Sketch the region enclosed by $$ y=6 x $$ and $$ y=8 x^{2} $$. Decide whether to integrate with respect to $$ x $$ or $$ y $$. Then find the area of the region.

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To find the area of the region enclosed by the curves $$ y=6x $$ and $$ y=8x^2 $$, we need to determine which curve is above the other in the region of interest.

Let's first find the intersection points of the two curves by setting them equal to each other:
$$ 6x = 8x^2 $$

Solving for $$ x $$, we get:
$$ 8x^2 - 6x = 0 $$
$$ 2x(4x - 3) = 0 $$
$$ x = 0 \text{ or } x = \frac{3}{4} $$

Now, we need to determine which curve is above the other in the region of interest. We can do this by choosing a point between the intersection points and checking which curve has a larger value.

Let's choose $$ x = \frac{1}{2} $$ and substitute it into both equations:
$$ y = 6 \left( \frac{1}{2} \right) = 3 $$
$$ y = 8 \left( \frac{1}{2} \right)^2 = 2 $$

Since $$ y = 3 $$ is greater than $$ y = 2 $$ at $$ x = \frac{1}{2} $$, the curve $$ y = 6x $$ is above the curve $$ y = 8x^2 $$ in the region of interest.

Therefore, we will integrate with respect to $$ x $$ to find the area of the region.

The area of the region can be found by integrating the difference between the two curves with respect to $$ x $$ from the lower intersection point to the upper intersection point:
$$ \text{Area} = \int_{0}^{\frac{3}{4}} (6x - 8x^2) \, dx $$

Now, let's calculate the integral to find the area of the region.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{0}^{\frac{3}{4}} \left(6x-8x^{2}\right) dx$$
- step1: Evaluate the integral:
  $$\int \left(6x-8x^{2}\right) dx$$
- step2: Use properties of integrals:
  $$\int 6x dx+\int -8x^{2} dx$$
- step3: Evaluate the integral:
  $$3x^{2}+\int -8x^{2} dx$$
- step4: Evaluate the integral:
  $$3x^{2}-\frac{8x^{3}}{3}$$
- step5: Return the limits:
  $$\left(3x^{2}-\frac{8x^{3}}{3}\right)\bigg |_{0}^{\frac{3}{4}}$$
- step6: Calculate the value:
  $$\frac{9}{16}$$
The area of the region enclosed by the curves $$ y=6x $$ and $$ y=8x^2 $$ is $$ \frac{9}{16} $$ or approximately 0.5625.
To find the area between $$ y = 6x $$ and $$ y = 8x^2 $$, first find their intersection points at $$ x = 0 $$ and $$ x = \frac{3}{4} $$. Since $$ y = 6x $$ is above $$ y = 8x^2 $$ between these points, integrate $$ 6x - 8x^2 $$ from 0 to $$ \frac{3}{4} $$. The area is $$ \frac{9}{16} $$ or 0.5625.

Sketch the region enclosed by and . Decide whether to integrate with respect to or .
Then find the area of the region.

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Sketch the region enclosed by and . Decide whether to integrate with respect to or . Then find the area of the region.