Problem 54 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the region under the curve \(y=f(x)\) over the interval \([a, b] .\) To do this, divide the interval \([a, b]\) into n equal subintervals, calculate the area of the corresponding circumscribed polygon, and then let \(n \rightarrow \infty .\) (See the example for \(y=x^{2}\) in the text. \()\) $$ y=\frac{1}{2} x^{2}+1 ; a=0, b=1 $$

Step-by-Step Solution

Answer

The area under the curve is \(\frac{7}{6}\).

Step 1: Divide the Interval into Subintervals

First, divide the interval \([a, b]\) into \(n\) equal subintervals. Here, we have \(a=0\) and \(b=1\), so the length of each subinterval is \(\Delta x = \frac{b-a}{n} = \frac{1}{n}\).

Step 2: Determine the Sample Points

Select a sample point in each subinterval, generally the right endpoint for simplicity. For the interval \([x_{i-1}, x_i]\), the right endpoint \(x_i\) is given by \(x_i = a + i\Delta x = \frac{i}{n}\).

Step 3: Calculate the Function Values at Sample Points

Evaluate the function \(f(x) = \frac{1}{2}x^2 + 1\) at each sample point \(x_i\). This gives \(f\left(\frac{i}{n}\right) = \frac{1}{2}\left(\frac{i}{n}\right)^2 + 1\).

Step 4: Sum the Areas of Rectangles

The area \(A\) of each rectangle is \(f(x_i)\Delta x\). Sum these areas:\[A_n = \sum_{i=1}^{n} \left(\frac{1}{2}\left(\frac{i}{n}\right)^2 + 1\right) \cdot \frac{1}{n}\]This simplifies to\[A_n = \sum_{i=1}^{n} \left(\frac{1}{2} \cdot \frac{i^2}{n^3} + \frac{1}{n}\right).\]

Step 5: Evaluate the Limit as n Approaches Infinity

Calculate the definite integral as a limit of the sum as \(n\to\infty\). Recognize the sums:\[A = \lim_{n\to\infty} \left( \frac{1}{2n^3} \sum_{i=1}^{n} i^2 + \frac{1}{n} \sum_{i=1}^{n} 1 \right)\]Using the formula for the sum of squares \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\) and the sum of ones \(\sum_{i=1}^{n} 1 = n\), you get:\[A = \frac{1}{2} \cdot \frac{1}{3} + 1 = \frac{1}{6} + 1 = \frac{7}{6}.\]

Step 6: Final Step: Conclude the Area Calculation

The area under the curve \(y = \frac{1}{2}x^2 + 1\) from \(x=0\) to \(x=1\) is \(\frac{7}{6}\).

Key Concepts

Area Under Curve

The concept of finding the "area under a curve" is fundamental in calculus, especially when analyzing functions graphically. Imagine a curve on a graph, such as the one defined by a function like \( y = \frac{1}{2}x^2 + 1 \). To find the area beneath this curve from point \(a\) to point \(b\) on the x-axis, we are essentially looking to measure the space enclosed between the curve and the x-axis over that interval.

This area is significant in many real-world applications, including physics and economics, as it can represent quantities such as distance covered, accumulated growth, or cost over time. To compute this area using calculus, we use the concept of integration, which lets us calculate this area precisely even when the shape under the curve is not a simple geometric figure like a rectangle or triangle.

Limit of Sum

The "limit of a sum" is a powerful idea that ties together sums and integrals. When calculating the area under a curve, we approximate the area using many narrow rectangles, each covering a subinterval of the full interval. The sum of the areas of these rectangles acts as an approximation of the total area under the curve.

For our function, \( y = \frac{1}{2}x^2 + 1 \), the interval \([0, 1]\) is divided into \(n\) parts. With each division, the width of each rectangle becomes \(\Delta x = \frac{1}{n}\). As we increase \(n\), making the rectangles narrower, our approximation improves.

The limit of this sum, as \(n\) approaches infinity, gives us the exact area. This step is what transitions us from a finite sum to the definite integral, providing an exact measure of the area under the curve.

Definite Integral Calculation

To compute a "definite integral," we utilize the limit process we discussed, which refines our approximation of the area. Specifically, a definite integral is defined as the limit of the sum of the areas of rectangles under a curve.

For our example, we calculate:

• \( \int_0^1 \left(\frac{1}{2}x^2 + 1\right) \, dx \)
• The integral symbol \(\int\) represents the process of integration.
• The numbers 0 and 1 are the limits of integration, specifying the interval on the x-axis.
• The expression \(\left(\frac{1}{2}x^2 + 1\right)\) is the function we are integrating.

This calculation results in an exact area value. In our case, it yields \( \frac{7}{6} \), which is the total area under the curve from \(x=0\) to \(x=1\). The power of integration lies in its ability to give exact results where simpler methods only approximate.

Integral of Polynomial Functions

Polynomials are expressions involving powers of a variable, such as \(x^2\). The integration of polynomial functions is particularly straightforward in calculus. When integrating a polynomial like \( \frac{1}{2}x^2 + 1 \), we apply specific rules that simplify the process.

For a polynomial \( ax^n \), the integral is \( \frac{a}{n+1}x^{n+1} + C \), where \( C \) is the constant of integration. This rule allows us to integrate each term separately and find the antiderivative. In the case of our function, the integral steps are as follows:

• Integrate \( \frac{1}{2}x^2 \) which yields \( \frac{1}{6}x^3 \).
• Integrate the constant \(1\), which yields \(x\).

Combining these results gives us the definite integral from 0 to 1, leading to our final area calculation. Working with polynomial functions in calculus is often seen as a foundational skill, due to both its simplicity and its wide range of applications.