Answer:
[tex]\int_{2}^{9}x^2 dx[/tex] so the limits are 2 and 9
Step-by-step explanation:
We want to express [tex]\lim_{n\rightarrow \infty} \sum_{k=1}^n\frac{7}{n}(2+\frac{7k}{n})^2[/tex] as a integral. To do this, we have to identify [tex]\sum_{k=1}^n\frac{7}{n}(2+\frac{7k}{n})^2[/tex] 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
[tex]\sum_{k=1}^n f(y_k) \frac{b-a}{n}[/tex], where [tex]y_k\in[a+(k-1)\frac{b-a}{n},a+k\frac{b-a}{n}][/tex] is a selected point of the subinterval.
In particular, if we select the ending point of each subinterval as the [tex]y_k[/tex], the Riemann sum is:
[tex]\sum_{k=1}^n f(a+k\frac{b-a}{n}) \frac{b-a}{n}[/tex].
Now, let's identify this in [tex]\sum_{k=1}^n\frac{1}{7n}(2+\frac{7k}{n})^2[/tex] .
The integrand is x² so this is our function f. When k=n, the summand should be [tex]\frac{b-a}{n}f(b)=\frac{b-a}{n}b^2[/tex] because the last selected point is b. The last summand is [tex]\frac{7}{n}(9)^2[/tex] 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.
