Determine a region whose area is equal to the given limit. Do not evaluate the limit. lim┬(n→∞)⁡∑_(i=1 )^n 2/n (1+2i/n)^18

Question

zarkon · Answer

$$\lim_{n\to\infty}\sum_{i=1}^{n}\frac{2}{n}\left(1+\frac{2i}{n}\right)^{18}$$

zarkon · Answer

$$\lim_{n\to\infty}\sum_{i=1}^{n}\frac{2}{n}\left(1+\frac{2i}{n}\right)^{18}$$

$$\int\limits_{a}^{b}f(x)dx=\lim_{n\to\infty}\sum_{i=1}^{n}f\left(a+i\frac{b-a}{n}\right)\frac{b-a}{n}$$

zarkon · Answer

$$a=1$$
$$b=3$$
$$f(x)=x^{18}$$

zarkon · Answer

$$=\int\limits_{1}^{3}x^{18}dx$$

zarkon · Answer

so it is the area under the graph of $$f(x)=x^{18}$$between the x=1 and x=3

anonymous · Answer

Thats what I thought but webassign says that the answer is wrong

zarkon · Answer

the limit of the sum gives the same value as the integral

zarkon · Answer

$$\frac{1162261466}{9}$$

zarkon · Answer

over 19

zarkon · Answer

$$\frac{1162261466}{19}$$

anonymous · Answer

Here are the choices from webassign

zarkon · Answer

2nd one is equivalent to mine...do the substitution u=1+x

anonymous · Answer

I'm sorry,  I'm not sure I understand

zarkon · Answer

$$\int\limits_{0}^{2}(1+x)^{18}dx$$

let u=1+x
du=dx
then 

$$x=0\rightarrow u=1$$
$$x=2\rightarrow u=3$$
and the integral becomes

$$\int\limits_{1}^{3}u^{18}du$$

which is equivalent to what I wrote before