91Ó°ÊÓ

The first way we can approximate the area under a curve using a Riemann sum is by employing the left-hand endpoint method. This technique involves selecting the left endpoint of each subinterval to determine the height of our approximating rectangles. Consider the function \( f(x) = x^2 - 1 \) over the interval \([0, 2]\), divided into four subintervals of equal length. These subintervals are \([0, 0.5]\), \([0.5, 1]\), \([1, 1.5]\), and \([1.5, 2]\).

For the left-hand endpoint Riemann sum, the function's value at the left endpoint of each subinterval specifies the rectangle's height:

• Subinterval \([0, 0.5]\): Height \(f(0) = -1\)
• Subinterval \([0.5, 1]\): Height \(f(0.5) = -0.75\)
• Subinterval \([1, 1.5]\): Height \(f(1) = 0\)
• Subinterval \([1.5, 2]\): Height \(f(1.5) = 1.25\)

The width of each rectangle, \(\Delta x\), is 0.5. The total approximate area, the left-hand Riemann sum, is calculated by summing these areas: \[ A_{\text{left}} = 0.5 \times (-1 + (-0.75) + 0 + 1.25) = -0.25 \]This calculation gives us an approximate value for the integral of \(f(x)\) over the interval \([0, 2]\). The result is negative because the area below the x-axis is also considered.

In contrast to the left-hand method, the right-hand endpoint method uses the right endpoints of each subinterval to establish the height of the rectangles. This approach provides another way to approximate an integral. For our example function, \( f(x) = x^2 - 1 \) on \([0, 2]\), this method utilizes the following right endpoints of each subinterval:

• Subinterval \([0, 0.5]\): Height \(f(0.5) = -0.75\)
• Subinterval \([0.5, 1]\): Height \(f(1) = 0\)
• Subinterval \([1, 1.5]\): Height \(f(1.5) = 1.25\)
• Subinterval \([1.5, 2]\): Height \(f(2) = 3\)

The width of each rectangle remains \(\Delta x = 0.5\). To find the total approximate area using the right-hand method, sum the areas of each rectangle:\[ A_{\text{right}} = 0.5 \times (-0.75 + 0 + 1.25 + 3) = 1.75 \]The right-hand endpoint Riemann sum typically gives different results than the left-hand endpoint method. This discrepancy arises because the right-hand method places rectangles based on future values of the function compared to the interval.

The midpoint method for Riemann sums provides yet another way of approximating the area under a curve. This technique selects the midpoint of each subinterval to define the heights of the rectangles. For the function \( f(x) = x^2 - 1 \) defined over \([0, 2]\), the midpoint calculations are based on the center point of each subinterval:

• Subinterval \([0, 0.5]\): Midpoint \(0.25\); Height \(f(0.25) = -0.9375\)
• Subinterval \([0.5, 1]\): Midpoint \(0.75\); Height \(f(0.75) = -0.4375\)
• Subinterval \([1, 1.5]\): Midpoint \(1.25\); Height \(f(1.25) = 0.5625\)
• Subinterval \([1.5, 2]\): Midpoint \(1.75\); Height \(f(1.75) = 2.0625\)

The rectangle width remains \(\Delta x = 0.5\). Summing the resultant areas gives the midpoint Riemann sum as:\[ A_{\text{mid}} = 0.5 \times (-0.9375 - 0.4375 + 0.5625 + 2.0625) = 0.625 \]The midpoint Riemann sum typically offers a more balanced and often closer approximation of the integral than either the left or right endpoint methods. This is because midpoints naturally average over the interval, smoothing out the effects of the curvature of \( f(x) \).