Find the area bounded by the curve and the lines
y = -x^2 - 4x; y = 1; x = -3; x = 1
To find the area bounded by the curve and the lines, we need to find the points where the curve intersects the lines and then integrate the curve between these points.
Step 1: Find the points of intersection.
To find the points where the curve intersects the lines, we need to set the equations equal to each other and solve for x.
Curve and line 1:
-y = x^2 + 4x
x^2 + 4x + y = 0
Curve and line 2:
y = 1
For line 2, y is always equal to 1. Hence, we can substitute y = 1 into line 1 equation to get:
x^2 + 4x + 1 = 0
We can solve this quadratic equation using the quadratic formula:
x = (-4 ± √(4^2 - 4 * 1 * 1)) / 2 * 1
Simplifying further, we get:
x = (-4 ± √(16 - 4)) / 2
x = (-4 ± √12) / 2
x = (-4 ± 2√3) / 2
x = -2 ± √3
So, the points of intersection are (-2 + √3, 1) and (-2 - √3, 1).
Step 2: Integrate the curve between the points of intersection.
To find the area bounded by the curve, we need to integrate the curve between the x-values of the points of intersection.
The integral of -x^2 - 4x with respect to x is given by:
∫[-2 - √3, -2 + √3] (-x^2 - 4x) dx
Integrating, we get:
∫[-2 - √3, -2 + √3] -x^2 - 4x dx
= [- (1/3)x^3 - 2x^2] ∣[-2 - √3, -2 + √3]
Evaluating this integral at the limits, we get:
[- (1/3)(-2 + √3)^3 - 2(-2 + √3)^2] - [-(1/3)(-2 - √3)^3 - 2(-2 - √3)^2]
Calculating further will give us the area enclosed by the curve and the lines.
Note: The above solution assumes that "enclosed area" is defined as the area above the curve and below the line y = 1 within the given x-interval of -3 to 1. If you wanted to find the area between the curve and the x-axis, the integral and limits would be different.