The area of the region {(x,y):∣x−y∣≤y≤4x} is:
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Step-by-Step Solution
Step 1: Break down the inequalities
The given region is defined by two inequalities: ∣x−y∣≤y and y≤4x. We first break down the absolute value inequality. The inequality ∣x−y∣≤y can be split into two separate inequalities: −(x−y)≤y and y≤x−y.
Step 2: Simplify the inequalities
From −(x−y)≤y, we get −x+y≤y, which simplifies to −x≤0, or x≥0. This is consistent with y≤4x, which requires x≥0. From y≤x−y, we get 2y≤x, which simplifies to y≤2x. However, the first part of the absolute value inequality −(x−y)≤y actually gives y≥x/2. Let's re-evaluate. −(x−y)≤y⟹−x+y≤y⟹−x≤0⟹x≥0. This is not correct. Let's re-evaluate the absolute value inequality. ∣x−y∣≤y means −y≤x−y≤y. The left part is −y≤x−y⟹0≤x, which is x≥0. The right part is x−y≤y⟹x≤2y⟹y≥2x. So the first inequality simplifies to y≥2x (and x≥0). The second inequality is y≤4x.
Step 3: Find intersection points
We need to find the intersection points of the two boundary curves y=2x and y=4x. Set the expressions for y equal to each other: 2x=4x. Squaring both sides gives 4x2=16x. Multiplying by 4 gives x2=64x. Rearranging, we get x2−64x=0, which factors as x(x−64)=0. This gives x=0 and x=64.
Step 4: Calculate the area using integration
The area of the region is given by the definite integral of the upper curve minus the lower curve, from the first intersection point to the second. In this case, y=4x is the upper curve and y=2x is the lower curve between x=0 and x=64.
Step 5: Evaluate the integral
Now we evaluate the definite integral. We integrate term by term: ∫4x1/2dx=4⋅3/2x3/2=38x3/2 and ∫2xdx=4x2. Then we apply the limits of integration.
Step 6: Calculate the final area
Substitute the upper limit x=64 and the lower limit x=0 into the antiderivative. (64)3/2=(64)3=83=512. (64)2=4096. Simplify the expression to find the final area.