The area of the region {(x,y):x2+4x+2≤y≤∣x+2∣} is equal to
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Step-by-Step Solution
Step 1: Analyze the inequalities
We are given a region defined by two inequalities. The lower bound for y is a parabola y1=x2+4x+2, and the upper bound for y is an absolute value function y2=∣x+2∣. We need to find the area between these two curves.
Step 2: Simplify the parabola equation
Let's simplify the equation of the parabola by completing the square. This makes it easier to see its vertex and how it relates to the absolute value function. The parabola y1=x2+4x+2 can be rewritten as y1=(x+2)2−4+2, which simplifies to y1=(x+2)2−2.
Step 3: Find intersection points
To find the intersection points of the two curves, we set their equations equal to each other: (x+2)2−2=∣x+2∣. Let u=x+2. Then the equation becomes u2−2=∣u∣. We consider two cases: u≥0 and u<0.
Step 4: Solve for intersection points (Case 1: u≥0)
For u≥0, the equation is u2−2=u, which rearranges to u2−u−2=0. Factoring this quadratic equation gives (u−2)(u+1)=0. The solutions are u=2 or u=−1. Since we assumed u≥0, we take u=2. Substituting back u=x+2, we get x+2=2, so x=0.
Step 5: Solve for intersection points (Case 2: u<0)
For u<0, the equation is u2−2=−u, which rearranges to u2+u−2=0. Factoring this quadratic equation gives (u+2)(u−1)=0. The solutions are u=−2 or u=1. Since we assumed u<0, we take u=−2. Substituting back u=x+2, we get x+2=−2, so x=−4.
Step 6: Set up the integral for the area
The intersection points are x=−4 and x=0. In the interval [−4,0], the absolute value function ∣x+2∣ is above the parabola (x+2)2−2. Therefore, the area is given by the definite integral of the upper function minus the lower function from x=−4 to x=0. We can simplify the integral by using the substitution u=x+2, so du=dx. When x=−4, u=−2. When x=0, u=2.
Step 7: Evaluate the integral
Due to the symmetry of the integrand (∣u∣−u2+2) around u=0, we can integrate from 0 to 2 and multiply the result by 2. For u≥0, ∣u∣=u. So the integral becomes 2∫02(u−u2+2)du. Evaluating the integral, we get 2[2u2−3u3+2u]02. Plugging in the limits, we find the area to be 2(222−323+2(2))=2(2−38+4)=2(6−38)=2(318−8)=2(310)=320.