Question

Use the midpoint rule to estimate \(\int_{0}^{1} x^{2} d x\) using four subintervals. Compare the result with the actual value of this integral.

Short answer

The estimated integral using the midpoint rule is 0.328125, very close to the actual 0.333333.

Step-by-step solution

Divide the Interval

First, determine the interval over which you want to estimate the integral, which is \( [0, 1] \) in this case. With four subintervals, each subinterval will have a length \( \Delta x = \frac{1-0}{4} = 0.25 \). Thus, the subintervals are \([0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1]\).

Find Midpoints

Next, find the midpoint of each subinterval. The midpoints for our subintervals will be:1. For \([0, 0.25]\), midpoint is \(x_1 = 0.125\)2. For \([0.25, 0.5]\), midpoint is \(x_2 = 0.375\)3. For \([0.5, 0.75]\), midpoint is \(x_3 = 0.625\)4. For \([0.75, 1]\), midpoint is \(x_4 = 0.875\)

Calculate Function Values at Midpoints

Calculate the value of the function \( f(x) = x^2 \) at each midpoint:1. \( f(0.125) = (0.125)^2 = 0.015625 \)2. \( f(0.375) = (0.375)^2 = 0.140625 \)3. \( f(0.625) = (0.625)^2 = 0.390625 \)4. \( f(0.875) = (0.875)^2 = 0.765625 \)

Apply the Midpoint Rule

The midpoint rule can be expressed as:\[ \int_{a}^{b} f(x) \, dx \approx \Delta x [f(x_1) + f(x_2) + \ldots + f(x_n)] \]Plug in the values to compute the approximation:\[ \approx 0.25 \times (0.015625 + 0.140625 + 0.390625 + 0.765625) = 0.25 \times 1.3125 = 0.328125 \]

Calculate Actual Integral Value

The actual value of the integral \( \int_{0}^{1} x^2 \, dx \) is found by calculating the antiderivative of \(x^2\) and evaluating at the bounds:\[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \approx 0.333333 \].

Compare Results

Finally, compare the estimated result (0.328125) from the midpoint rule with the actual value (0.333333). The estimation is very close but slightly less than the true value.

Key concepts

Integral Approximation

Integral approximation is a useful approach in calculus for estimating the area under a curve. When dealing with definite integrals, sometimes calculating the exact area can be complicated or impossible analytically. In such cases, approximation methods provide a practical solution. These methods are especially important in applied mathematics, physics, and engineering, when we need quick estimations without the luxury of time-consuming exact calculations.

To approximate integrals, different methods like the Trapezoidal Rule, Simpson’s Rule, and the Midpoint Rule can be applied. Each of these has its specifics in terms of ease of use and accuracy. By using these methods, one can break down the integral into smaller, more manageable computations, yielding results that are close to the actual integral value. Integral approximation enables us to deal with integrals in a more practical manner when exact solutions are difficult to obtain.

Midpoint Rule

The Midpoint Rule is a popular method for approximating the integral of a function. This technique involves dividing the interval into subintervals of equal length and using the midpoint of each subinterval to estimate the area under the curve.

Here's how it works:

• First, divide the entire interval into smaller, equal subintervals.
• Find the midpoint of each subinterval. This midpoint will be the point at which the function's value is calculated.
• Calculate the function value at each midpoint.
• Multiply each of these values by the width of the subintervals.

The sum of these products will give you the approximate integral.

The Midpoint Rule is particularly effective when the function is relatively smooth, offering a good balance between simplicity of calculation and accuracy of result. When applied to the integral from 0 to 1 of the function \(x^2\) using four subintervals, the Midpoint Rule provided an estimated result quite close to the actual value, illustrating its effectiveness.

Definite Integral

A definite integral is a fundamental concept in calculus representing the area under a curve between two specific points on the x-axis. The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration.

Definite integrals are used in various fields, such as physics and engineering, to calculate quantities like distance, area, and volume when the relationship between the variables is known. For example, integrating a velocity function over time gives the distance traveled. The neat part about definite integrals is they can provide the exact area without approximation when the function is integrable.

In our exercise, the calculation \( \int_{0}^{1} x^2 \, dx \) shows how the definite integral of \(x^2\) gives an exact value of \( \frac{1}{3} \). This capability makes definite integrals powerful tools in mathematical analysis.

Numerical Integration

Numerical integration refers to a broad set of algorithms for obtaining the numerical value of a definite integral. This process is crucial when an analytical solution is not possible.

There are several common methods for numerical integration, among which the Midpoint Rule, the Trapezoidal Rule, and Simpson's Rule stand out based on their simplicity and implementation. These methods approximate the integral by summing up areas of shapes that closely resemble the area under the curve.

In practical applications, numerical integration is used extensively in data analysis and simulation models. It allows us to work with functions that are defined by discrete data points without a clear analytical formula. By employing methods of numerical integration, such as the Midpoint Rule used in our exercise, one can achieve a remarkable approximation to the actual integral value, proving incredible utility, especially in computing integral values from complex or non-standard functions.