Problem 50 - Essential Calculus

Question

In Exercises \(49-52,\) use the limit process to find the area of the region between the graph of the function and the \(y\) -axis over the given \(y\) -interval. Sketch the region. $$ f(y)=4 y-y^{2}, 1 \leq y \leq 2 $$

Step-by-Step Solution

Answer

The area \( A \) is equal to 3/2.

Step 1: Definition of the problem in terms of limits

The area can be found by integrating the function over the given limits, which corresponds to taking the limit as the width of the rectangles approaches zero in a Riemann sum. Hence the problem can be expressed as follows in terms of limits: \[A=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(y_{i}\right) \Delta y\] where \( \Delta y=\frac{b-a}{n} \) and \( y_{i}=a+i \Delta y \). Here \( a=1 \) and \( b=2 \).

Step 2: Simplify the Limit

Substituting the values of \( a \), \( b \), and \( f(y) \) into the equation we get: \[A=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left[4\left(1+\frac{i}{n}\right)-\left(1+\frac{i}{n}\right)^{2}\right] \frac{1}{n}\] After simplifying the equation, the limit will be expressed as: \[A=\lim _{n \rightarrow \infty} \left[\frac{4}{n} \sum_{i=1}^{n} i-\frac{1}{n^{2}} \sum_{i=1}^{n} i^{2}-1\right]\] The problem has thus transformed into the determination of the limits of two series: \[ \frac{4}{n} \sum_{i=1}^{n} i \] and \[ -\frac{1}{n^{2}} \sum_{i=1}^{n} i^{2} \].

Step 3: Evaluate the Limit

The above expressions can be replaced with \[ \frac{n(n+1)}{2} \] and \[ \frac{n(n+1)(2n+1)}{6} \] respectively which are formulas for the sum of first \( n \) natural numbers and the sum of squares of first \( n \) natural numbers respectively. Consequently the final limit can be expressed as: \[ A=\lim _{n \rightarrow \infty} \left[\frac{4(n+1)}{2}-\frac{(2n+1)}{6}-1\right] \] This limit can be easily computed by applying L'Hopital Rule or simple limit properties. The answer should be the required area.

Key Concepts

Riemann Sum

A Riemann sum is a method for approximating the total area under a curve on a graph, which is the foundation of integral calculus. Imagine trying to find the area under the function curve between two points. Instead of using a complex shape that matches the curve perfectly, a Riemann sum uses simpler shapes, like rectangles, to estimate this area.

• Split the interval into smaller, equal segments.
• Over each segment, form a rectangle whose height matches the function's value at a certain point within it, often the midpoint or either end.
• The area of each rectangle is its width times its height, contributing to the sum.
• As the number of rectangles increases, their width becomes smaller and the sum approaches the curve more closely.

By taking the limit where the number of rectangles goes to infinity, the Riemann sum converges to the exact area.

Definite Integral

The definite integral of a function over a specified interval gives the exact total area under the curve between two points. If you've nailed down

• The starting point (lower limit) and the ending point (upper limit) of an interval.
• The function itself that defines your curve.

you're set to use the definite integral to find the total area beneath that curve.
The Riemann sum's limit, as the width of the subintervals approaches zero, becomes the definite integral. Mathematically, it's given by:\[\int_{a}^{b} f(y) \, dy\]where

• \(a\) is the lower limit.
• \(b\) is the upper limit.
• \(f(y)\) is the function.

This transformation simplifies the process of finding the area by providing a straightforward formula.

Area Under a Curve

The area under a curve in calculus represents all the accumulated sums of the values of a function over a certain interval. It's a key notion that allows for measurements like distance, space, or probability, depending on the context.

• It represents a collection of results obtained from summing up the function's value across a continuous range.
• Visualize this as the sum of infinitely many infinitesimally thin rectangles, stacking from the start to the end of the interval.

In the original exercise, determining the area under the curve involves evaluating the function between given boundaries, specifically between \(y=1\) and \(y=2\). By applying the definite integral, one can precisely compute the region underneath the curve \(f(y)=4y-y^2\), which translates to the sum of the function's continuous distribution along this axis.

Limit of a Sequence

The limit of a sequence in the context of calculus is a concept describing the behavior of numbers in a sequence as the sequence progresses to infinity. In simpler terms, it's where the numbers in your series are heading as you pile on more terms.
In relation to our problem:

• As the number \(n\) in the Riemann sum increases to infinity, the sequence of sums approaches a specific boundary value called the limit.
• This limit represents the exact area under the curve we're trying to find.

The beauty here is that while individual sequence terms might seem scattered or irregular, together they converge towards a single number—the limit. This number helps us translate the approximate sum from the Riemann sum into a precise area, effectively completing the calculus puzzle we've been assembling.