(1 point) the following sum (sqrt{49 - (\frac{7}{n})^2}cdot\frac{7}{n}+sqrt{49 - (\frac{14}{n})^2}cdot\frac{7…

(1 point) the following sum (sqrt{49 - (\frac{7}{n})^2}cdot\frac{7}{n}+sqrt{49 - (\frac{14}{n})^2}cdot\frac{7}{n}+cdots+sqrt{49 - (\frac{7n}{n})^2}cdot\frac{7}{n}) is a right riemann sum for the definite integral (int_{0}^{b}f(x)dx) where (b =) and (f(x)=) the limit of these riemann sums as (n\rightarrowinfty) is

(1 point) the following sum (sqrt{49 - (\frac{7}{n})^2}cdot\frac{7}{n}+sqrt{49 - (\frac{14}{n})^2}cdot\frac{7}{n}+cdots+sqrt{49 - (\frac{7n}{n})^2}cdot\frac{7}{n}) is a right riemann sum for the definite integral (int_{0}^{b}f(x)dx) where (b =) and (f(x)=) the limit of these riemann sums as (n\rightarrowinfty) is

Answer

Explanation:

Step1: Recall right - Riemann sum formula

The right - Riemann sum for $\int_{a}^{b}f(x)dx$ with $n$ subintervals of equal width $\Delta x=\frac{b - a}{n}$ and $x_i=a + i\Delta x$ is $R_n=\sum_{i = 1}^{n}f(x_i)\Delta x$. Here $a = 0$, and $\Delta x=\frac{b-0}{n}=\frac{b}{n}$. The general term of the sum is $\sqrt{49-(i\frac{b}{n})^2}\cdot\frac{b}{n}$. Comparing with the given sum terms $\sqrt{49 - (\frac{7i}{n})^2}\cdot\frac{7}{n}$, we can see that $\frac{b}{n}=\frac{7}{n}$, so $b = 7$. The function $f(x)=\sqrt{49 - x^2}$.

Step2: Write the definite integral

The limit of the right - Riemann sum as $n\rightarrow\infty$ is the definite integral $\int_{0}^{7}\sqrt{49 - x^2}dx$.

Answer:

$b = 7,f(x)=\sqrt{49 - x^2}$