Question

A midpoint Riemann sum Approximate the area of the region bounded by the graph of \(f(x)=100-x^{2}\) and the \(x\) -axis on [0,10] with \(n=5\) subintervals. Use the midpoint of each subinterval to determine the height of each rectangle (check your book to see figure).

Short answer

Answer: The approximate area is 670.

Step-by-step solution

Divide the interval into equal subintervals

We will divide the interval [0, 10] into 5 equal subintervals. To do this, first find the width of each subinterval: $$\Delta x = \frac{10 - 0}{5} = 2$$ So each subinterval has a width of 2.

Calculate the midpoint of each subinterval

Now we need to find the midpoint of each subinterval. These midpoints are: $$x_1 = 1, x_2 = 3, x_3 = 5, x_4 = 7, x_5 = 9$$

Evaluate the function at the midpoint

Now that we have the midpoints of each subinterval, we need to evaluate the function \(f(x)\) at each of these points to determine the height of each rectangle: $$f(x_1) = f(1) = 100 - 1^2 = 99$$ $$f(x_2) = f(3) = 100 - 3^2 = 91$$ $$f(x_3) = f(5) = 100 - 5^2 = 75$$ $$f(x_4) = f(7) = 100 - 7^2 = 51$$ $$f(x_5) = f(9) = 100 - 9^2 = 19$$ These values will be used as the heights of the rectangles.

Calculate the area of each rectangle

Now, let's find the area of each rectangle by multiplying the height by the width (2) for each subinterval: $$A_1 = f(x_1) \times \Delta x = 99 \times 2 = 198$$ $$A_2 = f(x_2) \times \Delta x = 91 \times 2 = 182$$ $$A_3 = f(x_3) \times \Delta x = 75 \times 2 = 150$$ $$A_4 = f(x_4) \times \Delta x = 51 \times 2 = 102$$ $$A_5 = f(x_5) \times \Delta x = 19 \times 2 = 38$$

Sum up the areas of all the rectangles

Finally, we add up the areas of all the rectangles to find the approximate total area under the curve: Total area ≈ \(A_1 + A_2 + A_3 + A_4 + A_5 = 198 + 182 + 150 + 102 + 38 = 670\) So, the approximate area of the region bounded by the graph of \(f(x) = 100 - x^2\) and the x-axis on [0, 10] using a midpoint Riemann sum with 5 subintervals is 670.

Key concepts

Approximate Area

In calculus, one common task is to find the area under a curve, which can represent various quantities such as distance, volume, or total cost. When the exact area is challenging to compute directly, we use methods like the midpoint Riemann sum for approximation. This technique involves dividing a region into rectangles and calculating the sum of their areas. Although these rectangles form only an estimate, they provide a convenient way to get a close value of the true area under the curve. Especially in the case of complex functions, this method can simplify the process significantly. The approximate area is calculated by adding the areas of each rectangle, determined by their height and width. The height corresponds to the value of the function at the midpoint of each subinterval. The midpoint Riemann sum specifically offers improved accuracy over other Riemann sums by using midpoints rather than endpoints to determine these heights. This technique balances the inevitable errors caused by the discrepancies between the rectangles and the actual curve.

Subintervals

When using Riemann sums, subintervals are critical because they help divide the interval into manageable sections. Our original problem involves the interval [0, 10] being divided into 5 equal subintervals.

• The formula used here for finding the width of each subinterval is equation eforementioned: \( \Delta x = \frac{b-a}{n} \) where \( b \) and \( a \) are the bounds of the interval, and \( n \) is the number of subintervals. For our task, the calculation provided a width of 2 for each subinterval— an important parameter for determining both the height and area of each rectangle.

These subdivisions are essential because they allow for a more precise estimation of the area by breaking the curve into smaller, more easily analyzed sections. Increasing the number of subintervals generally leads to a more accurate approximation, as narrower rectangles fit more closely under the curve.

Midpoints

The midpoint is a central concept in constructing midpoint Riemann sums, as the name suggests. For each subinterval, the midpoint needs to be identified as it is used to calculate the height of the rectangle associated with that subinterval.

• In our exercise, the key midpoints for the subintervals [0,2], [2,4], [4,6], [6,8], and [8,10] are calculated as follows: 1, 3, 5, 7, and 9.
• The formula for finding a midpoint is straightforward. For a subinterval [a, b], the midpoint is calculated as \( x = \frac{a+b}{2} \).

Midpoints provide a balance between underestimating and overestimating the area of each subinterval, which offers a more accurate representation compared to using either endpoint of each subinterval.

Function Evaluation

Once we've established the midpoints of our subintervals, the next step is to evaluate the function at these points. This gives the height of the rectangles used in the estimation process.We take each midpoint, substitute it into our function, and evaluate it. For example, our function \( f(x) = 100 - x^2 \) is evaluated at each midpoint:

• At \( x_1 = 1 \), \( f(1) = 100 - 1^2 = 99 \)
• At \( x_2 = 3 \), \( f(3) = 100 - 3^2 = 91 \)
• And so forth with the midpoints 5, 7, and 9.

The results of these evaluations give us the heights, which, when multiplied by the width of the subintervals, yield the area of each rectangle. Evaluating a function at midpoints rather than endpoints generally provides a more realistic approximation since they are more representative of the whole subinterval.