Understanding the concept of the area bounded by the curve and the x-axis is crucial in calculus. This article provides an in-depth analysis of the specific function y x^3 - 4x^2 6x, discussing its graph, integration techniques, and different approaches to evaluate the total area covered. The content is structured to cater to students, educators, and professionals in mathematics and related fields.

Introduction

The function y x^3 - 4x^2 6x can be rewritten as y x(x-2)(x-3). Analyzing the behavior of this function and computing the area bounded by its curve and the x-axis are fundamental tasks in calculus.

Graphical Analysis

The curve representing the function y x(x-2)(x-3) lies above the x-axis for -1 ≤ x ≤ 2 and below the x-axis for 2 ≤ x ≤ 3. This behavior is due to the nature of the roots of the polynomial, which are x 0, x 2, and x 3.

Integration Techniques

To find the area under the curve, we need to evaluate the definite integral of the function over specific intervals. The primary methods include shifting the function and integrating piecewise.

Shifting the Function

One useful trick is to shift the function. For instance, setting u x - 2, the function can be transformed into a simpler form:

y x(x-1)(x-3) becomes y u(u-1)(u 1) u^3 - u. This simplification helps in finding the area by integrating a simpler polynomial function.

Integration Approach

The integral of the function ( y x^3 - 4x^2 6x ) from different intervals can be expressed as follows:

For the interval x -1 to x 2:

A int_{-1}^{0} (u^3 - u) du int_{0}^{2} (u - u^3) du
left[ frac{u^4}{4} - frac{u^2}{2} right]_{-1}^{0} left[ frac{u^2}{2} - frac{u^4}{4} right]_{0}^{2}
0 - (-frac{1}{4}) - (0 - (-frac{1}{2})) (-frac{1}{2} 0) - 0
frac{1}{4} - frac{1}{2} frac{1}{2} - frac{1}{2} frac{1}{2})

Alternative Integration Method

The alternative method involves breaking the function into intervals where it is above or below the x-axis:

A left| int_{-1}^{1} (x^3 - 4x^2 6x) dx right| left| int_{1}^{2} (x^3 - 4x^2 6x) dx right| left| int_{2}^{3} (x^3 - 4x^2 6x) dx right|

For the interval x 1 to x 2 and x 2 to x 3, the above method simplifies to the sum of the absolute values of the definite integrals in these intervals.

Calculating the integrals separately:
( A left| int_{-1}^{0} x^3 dx - 4 int_{-1}^{0} x^2 dx 6 int_{-1}^{0} x dx right| left| int_{1}^{2} (x^3 dx - 4 x^2 dx 6 x dx) - int_{1}^{2} (x^3 dx - 4 x^2 dx 6 x dx) right| left| int_{2}^{3} (x^3 dx - 4 x^2 dx 6 x dx) - int_{2}^{3} (x^3 dx - 4 x^2 dx 6 x dx) right|)

After evaluation, the total area is found to be 5 square units.

Conclusion

The detailed analysis and integration of the function ( y x^3 - 4x^2 6x ) demonstrate the practical application of integration techniques and the importance of area considerations above and below the x-axis. This knowledge is pivotal for a deeper understanding of calculus and various fields requiring quantitative analysis.