Question

For the given regions R₁ and R₂, complete the following steps.

b. Find the area of region R₂ using geometry and the answer to part (a).

R₁is the region in the first quadrant bounded by the line x=1 and the curve y=6x(2−x^2)^2; R₂ is the region in the first quadrant bounded the curve y=6x(2−x^2)^2and the line y=6x.

Accepted Answer

First, clearly identify the boundaries of region R₂. It is bounded above and below by the curves $$ y = 6x(2 - x^2)^2 $$ and $$ y = 6x $$, respectively, in the first quadrant. Determine the interval of $$ x $$ over which these two curves intersect to define the horizontal limits of R₂. Find the points of intersection between the two curves by setting $$ 6x(2 - x^2)^2 = 6x . $$Since $$ 6x  is $$common on both sides, consider the cases where $$ x = 0  or$$ $$ (2 - x^2)^2 = 1 . $$Solve for $$ x  to $$find the exact intersection points that bound R₂. Use the result from part (a), which likely gives the area of region R₁ bounded by $$ x = 1 $$ and $$ y = 6x(2 - x^2)^2 $$, to help express the area of R₂. Since R₂ lies between the two curves, the area of R₂ can be found by subtracting the area under $$ y = 6x $$ from the area under $$ y = 6x(2 - x^2)^2 $$ over the interval determined in step 2. Set up the integral expression for the area of R₂ as $$ \int_{a}^{b} \left[6x(2 - x^2)^2 - 6x\right] \, dx $$, where $$ a $$ and $$ b $$ are the intersection points found earlier. This integral represents the vertical distance between the two curves integrated over the interval where R₂ exists. Evaluate the integral expression or use the known area from part (a) to compute the area of R₂. Remember, the integral of $$ 6x $$ over the interval can be subtracted from the integral of $$ 6x(2 - x^2)^2  to $$find the exact area of R₂.

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Key Concepts

Area of a Region Bounded by Curves

Using Geometry to Find Areas

Interpreting and Using Given Functions and Boundaries