Question

Write the definite integral for the summation: the limit as n goes to infinity of the summation from k equals 1 to n of the product of the square of the quantity 1 plus k over n squared and 1 over n.
the integral from x equals 0 to 1 of x squared, dx
the integral from x equals 1 to 2 of the quantity x plus 1 squared, dx
the integral from x equals 1 to 2 of x squared, dx
the integral from x equals 2 to 1 of x squared, dx

Answer 1

Sounds like you have

$\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\left(1+\frac kn\right)^2\frac1n$

which translates to the sum of the areas of $n$ rectangles with dimensions $\left(1+\dfrac kn\right)^2$ (height) and $\dfrac1n$ (width). This is the right-endpoint Riemann sum for approximating the area under $x^2$ over the interval [1, 2].

Answer 2

Answer:

Option C

Step-by-step explanation:

We are given that

$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}(1+\frac{k}{n})^2\times \frac{1}{n}$

We have to find the definite integral for the given summation.

We know that

$\lim_{n\rightarrow \infty}\sum_{k=a}^{n}f(a+k\frac{b-a}{n})(\frac{b-a}{n}=\int_{a}^{b}f(x)dx$

Using the formula

a=1

$\frac{b-a}{n}=\frac{b-1}{n}$

$\frac{b-1}{n}=\frac{1}{n}$

$b-1=1$

$b=1+1=2$

$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}(1+\frac{k}{n})^2\times \frac{1}{n}=\int_{1}^{2}x^2 dx$

Option C is true.