Questions: y=x^3-8x, 8x Find the area A of the region enclosed by the graphs. 1. (Give an exact answer. Use symbolic notation and fractions where needed.)

Solution Steps

The given functions are: \[ y = x^3 - 8x \] \[ y = 8x \]

To find the area enclosed by these graphs, we first need to determine their points of intersection by setting the equations equal to each other: \[ x^3 - 8x = 8x \] \[ x^3 - 16x = 0 \] \[ x(x^2 - 16) = 0 \] \[ x(x - 4)(x + 4) = 0 \]

So, the points of intersection are \( x = 0 \), \( x = 4 \), and \( x = -4 \).

The area \( A \) enclosed by the curves can be found by integrating the difference of the functions from \( x = -4 \) to \( x = 4 \): \[ A = \int_{-4}^{4} [(8x) - (x^3 - 8x)] \, dx \] \[ A = \int_{-4}^{4} [8x - x^3 + 8x] \, dx \] \[ A = \int_{-4}^{4} [16x - x^3] \, dx \]

Evaluate the integral: \[ A = \int_{-4}^{4} (16x - x^3) \, dx \]

First, find the antiderivative: \[ \int (16x - x^3) \, dx = 8x^2 - \frac{x^4}{4} + C \]

Now, evaluate this from \( x = -4 \) to \( x = 4 \): \[ A = \left[ 8x^2 - \frac{x^4}{4} \right]_{-4}^{4} \] \[ A = \left( 8(4)^2 - \frac{(4)^4}{4} \right) - \left( 8(-4)^2 - \frac{(-4)^4}{4} \right) \] \[ A = \left( 8 \cdot 16 - \frac{256}{4} \right) - \left( 8 \cdot 16 - \frac{256}{4} \right) \] \[ A = \left( 128 - 64 \right) - \left( 128 - 64 \right) \] \[ A = 64 - 64 \] \[ A = 0 \]

Final Answer