find the area bounded by the curves y^2=2x+6 and x=y+1
The curves intersect at (-1,-2) and (5,4)
Horizontal strips of width dy are the best here, so
A = ∫[-2,4] (y+1)-(y^2-6)/2 dy = 18
Using vertical strips is a bit trickier, because the boundary changes at x = -1
A = ∫[-3,-1] √(2x+6) - (-√(2x+6)) dx
+ ∫[-1,5] √(2x+6)-(x-1) dx
= 38/3 + 16/3 = 54/3 = 18
Thank you!
To find the area bounded by the curves y^2=2x+6 and x=y+1, follow these steps:
1. Graph the curves on a coordinate plane to visualize the region bounded by them.
The first equation, y^2=2x+6, is a parabolic curve opening to the right, with the vertex at (-3,0). Plot some points to draw the graph.
The second equation, x=y+1, represents a straight line, which passes through (0,1) and (1,2).
2. Determine the points of intersection between the curves.
Substitute the value of x from the second equation into the first equation: y^2 = 2(y+1) + 6
Simplify: y^2 = 2y + 2 + 6
y^2 - 2y - 8 = 0
Now, factorize the quadratic equation: (y - 4)(y + 2) = 0
So, y = 4 and y = -2.
Substitute these y-values back into the second equation to find the corresponding x-values.
When y = 4, x = y + 1 = 4 + 1 = 5
When y = -2, x = y + 1 = -2 + 1 = -1
So, the points of intersection are (5, 4) and (-1, -2).
3. Determine the limits of integration.
The area can be found by integrating the difference of the curves with respect to x. The limits of integration are the x-values of the points of intersection, which are -1 and 5.
4. Integrate the difference of the curves with respect to x.
To find the area, set up the integral as follows:
∫[from -1 to 5] (y+1) - (1/2)(y^2 - 6) dx
5. Convert the integral with respect to x to an integral with respect to y.
Since the given equations are defined in terms of y, you need to convert the integral from dx to dy.
Using the property dx = dy / (dy/dx), where dy/dx is the derivative of x with respect to y, the integral becomes:
∫[from -2 to 4] (y+1) - (1/2)(y^2 - 6) dy / (dy/dx).
To find dy/dx, rearrange the second equation x = y + 1 to solve for y in terms of x: y = x - 1.
Differentiate it with respect to x: dy/dx = 1.
So, the integral becomes:
∫[from -2 to 4] (y+1) - (1/2)(y^2 - 6) dy / 1
6. Evaluate the integral and find the area.
∫[from -2 to 4] (y+1) - (1/2)(y^2 - 6) dy
= ∫[from -2 to 4] y + 1 - (1/2)y^2 + 3 dy
= [ (1/2)y^2 + y - (1/6)y^3 + 3y ] |_from -2 to 4
= [ (1/2)(4)^2 + 4 - (1/6)(4)^3 + 3(4) ] - [ (1/2)(-2)^2 + (-2) - (1/6)(-2)^3 + 3(-2) ]
= 32/3
Therefore, the area bounded by the curves y^2 = 2x + 6 and x = y + 1 is 32/3 square units.