( (-1 + √9) / 4 , 3/4 ).
Now let's find the area using integration. The area will be the integral of the curve y=1-2x^2 minus the curve y=|x| from the negative intersection point to the positive intersection point.
Using the given equations, we have:
∫(1-2x^2 - |x|) dx
Integrating each term separately, we get:
∫(1-2x^2) dx - ∫|x| dx
The first integral, ∫(1-2x^2) dx, can be calculated as:
x - (2/3)x^3
The second integral, ∫|x| dx, can be calculated as:
x^2/2 if x is positive, or -x^2/2 if x is negative
Now, we can calculate the area by substituting the limits of integration:
Area = ∫(1-2x^2) dx - ∫|x| dx
= [( -1 + √9 ) / 4 - (2/3)( (-1 + √9 ) / 4 )^3] - [ ( (-1 + √9 ) / 4 )^2 / 2 - ( 0 )^2 / 2]
Simplifying, we have:
Area = [ ( -1 + √9 ) / 4 - (2/3)( (-1 + √9 ) / 4 )^3 ] - [ ( (-1 + √9 )^2 / 4 ) / 2 ]
Area = [ ( -1 + √9 ) / 4 - ( (-1 + √9 ) / 4 )^3 / (3/2) ] - [ ( (-1 + √9 )^2 / 4 ) / 2 ]
Area ≈ 0.458
Therefore, the area of the region bounded by y = 1 - 2x^2 and y = |x| is approximately 0.458 square units.
To find the area of the region bounded by the curves y = 1 - 2x^2 and y = |x|, we first need to identify the x-values at which these curves intersect.
Setting y = 1 - 2x^2 equal to y = |x|, we can solve for x:
1 - 2x^2 = |x|
We can split this equation into two cases, depending on the value of x:
Case 1: x ≥ 0
In this case, we have 1 - 2x^2 = x. Rearranging the equation, we get:
2x^2 + x - 1 = 0
Solving this quadratic equation using the quadratic formula, we find two possible x-values: x = 0.5 and x = -1.
Case 2: x < 0
In this case, we have 1 - 2x^2 = -x. Rearranging the equation, we get:
2x^2 - x - 1 = 0
Solving this quadratic equation using the quadratic formula, we find two possible x-values: x = -0.5 and x = 1.
So, the x-values at which the curves intersect are: x = -1, x = -0.5, x = 0.5, and x = 1.
To find the area of the region bounded by the curves, we can split it into two separate regions and calculate the area of each separately.
Region 1: To the left of the y-axis (x < 0)
In this region, the curves are y = 1 - 2x^2 and y = -x.
To find the area, we need to integrate the difference between the two curves with respect to x, from x = -1 to x = -0.5. The integral expression for this region is:
A1 = ∫[from -1 to -0.5] [(1 - 2x^2) - (-x)] dx
Region 2: To the right of the y-axis (x ≥ 0)
In this region, the curves are y = 1 - 2x^2 and y = x.
To find the area, we need to integrate the difference between the two curves with respect to x, from x = 0.5 to x = 1. The integral expression for this region is:
A2 = ∫[from 0.5 to 1] [(1 - 2x^2) - x] dx
Finally, to find the total area of the region bounded by the curves, we add the areas of Region 1 and Region 2:
Total Area = A1 + A2