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. $$ f(y)=4 y-y^{2}, 1 \leq y \leq 2 $$

Short answer

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

Step-by-step solution

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 \( b=2 \).

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 \[ -\frac{1}{n^{2}} \sum_{i=1}^{n} i^{2} \].

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.