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?

so the limits are 2 and 9

Step-by-step explanation:

We want to express as a integral. To do this, we have to identify 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

, where is a selected point of the subinterval.

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

.

Now, let's identify this in .

The integrand is x² so this is our function f. When k=n, the summand should be because the last selected point is b. The last summand is 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.

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42