Question

Sketch the graph of the given function over the interval \([a, b] ;\) then divide \([a, b]\) into \(n\) equal subintervals. Finally, calculate the area of the corresponding circumscribed polygon. $$ f(x)=x^{2}-1 ; a=2, b=3, n=6 $$

Short answer

The area of the circumscribed polygon is approximately 5.754.

Step-by-step solution

Understand the Function and Interval

The function given is \( f(x) = x^2 - 1 \). We need to sketch this over the interval \([2, 3]\). The function is a parabola opening upwards, shifted down by 1.

Divide the Interval into Subintervals

The interval \([a, b]\) is \([2, 3]\). We divide this interval into \( n = 6 \) equal parts. Calculate the width of each subinterval: \( \Delta x = \frac{3 - 2}{6} = \frac{1}{6} \).

Determine the Subinterval Points

The points dividing the interval from \(2\) to \(3\) are: \(2, 2 + \frac{1}{6}, 2 + \frac{2}{6}, \ldots, 3\). Simplifying, these are \(2, 2.1667, 2.3333, 2.5, 2.6667, 2.8333, 3\).

Evaluate the Function at Right Endpoints

Evaluate \( f(x) = x^2 - 1 \) at each right endpoint of the subintervals: \(f(2.1667), f(2.3333), f(2.5), f(2.6667), f(2.8333), f(3)\).

Calculate the Function Values

Compute the function values: - \( f(2.1667) \approx 3.69445 \) - \( f(2.3333) \approx 4.44445 \) - \( f(2.5) = 5.25 \) - \( f(2.6667) \approx 6.11115 \) - \( f(2.8333) \approx 7.02775 \) - \( f(3) = 8 \)

Calculate Area of Circumscribed Polygon

The area of each rectangle is \( \text{height} \times \text{width} = f(x_i) \cdot \Delta x\). Sum the areas for all rectangles: \[\text{Total Area} = \Delta x \left( f(2.1667) + f(2.3333) + f(2.5) + f(2.6667) + f(2.8333) + f(3) \right).\]Substitute the values, then calculate: \[\text{Total Area} = \frac{1}{6}(3.69445 + 4.44445 + 5.25 + 6.11115 + 7.02775 + 8) \approx 5.75437.\]

Key concepts

Parabolic Curves

In integral calculus, understanding the shape and properties of the function being analyzed is crucial, especially when dealing with parabolas. Parabolas are distinct curves represented by quadratic functions. Our function, \( f(x) = x^2 - 1 \), is a classic example. In this case, the parabola opens upwards because of the positive coefficient of \( x^2 \). The graph is also shifted down by 1 unit due to the \(-1\) constant term. This means that the basic \(y = x^2\) curve is moved down along the y-axis by one unit, altering its vertex to a lower point at \((0, -1)\). This downward shift affects not only the graph aesthetically but influences calculations like area approximation, making understanding the shape fundamental.

Subintervals

When addressing problems involving integrals, partitioning the interval into subintervals is a key step. Given the interval \([a, b] = [2, 3]\), we divide it into \(n = 6\) equal parts or subintervals. Understanding subintervals helps in approximating the area under the curve more accurately. Each subinterval has a width denoted by \( \Delta x \).
This width is calculated as \( \Delta x = \frac{b-a}{n} = \frac{1}{6} \). Essentially, from 2 to 3, every \( \frac{1}{6} \) unit is a critical point for approximation. Knowing these points guarantees that each part is identical in width, which simplifies subsequent calculations such as function evaluation at strategic points. These points include the start and end of each subinterval, leading to precise area calculations.

Function Evaluation

To approximate areas under a curve using rectangles, evaluating the function at specific points is crucial. These evaluations are often done at the endpoints of subintervals. In our scenario, the right endpoints are chosen for calculations. Therefore, we evaluate \( f(x) = x^2 - 1 \) at the points \(2.1667, 2.3333, 2.5, 2.6667, 2.8333, 3\). Calculating the function's output at these precise intervals provides heights for the rectangles that approximate the area under the curve. This evaluation provides distinct values, for instance, \( f(2.5) = 5.25 \) and \( f(3) = 8 \), which directly relate to the polygonal area's height. Thus, function evaluation at strategic endpoints is integral for correctly approximating area under a parabolic curve.

Area Approximation

Approximating the area under a curve using rectangles involves adding up the areas of multiple sub-region rectangles. Each rectangle's area is calculated by multiplying the function's height value at a particular point by the subinterval width \( \Delta x \). Hence, the area of one rectangle is \( f(x_i) \cdot \Delta x \). For our function, we calculate repetitive areas for each subinterval and sum them to estimate the total area under the curve between \(a\) and \(b\).

• For instance, the rectangles' heights are determined by \( f(x) \)'s values at the points \(2.1667, 2.3333, 2.5, 2.6667, 2.8333, 3\).
• Then these are multiplied by the subinterval width of \( \frac{1}{6} \).

This method, often termed a Riemann sum, uses these rectangular estimates to approach the actual area under a curve, which might otherwise be calculated exactly via integration. Even though this approach gives an approximation, it's particularly significant when exact computation is complex, thus highlighting area approximation's utility in integral calculus.