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. \()\) $$ y=x+2 ; a=0, b=1 $$

Short answer

The area under the curve is \( \frac{5}{2} \).

Step-by-step solution

Define the Function and the Interval

The function given is \( y = x + 2 \), and we need to find the area under this curve over the interval \([0, 1]\).

Divide the Interval into Subintervals

Divide the interval \([0, 1]\) into \(n\) equal subintervals. The width of each subinterval, \( \Delta x \), is given by \( \Delta x = \frac{1}{n} \).

Evaluate the Function at Right Endpoints

To find the area under the curve, use the sum of rectangles approach using the right endpoint. The height of each rectangle is given by \( f(x_i) = x_i + 2 \), where \( x_i = \frac{i}{n} \) for \( i = 1, 2, ... , n \).

Set up the Riemann Sum

The total area, \( A \), can be approximated by the Riemann sum \( A \approx \sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \left( \frac{i}{n} + 2 \right) \cdot \frac{1}{n} \).

Simplify the Riemann Sum

Simplify the sum to \( A = \frac{1}{n} \sum_{i=1}^{n} \left( \frac{i}{n} + 2 \right) = \sum_{i=1}^{n} \frac{i}{n^2} + 2 \cdot \frac{1}{n} \sum_{i=1}^{n} 1 \). This reduces to \( \frac{1}{n^3} \sum_{i=1}^{n} i + \frac{2}{n} \cdot n \).

Use the Formula for the Sum of Integers

The formula for the sum of the first \(n\) integers is \( \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \). Substitute this into the Riemann sum to get \( A = \frac{1}{n^3} \cdot \frac{n(n+1)}{2} + 2 = \frac{n(n+1)}{2n^3} + 2 \).

Evaluate the Limit as n Approaches Infinity

Take the limit of the Riemann sum as \( n \to \infty \): \( A = \lim_{n \to \infty} \left( \frac{1}{2n} + \frac{1}{2n^2} + 2 \right) = \frac{1}{2} + 2 = \frac{5}{2} \).

Conclusion

The area under the curve \( y = x + 2 \) over the interval \([0, 1]\) is \( \frac{5}{2} \).

Key concepts

Definite Integral

The concept of a definite integral is essential when we talk about finding the area under a curve. It is formally defined as the limit of a Riemann sum as the number of subintervals approaches infinity. This process turns the sum of infinitesimally small areas into a precise measurement. To understand this, consider the function given, which is the line \( y = x + 2 \). Our aim is to calculate the area under this line from \( a = 0 \) to \( b = 1 \).

The definite integral is symbolized using the integral sign with limits, \( \int_{a}^{b} f(x) \, dx \), which translates the problem of finding areas into a mathematical problem of Calculus. The values \( a \) and \( b \) represent the interval's bounds, while \( f(x) \) is our function. The solution involves calculating the total area between the curve and the x-axis over this interval.

Thus, through the definite integral, the concept of area under a line or curve is formalized, enabling us to solve a broad range of real-world problems involving areas and accumulations.

Limit of a Sum

The idea of the 'limit of a sum' is integral in calculus, especially when working with Riemann sums. As per the step-by-step solution, calculating the area under \( y = x + 2 \) starts by breaking the interval \([0,1]\) into \( n \) equal parts. Each part is tiny and has a width of \( \Delta x = \frac{1}{n} \).

To get an approximate area, the function \( f(x_i) = x_i + 2 \) is evaluated at each subinterval's right endpoint. Adding up all these rectangles gives us \( \sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \left( \frac{i}{n} + 2 \right) \cdot \frac{1}{n} \).

But this isn't precise until we deal with limits. The final area is exact only when \( n \to \infty \), provided by the limit of the sum. This means we're considering infinite rectangles with infinitesimal widths, ultimately leading to an accurate calculation of the area, as highlighted in the solution's conclusion, where the height and width converge into an ideal value of \( \frac{5}{2} \).

By understanding limits, we can move from rough estimation to accurate computation, which is the crux of using limits with sums in calculus.

Area Under a Curve

Finding the area under a curve is quintessential in analyzing physical spaces and significant for many applications, from physics to economics. In the given exercise for \( y = x + 2 \), the process begins by visualizing this area.

This is the space between the curve and the x-axis along the given interval. The method begins with approximations using rectangles whose heights are determined by the function value at certain points, either left, right, or midpoints of the intervals. These rectangles form what we call a circumscribed polygon.

The combined area of these rectangles will approach the actual area under the curve as the number of subdivisions increases. This is why the exercise employs a Riemann sum approach. As \( n \to \infty \), this sum converges to the exact total area under \( y = x + 2 \) over \([0,1]\), which turns out to be \( \frac{5}{2} \).

Such calculations are not only about obtaining a number but also offer insights into how functions behave over intervals. Understanding the area under a curve provides a mathematical and graphical interpretation of integration.