Question

Approximate the following integrals using either the midpoint rule, trapezoidal rule, or Simpson's rule as indicated. (Round answers to three decimal places.) \(\int_{0}^{1} \sin ^{2}(\pi x) d x ;\) midpoint rule; \(n=3\)

Short answer

The approximate integral is 0.500.

Step-by-step solution

Determine the interval and subinterval width

We are given the integral \( \int_{0}^{1} \sin^2(\pi x) \, dx \) and we need to use the midpoint rule with \( n = 3 \) subintervals. The interval \([0, 1]\) will be divided into 3 equal subintervals. The width of each subinterval, \( \Delta x \), is calculated as follows: \[ \Delta x = \frac{b-a}{n} = \frac{1-0}{3} = \frac{1}{3} \]

Identify midpoints of subintervals

The midpoints of each subinterval are used to approximate the integral. Calculate the midpoints: \( x_1, x_2, \) and \( x_3 \). For each subinterval, the midpoint \( x_i \) is found using the formula: \( x_i = a + (i-0.5) \Delta x \). \[ x_1 = 0 + (1-0.5)\frac{1}{3} = \frac{1}{6}, \quad x_2 = 0 + (2-0.5)\frac{1}{3} = \frac{1}{2}, \quad x_3 = 0 + (3-0.5)\frac{1}{3} = \frac{5}{6} \]

Evaluate the function at the midpoints

Evaluate \( \sin^2(\pi x) \) at each of the midpoints calculated in Step 2. \[ f\left(\frac{1}{6}\right) = \sin^2\left(\frac{\pi}{6}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \] \[ f\left(\frac{1}{2}\right) = \sin^2\left(\frac{\pi}{2}\right) = 1^2 = 1 \] \[ f\left(\frac{5}{6}\right) = \sin^2\left(\frac{5\pi}{6}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

Apply the Midpoint Rule

Using the midpoint rule, approximate the integral as follows: \[ \int_{0}^{1} \sin^2(\pi x)dx \approx \Delta x \left( f\left(x_1\right) + f\left(x_2\right) + f\left(x_3\right) \right) \] \[ \approx \frac{1}{3} \left( \frac{1}{4} + 1 + \frac{1}{4} \right) \] \[ \approx \frac{1}{3} \times \frac{3}{2} \] \[ \approx \frac{1}{2} \] So the approximate integral, rounded to three decimal places, is 0.500.

Key concepts

Midpoint Rule

The Midpoint Rule is a method used for approximating the value of an integral. It is particularly useful when the function is very complex or when an exact solution is difficult to obtain. Here's how it works:

• First, you divide the interval over which you want to integrate into smaller, equal parts called subintervals.
• Then, for each subinterval, you find its midpoint.
• Next, you calculate the function value at each of these midpoints.
• Finally, you sum these values and multiply by the width of each subinterval to get the approximate value of the integral.

Think of the Midpoint Rule as a way to replace the actual curve of the function with a series of flat segments, using the midpoints for a better approximation. This method helps achieve a balance between computational simplicity and accuracy.

Integral Approximation

Integral approximation is the process of estimating the value of a definite integral, especially when an exact calculation is impractical. This is essential in many fields such as physics, engineering, and economics, where exact solutions are difficult to find.

Methods for approximating integrals include the Midpoint Rule, Trapezoidal Rule, and Simpson's Rule. Each has its strengths:

• Midpoint Rule: This uses the midpoint of each subinterval, providing a fairly accurate approximation.
• Trapezoidal Rule: This uses the ends of each subinterval to form trapezoids, typically providing smoother approximations for linear functions.
• Simpson's Rule: This applies quadratic polynomials through each subinterval, which usually gives a high precision for smooth functions.

The choice of method often depends on the nature of the function and the desired accuracy for the application you are working on. Always consider the balance between computational efficiency and approximation precision.

Subintervals

Subintervals are smaller divisions of the larger interval you are integrating over. In numerical integration, dividing the total interval into subintervals allows for more manageable calculations.

To determine the width of each subinterval, use the formula: \[ \Delta x = \frac{b-a}{n} \]

where \(b\) is the upper limit of the interval, \(a\) is the lower limit, and \(n\) is the number of subintervals. This equation gives you the width of each part, which is consistently used in methods like the Midpoint Rule to approximate integrals.

Subintervals help in breaking down complex calculations into simpler units. The more subintervals you use, typically the more accurate the approximation becomes, as it captures more variations of the function over smaller sections. However, more subintervals also mean more calculations, so finding a balance is key.