Question

Midpoint Rule approximations Find the indicated Midpoint Rule approximations to the following integrals. $$\int_{2}^{10} 2 x^{2} d x \text { using } n=1,2, \text { and } 4 \text { sub-intervals }$$

Short answer

Answer: The Midpoint Rule approximations for the definite integral of \(f(x) = 2x^2\) over the interval \([2, 10]\) using \(n=1, 2, \text{and } 4\) sub-intervals are approximately: 1. For \(n=1\): \(M_1 \approx 576\) 2. For \(n=2\): \(M_2 \approx 640\) 3. For \(n=4\): \(M_4 \approx 656\)

Step-by-step solution

Identify the function and the interval

We have the function \(f(x) = 2x^2\) and the interval \([2, 10]\). We will approximate the definite integral of this function over the interval using the Midpoint Rule.

Calculate the width of a single sub-interval

The width of a single sub-interval can be found using the formula: $$\Delta x = \frac{b-a}{n}$$ where \(a\) and \(b\) are the endpoints of the interval, and \(n\) is the number of sub-intervals. We will compute the width of a sub-interval for each \(n\) value. \(1.\) For \(n=1\): $$\Delta x = \frac{10-2}{1} = 8$$ \(2.\) For \(n=2\): $$\Delta x = \frac{10-2}{2} = 4$$ \(3.\) For \(n=4\): $$\Delta x = \frac{10-2}{4} = 2$$

Calculate the midpoints of each sub-interval

Using the calculated sub-interval width, we will find the midpoints of each sub-interval. The midpoint of a sub-interval is obtained by adding half of the width to the left endpoint value of the sub-interval. \(1.\) For \(n=1\): Midpoint = \(2 + \frac{8}{2} = 6\) \(2.\) For \(n=2\): Midpoints = \(2 + \frac{4}{2} = 4\) and \(6 + \frac{4}{2} = 8\) \(3.\) For \(n=4\): Midpoints = \(2 + \frac{2}{2} = 3, 4 + \frac{2}{2} = 5, 6 + \frac{2}{2} = 7, \text{and } 8 + \frac{2}{2} = 9\)

Calculate the Midpoint Rule approximation

We will now calculate the Midpoint Rule approximation by plugging the midpoints into the function, multiplying the result by the width of the sub-interval, and summing the products: \(1.\) For \(n=1\): $$M_1 = 8(2(6^2)) = 8(72) = 576$$ \(2.\) For \(n=2\): $$M_2 = 4(2(4^2)) + 4(2(8^2)) = 4(32) + 4(128) = 128+512 = 640$$ \(3.\) For \(n=4\): $$M_4 = 2(2(3^2)) + 2(2(5^2)) + 2(2(7^2)) + 2(2(9^2)) = 2(18) + 2(50) + 2(98) + 2(162) = 36+100+196+324 = 656$$

Summarize the results

The Midpoint Rule approximations for the definite integral of \(f(x) = 2x^2\) over the interval \([2, 10]\) using \(n=1, 2, \text{and } 4\) sub-intervals are as follows: \(1.\) For \(n=1\): $$M_1 \approx 576$$ \(2.\) For \(n=2\): $$M_2 \approx 640$$ \(3.\) For \(n=4\): $$M_4 \approx 656$$

Key concepts

Numerical Integration

Numerical integration is a method used to approximate the value of a definite integral, especially when an antiderivative cannot be easily found or when working with data rather than functions.
Instead of finding the exact area under a curve, which can sometimes be complex or impossible to calculate analytically, numerical integration provides a practical solution. Here, the integral is approximated by summing the areas of geometric shapes – like rectangles, trapezoids, or other shapes – that fit under the curve.
Some popular methods of numerical integration include the Trapezoidal Rule, Simpson's Rule, and the Midpoint Rule.

• Trapezoidal Rule: Approximates the area by dividing it into trapezoids.
• Simpson's Rule: Uses parabolic arcs instead of straight lines or constant height.
• Midpoint Rule: Uses rectangular approximations where the midpoint is used for the height of the rectangle.

Each method has different applications, and their accuracy can vary based on the function being integrated and the number of sub-intervals used.

Definite Integral

The definite integral is a fundamental concept in calculus, representing the accumulation of quantities, such as the area under a curve within a certain interval.
It is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the boundaries of the interval and \( f(x) \) is the integrand function.
Definite integrals have a wide range of real-world applications, such as calculating distances, areas, volumes, and other physical quantities where values change continuously. When applying the definite integral, one obtains a number, unlike the indefinite integral which results in a family of functions (an antiderivative).

• To solve a definite integral, one can use analytical methods (finding an antiderivative) if possible.
• If analytical solutions are challenging, numerical integration methods, like the Midpoint Rule, provide an approximation.

The definite integral not only helps in mathematical theory but also in practical applications such as physics and engineering.

Sub-intervals

Sub-intervals are smaller sections into which the overall interval of integration is divided.
They are critical in the numerical integration process because they determine the granularity of the approximation. The number of sub-intervals, denoted by \( n \), affects the width of each sub-interval and thus the accuracy of the estimation of the integral.
When using the Midpoint Rule to approximate an integral, the larger the number of sub-intervals, the narrower each one is, resulting in a more accurate approximation:

• The width of each sub-interval is calculated as \( \Delta x = \frac{b-a}{n} \).
• More sub-intervals mean narrower widths, leading to less approximation error and fitting the function more closely.
• The choice of \( n \) can balance complexity and precision based on the specific requirements.

Choosing the appropriate number of sub-intervals is crucial for achieving the desired accuracy in numerical integration techniques.