Question

Find \(\int_{0}^{2} f(x) d x\), where \(f(x)=\left\\{\begin{array}{c}\frac{1}{\sqrt[4]{x^{3}}} \quad \text { for } 0 \leq x \leq 1 \\ \frac{1}{\sqrt[4]{(x-1)^{3}}}\end{array}\right.\) for \(1

Short answer

Answer: The definite integral of the given function over the interval [0, 2] is \(\frac{8}{3}\).

Step-by-step solution

Determine the integral for each case

For 0 ≤ x ≤ 1, \(f(x)=\frac{1}{\sqrt[4]{x^{3}}}\). Then for 1 < x < 2, \(f(x)=\frac{1}{\sqrt[4]{(x-1)^{3}}}\). First, let's find the integral of each case: 1. Case 1: \(f(x)=\frac{1}{\sqrt[4]{x^{3}}}\). Rewrite this function as \(f(x)=x^{-\frac{3}{4}}\). Then, the integral of this function is: $ \int x^{-\frac{3}{4}}d x = \frac{4}{3}x^{\frac{1}{4}}+\mathcal{C} $ 2. Case 2: \(f(x)=\frac{1}{\sqrt[4]{(x-1)^{3}}}\). Rewrite this function as \(f(x)=(x-1)^{-\frac{3}{4}}\). Then, the integral of this function is: $ \int (x-1)^{-\frac{3}{4}} d x = \frac{4}{3}(x-1)^{\frac{1}{4}}+\mathcal{C} $

Split the definite integral into two terms

Since our function f(x) is defined piecewise, we can split the definite integral from 0 to 2 into two terms: (0 ≤ x ≤ 1) and (1 < x < 2). $ \int_{0}^{2} f(x) d x = \int_{0}^{1} \frac{1}{\sqrt[4]{x^{3}}} d x+\int_{1}^{2} \frac{1}{\sqrt[4]{(x-1)^{3}}} d x $

Evaluate the definite integral for each term

Now, substitute the integrals we found in Step 1 into each term, and evaluate the definite integral for each term: 1. \(\int_{0}^{1} \frac{1}{\sqrt[4]{x^{3}}} d x = \int_{0}^{1} x^{-\frac{3}{4}} d x = \left[\frac{4}{3}x^{\frac{1}{4}}\right]_{0}^{1}= \frac{4}{3}(1)^{\frac{1}{4}} - \frac{4}{3}(0)^{\frac{1}{4}} = \frac{4}{3}-0 =\frac{4}{3}\) 2. \(\int_{1}^{2} \frac{1}{\sqrt[4]{(x-1)^{3}}} d x = \int_{1}^{2} (x-1)^{-\frac{3}{4}} d x = \left[\frac{4}{3}(x-1)^{\frac{1}{4}}\right]_{1}^{2}= \frac{4}{3}(1)^{\frac{1}{4}} - \frac{4}{3}(0)^{\frac{1}{4}} = \frac{4}{3}-0 =\frac{4}{3}\)

Add both terms to find the required definite integral

Finally, add both definite integrals together: $ \int_{0}^{2} f(x) d x = \frac{4}{3}+\frac{4}{3}=2\cdot\frac{4}{3}=\frac{8}{3} $ Thus, the definite integral of the given function over the interval [0, 2] is \(\frac{8}{3}\).

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on the concepts of integration and its applications. This field plays a crucial role in science, engineering, and mathematics because it allows for the computation of areas, volumes, and the accumulation of quantities, among other things.

In a more technical sense, the process of integration is the reverse operation of differentiation, which means finding the antiderivative or integral of a function can tell us a lot about the original function's behavior over a certain interval. For example, in physics, integration enables us to calculate quantities such as work done by a force or the total charge in an electric field.

Piecewise Function Integration

Integration of piecewise functions requires understanding the concept of a function that is defined by different expressions over different intervals. These functions often represent systems where the rules change under different circumstances, such as an object moving at different speeds during different time intervals.

When integrating a piecewise function, it's important to handle each piece separately over its specific interval. In the exercise provided, the function is different for the intervals \(0 \leq x \leq 1\) and \(1 < x < 2\), requiring separate evaluations of their integrals. Often, to make integration simpler, these functions can be rewritten in a more familiar form, such as using exponents, to apply the standard rules of integration more easily.

Definite Integral Evaluation

The definite integral evaluation is a crucial process in integral calculus. It determines the exact value of the area under a curve or the accumulation of a quantity across an interval. The definite integral has specific limits, which in this case are 0 and 2 for the exercise provided.

To evaluate a definite integral, you substitute the upper and lower limits of integration into the indefinite integral, which has been calculated without considering the limits, and then subtract the resulting values. This process, known as the 'Fundamental Theorem of Calculus,' enables mathematicians and scientists to determine the accumulated quantity over an interval with precision.