Question

What are the limits of integration if the summation the limit as n goes to infinity of the summation from k equals 1 to n of the product of the quantity of the square of 2 plus 7 times k over n and the quotient of 7 and n is written as a definite integral with integrand x2?

Answer 1

Answer:

$\int_{2}^{9}x^2 dx$ so the limits are 2 and 9

Step-by-step explanation:

We want to express $\lim_{n\rightarrow \infty} \sum_{k=1}^n\frac{7}{n}(2+\frac{7k}{n})^2$ as a integral. To do this, we have to identify $\sum_{k=1}^n\frac{7}{n}(2+\frac{7k}{n})^2$ as a Riemann Sum that approximates the integral. (taking the limit makes the approximation equal to the value of the integral)

In general, to find a Riemann sum that approximates the integral of a function f over an interval [a,b] we can the interval in n subintervals of equal length and approximate the area (integral) with rectangles in each subinterval and them sum the areas. This is equal to

$\sum_{k=1}^n f(y_k) \frac{b-a}{n}$, where $y_k\in[a+(k-1)\frac{b-a}{n},a+k\frac{b-a}{n}]$ is a selected point of the subinterval.

In particular, if we select the ending point of each subinterval as the $y_k$, the Riemann sum is:

$\sum_{k=1}^n f(a+k\frac{b-a}{n}) \frac{b-a}{n}$.

Now, let's identify this in $\sum_{k=1}^n\frac{1}{7n}(2+\frac{7k}{n})^2$ .

The integrand is x² so this is our function f. When k=n, the summand should be $\frac{b-a}{n}f(b)=\frac{b-a}{n}b^2$ because the last selected point is b. The last summand is $\frac{7}{n}(9)^2$ thus b=9 and b-a=7, then 9-a=7 which implies that a=2.

To verify our answer, note that if we substitute a=2, b=9 and f(x)=x² in the general Riemann Sum, we obtain the sum inside the limit as required.