91Ó°ÊÓ

Approximating the area under a curve is an essential concept in calculus. It helps us estimate the integral of a function, especially when an exact integration is complex or impossible. This technique is vital for understanding how functions behave across a certain interval.

In the given problem, we are focusing on the function defined as:

• \( f(x) = x^2 + 2 \)

We want to approximate the area between this function and the x-axis over the interval

• \([0, 2] \)

Using different Riemann sums allows us to break this interval into smaller sub-intervals and calculate approximations through the summation of the areas of geometric shapes—usually rectangles—in various strategic points along the curve.

Each type of Riemann sum uses a different approach to approximating these areas, and thus each provides a slightly different result. The three main types involve taking these rectangle heights from the left endpoint, midpoint, or right endpoint of these sub-intervals.

The left Riemann sum is a popular technique for approximating the area under a curve. It involves taking the height of each rectangle from the left side of each sub-interval in your defined domain.

For our function, when we divide the interval

• \([0, 2] \) into \(n = 4\)

Equal sub-intervals, each of these has a width calculated as:

• \( \Delta x = \frac{2 - 0}{4} = 0.5 \)

The left endpoints are

• \(0, 0.5, 1, 1.5\)

Calculating the function values at these points gives us *rectangle heights* which we multiply by the sub-interval width to find the area:

• \(f(0) = 2, f(0.5) = 2.25, f(1) = 3, f(1.5) = 4.25\)

The left Riemann sum is thus:

• \( (0.5)(2) + (0.5)(2.25) + (0.5)(3) + (0.5)(4.25) = 5.75 \)

Generally, the rectangles will tend to overestimate the area if the function is increasing, since all endpoints are at a lower x-value compared to most of the sub-interval.

When using the midpoint Riemann sum, we consider the value of the function at the center or midpoint of each sub-interval. This method often provides a more balanced approximation as it centers each rectangle around the curve, sometimes compensating for the changes of the function better than left or right sums.

In our calculation with

• \( = 4 \)

Sub-intervals, we calculate the midpoints as

• \(0.25, 0.75, 1.25, 1.75\)

Using the function to find the heights of rectangles gives us these values:

• \(f(0.25) = 2.0625, f(0.75) = 2.5625, f(1.25) = 3.5625, f(1.75) = 5.0625\)

Thus, the midpoint Riemann sum is:

• \( (0.5)(2.0625) + (0.5)(2.5625) + (0.5)(3.5625) + (0.5)(5.0625) = 6.625 \)

This method tends to give a more accurate estimation especially when the function has a more uniform curve over the interval.

In the right Riemann sum approach, we select the height of each rectangle according to the value of the function at the *right* endpoint of each sub-interval. This method can skew the summation to overestimate (on increasing functions) or underestimate (on decreasing functions).

For the interval

• \([0, 2] \) split into\( n=4 \)

Sub-intervals with:

• Width as\( \Delta x = 0.5 \)

We use the right endpoints

• \(0.5, 1, 1.5, 2\)

The rectangular heights from the function will be:

• \(f(0.5) = 2.25, f(1) = 3, f(1.5) = 4.25, f(2) = 6\)

The calculated right Riemann sum is:

• \( (0.5)(2.25) + (0.5)(3) + (0.5)(4.25) + (0.5)(6) = 7.75 \)

This technique illustrates an overestimation of the area especially for a function \((f(x) = x^2 + 2)\) which is increasing over the interval.