Exploring the Function y x3 - 4x2 - 3x and Finding the Area between the Curve and x-axis

Introduction to the Problem

This article explores the problem of determining the area between the cubic function y x3 - 4x2 - 3x and the x-axis from x 0 to x 3.

Identifying the Zeros

To find where the curve intersects the x-axis, we set y 0 and solve for x: [ x^3 - 4x^2 - 3x 0 ]

Solving this equation involves factoring and identifying the roots:

[ x(x - 1)(x 3) 0 ]

The roots are x 0, x 1, x -3. However, x -3 is outside the interval [0, 3], so we only consider x 0, x 1, and x 3.

Determining the Sign of the Function

To find the area, we need to determine where the function is above and below the x-axis. This requires analyzing the sign of the function in the interval [0, 3].

Applying Integration to Find the Area

The function changes sign at x 1, so we need to calculate the area in two parts: from x 0 to x 1 and from x 1 to x 3.

Area from x 0 to x 1

For the interval [0, 1], the function is above the x-axis. We find the integral:

[ int_{0}^{1} (x^3 - 4x^2 - 3x) , dx ]

Integrating term by term:

[ left[ frac{x^4}{4} - frac{4x^3}{3} - 3x right]_{0}^{1} ]

Evaluating this at the bounds:

[ left( frac{1}{4} - frac{4}{3} - 3 right) - 0 frac{3 - 16 - 36}{12} frac{-49}{12} implies frac{23}{12} ]

Area from x 1 to x 3

For the interval [1, 3], the function is below the x-axis. We need to take the negative of the integral:

[ -int_{1}^{3} (x^3 - 4x^2 - 3x) , dx ]

First, evaluate the integral:

[ int (x^3 - 4x^2 - 3x) , dx frac{x^4}{4} - frac{4x^3}{3} - 3x ]

Evaluating from 1 to 3:

[ left( -left( frac{3^4}{4} - frac{4 cdot 3^3}{3} - 3 cdot 3 right) right) - left( -left( frac{1^4}{4} - frac{4 cdot 1^3}{3} - 3 cdot 1 right) right) ]

This simplifies to:

[ -left( frac{81}{4} - 36 - 9 right) - left( -left( frac{1}{4} - frac{4}{3} - 3 right) right) ]

Further simplification:

[ -left( frac{81}{4} - 45 right) left( frac{23}{12} right) -frac{81}{4} frac{23}{12} 45 ]

Combining all terms:

[ frac{23}{12} - frac{81}{4} 45 frac{127}{12} approx 10.583 text{ square units} ]

Conclusion

The area between the curve y x3 - 4x2 - 3x and the x-axis from x 0 to x 3 is approximately 10.583 square units.