Question

Evaluate the definite integral. $$ \int_{0}^{3}(x-2)^{3} d x $$

Short answer

The result of the definite integration is -9.

Step-by-step solution

Expand the Integrant

First, expand the cubic expression in the integrand. The expansion of \( (x-2)^3 \) gives: \( x^3 - 6x^2 + 12x - 8 \)

Perform the Integration

Next, it's time to integrate term by term. The integral of each terms are as follows: \n\n - The integral of \(x^3\) is \(\frac{1}{4}x^4\)\n - The integral of \(6x^2\) is \(\frac{2}{3}x^3\)\n - The integral of \(12x\) is \(6x^2\)\n - The integral of \(8\) is \(8x\)\n\nSo, we have \(\int (x-2)^3 dx = \frac{1}{4}x^4 - \frac{2}{3}x^3 + 6x^2 - 8x\).

Apply Definite Integration

Finally, the integral is evaluated using the limits from 0 to 3. It is done by substituting each limit into the integral expression and subtracting them.\n\nEvaluating at 3: \[\frac{1}{4}(3)^4 - \frac{2}{3}(3)^3 + 6*(3)^2 - 8*(3)\]\n\nEvaluating at 0: \[\frac{1}{4}(0)^4 - \frac{2}{3}(0)^3 + 6*(0)^2 - 8*(0)\]\n\nSubtracting these gives the final result.

Key concepts

Integration Techniques

One of the fundamental tools in calculus is integration, a process that can be thought of as the inverse of differentiation, or as calculating the area under a curve. Integration techniques allow us to find the antiderivative of functions. There are several integration tactics including power rule, substitution, integration by parts, partial fraction decomposition, and trigonometric integration. In our specific exercise, the power rule applies directly since the function to integrate, after polynomial expansion, consists of simple power functions of the form \(x^n\). Mastering these techniques is crucial, as it enables the evaluation of more complex integrals that arise in various fields of science and engineering.

When faced with an integrand like \( (x-2)^3 \) at the first step of the textbook solution, the power rule becomes very practical. This rule states that the integral of \( x^n \) with respect to \( x \) is \( \frac{1}{n+1}x^{n+1} \) followed by the constant of integration. In practice, this technique greatly simplifies the integration process, as seen in the step-by-step solution provided.

Polynomial Expansion

In calculus, before we can integrate a complex expression like \( (x-2)^3 \) presented in our exercise, we may need to perform a polynomial expansion. This is the process of rewriting a raised power expression, which usually appears as a binomial like \( (a+b)^n \) or \( (a-b)^n \) into its extended form. We use the binomial theorem or simple algebraic multiplication for this purpose.

The expansion of \( (x-2)^3 \) is carried out by multiplying the binomial by itself three times (as indicated by the exponent), leading to \( x^3 - 6x^2 + 12x - 8 \). This expanded form then readily allows for term-by-term integration using basic rules of integration, as demonstrated in the solution. Understanding polynomial expansion is essential as it provides the simplified form required for direct integration and sets the stage for accurate calculations of areas and volumes, among other applications.

Calculus

Calculus, the mathematical study of continuous change, is split mainly into two branches: differential calculus and integral calculus, which are connected by the fundamental theorem of calculus. In the context of our exercise, we are delving into integral calculus, which is concerned with the accumulation of quantities, such as areas under curves.

Integral calculus plays a vital role in numerous areas of mathematics and applied science. From computing the trajectory of a spacecraft, analyzing the growth of populations in biology, to calculating the distribution of electric charge in physics, the application of integral calculus is versatile and widespread. The key to mastering calculus is to understand its concepts thoroughly and practice problems variably, starting from simple ones, like integrals of polynomials, to more complex multi-variable integrals.

Definite Integral Evaluation

Definite integrals have both a starting and an ending point, which are known as the limits of integration. To evaluate a definite integral, we find the antiderivative of the function with respect to the variable of integration, just like in an indefinite integral, but then we also take the difference of the antiderivative's values at the upper and lower limits.

In the exercise at hand, the definite integral \( \int_{0}^{3}(x-2)^{3} dx \) is evaluated by first finding the antiderivative of the expanded polynomial and then applying the limits 0 and 3 to the result, as illustrated in the solution steps. The values obtained after putting in the limits are subtracted, effectively calculating the 'net area' under the curve of the integrand between the two points. The ability to evaluate definite integrals accurately is critical for applications in physics, economics, engineering, and more, where precise quantitative analysis is essential.