Problem 100 - Calculus Volume 1

Question

In the following exercises, given that \(\int_{0}^{1} x d x=\frac{1}{2}, \int_{0}^{1} x^{2} d x=\frac{1}{3}, \quad\) and \(\quad \int_{0}^{1} x^{3} d x=\frac{1}{4}\) compute the integrals. \(\int_{0}^{1}(1-x)^{2} d x\)

Step-by-Step Solution

Answer

The integral \( \int_{0}^{1}(1-x)^{2} \, dx \) equals \( \frac{1}{3} \).

Step 1: Expand the Integrand

First, expand the expression \( (1-x)^2 \). This gives us: \[(1-x)^2 = 1 - 2x + x^2\]

Step 2: Separate the Integral

Now, express the given integral \( \int_{0}^{1}(1-x)^{2} \ dx \) as the sum of simpler integrals:\[\int_{0}^{1}(1-x)^{2} \, dx = \int_{0}^{1} 1 \, dx - 2 \int_{0}^{1} x \, dx + \int_{0}^{1} x^2 \, dx\]

Step 3: Calculate Each Integral

Calculate each part separately:1. \( \int_{0}^{1} 1 \, dx \): The integral of 1 from 0 to 1 is simply the length of the interval, which is 1.2. \( 2 \int_{0}^{1} x \, dx \): Based on the given information, \( \int_{0}^{1} x \, dx = \frac{1}{2} \), so \( 2 \times \frac{1}{2} = 1 \).3. \( \int_{0}^{1} x^2 \, dx \): From the given data, \( \int_{0}^{1} x^2 \, dx = \frac{1}{3} \).

Step 4: Combine the Results

Combine the results from the previous step to get the final answer:\[1 - 1 + \frac{1}{3} = \frac{1}{3}\]

Key Concepts

Integration Techniques

Integration, in basic terms, is the process of finding the integral of a function. It's much like summing up infinitesimal parts to find the whole.
One common integration technique involves breaking down complex expressions into simpler parts. This was demonstrated when solving the integral of \( (1-x)^2 \), where we expanded it to \( 1 - 2x + x^2 \). These simpler terms can then be individually integrated.
Other techniques include substitution, which involves changing the variable of integration to simplify the integral; and integration by parts, which breaks down products of functions. While not used in this problem, these methods form the foundational tools to approach different types of integrals effectively.

Polynomial Integration

Every polynomial can be integrated term by term to find an integral solution. The integral of a polynomial encompasses integrating each term separately and then adding the results.
For each simple power function, such as \( x^n \), we can find its integral using the formula:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

For the exercise, once \( (1-x)^2 \) was expanded, we could compute the integrals for \( x \) and \( x^2 \) using the given values. This highlights how polynomial integration simplifies each term individually, making the integration process straightforward.

Basic Calculus Concepts

Calculus is essential for understanding changes in and around natural phenomena. In this context, we focus on the definite integrals which measure the accumulated area under curves between two points.
The given exercise required computing the definite integral of a function from 0 to 1. A key property of definite integrals is that they are additive. This was used when the expanded form \( 1 - 2x + x^2 \) was written as separate integrals that could be summed up.
Knowing the results of basic integrals such as \( \int x \, dx \) or \( \int x^2 \, dx \) allows us to tackle more complex functions by using these foundational results. It reflects the essence of calculus: building complex solutions from simple, fundamental principles.