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

`\int_(2)^(9)x^2 dx` so the limits are 2 and 9

Explanation:

We want to express
`\lim_(n\rightarrow \infty) \sum_(k=1)^n(7)/(n)(2+(7k)/(n))^2` as a integral. To do this, we have to identify
`\sum_(k=1)^n(7)/(n)(2+(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) (b-a)/(n)`, where
`y_k\in[a+(k-1)(b-a)/(n),a+k(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(b-a)/(n)) (b-a)/(n)`.

Now, let's identify this in
`\sum_(k=1)^n(1)/(7n)(2+(7k)/(n))^2` .

The integrand is x² so this is our function f. When k=n, the summand should be
`(b-a)/(n)f(b)=(b-a)/(n)b^2` because the last selected point is b. The last summand is
`(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.