Question

Use the Midpoint Rule with \(n=4\) to approximate the area of the region. Compare your result with the exact area obtained with a definite integral. $$ f(x)=\frac{1}{x}, \quad[1,5] $$

Short answer

The approximate area using the Midpoint Rule is 2.083, while the exact area using definite integral is 1.609. The Midpoint approximation is over by about 0.474.

Step-by-step solution

Calculation of Δx

First, calculate the width of the rectangles that make up the Riemann Sum. For the interval \([1,5]\) and \(n=4\), the width \(Δx\) can be calculated as: \[ Δx = \frac{5-1}{4} = 1 \]

Setup and Solve the Midpoint Rule

Midpoint Rule is given as: \[M_n = Δx [ f(x_1) + f(x_2) + .... + f(x_n) ]\] where \(x_i = a + \frac{Δx}{2} + (i-1) Δx\) is the point in the middle of each subinterval. Hence compute \(f(x_i)\) for \(i = 1,2,3,4\), and insert it into the Midpoint Rule formula to obtain the approximate area. For \(i = 1,2,3,4\) we get \(x_i = 1, 2, 3, 4\) respectively. The function \(f(x)\) is \(1/x\), so the values of the terms are \[f(x_1) = 1, f(x_2) = 0.5, f(x_3) = 0.333, f(x_4) = 0.25\]. The Midpoint Rule approximation becomes: \(M_n = 1 [1 + 0.5 + 0.333 + 0.25] = 2.083\)

Calculate Exact Area With Integration

We know the function we want to integrate is \(f(x) = 1/x\), and we want to integrate from 1 to 5. This integral is performed as follows: \[ ∫_{1}^{5} \frac{1}{x} dx = [ ln |x| ]_{1}^{5} = ln|5| - ln|1| = ln(5)\]

Compare Midpoint Rule with Integration

Now the problem requires to compare the result of the midpoint rule with the exact solution by integration. The Midpoint Rule gave an approximate area of 2.083 under the curve, while the true area under the curve from 1 to 5 is \(ln(5) = 1.609\). So, the result from the Midpoint Rule is a little higher than the true value by about 0.474.

Key concepts

Riemann Sum

The Riemann Sum is an important mathematical concept used to approximate the area under a curve. It is a method of constructing a sum from values of a function at certain points within a specified interval. This sum approximates the value of a definite integral.In our exercise, the Midpoint Rule is a form of Riemann Sum. Imagine placing a series of rectangles under a curve. The base of each rectangle fits exactly into subintervals of your chosen interval - in this case, the interval [1,5]. These rectangles' heights are determined by the value of the function at a point, usually the midpoint, inside each subinterval.To calculate a Riemann Sum, follow these steps:

• Divide the interval into equal subintervals, each with a width of \(\Delta x\).
• Choose a point within each subinterval, such as the midpoint.
• Calculate the function value at each chosen point to determine the height of the rectangles.
• Multiply the width \(\Delta x\) by each function value, and sum these products to get the total approximate area.

The precision of the Riemann Sum improves with more subdivisions, capturing the curve's area more accurately.

definite integral

A definite integral is a fundamental concept in calculus that represents the exact area under a curve within a specific interval. Unlike approximation methods, definite integrals provide the exact value without error, assuming the function is known and can be integrated.When calculating a definite integral, the area between the curve of a function \(f(x)\) and the x-axis over the interval \([a, b]\) is calculated using the integral symbol \(\int\) as follows:\[\int_{a}^{b} f(x) \, dx\]The process involves finding the antiderivative of \(f(x)\) and then evaluating it at the bounds of the interval. The result gives the total area, some of which might be negative if parts of the curve are below the x-axis.Using our exercise example, integrating \(f(x) = \frac{1}{x}\) from 1 to 5, we find:\[\int_{1}^{5} \frac{1}{x} \, dx = \left[ \ln |x| \right]_{1}^{5} = \ln(5) - \ln(1) = \ln(5)\]This integration shows that the true area under the curve, from 1 to 5, is \(\ln(5)\), exact and with minimal computation error, illustrating how powerful a tool definite integration is.

approximation methods

Approximation methods in mathematics are techniques used to estimate values that are difficult or impossible to measure exactly. These methods are particularly useful when dealing with functions that are complex or when an exact solution is not feasible due to time constraints or computational limitations.In calculus, several approximation methods estimate the area under a curve:

• The Left Riemann Sum uses the left endpoints to calculate areas.
• The Right Riemann Sum utilizes the right endpoints.
• The Trapezoidal Rule calculates average values of left and right endpoints, essentially forming trapezoids instead of rectangles.
• Lastly, the Midpoint Rule, as used in our original exercise, takes the function value at the midpoint of each subinterval.

Each method has varying levels of accuracy and efficiency. The Midpoint Rule is particularly effective because it balances the tendency of the curve to under-estimate or over-estimate area, provided the function is reasonably smooth.Approximations are revised constantly to deliver better accuracy. Increasing the number of subintervals \(n\) is a common strategy to enhance precision. As \(n\) increases, the approximation typically converges closer to the true value, reflecting the integral's exact area adequately.