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+\frac kn\right)^2` (height) and
`\frac1n` (width). This is the right-endpoint Riemann sum for approximating the area under
`x^2` over the interval [1, 2].

Answer 2

Option C

Explanation:

We are given that

`\lim_(n\rightarrow\infty)\sum_(k=1)^(n)(1+(k)/(n))^2* (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(b-a)/(n))((b-a)/(n)=\int_(a)^(b)f(x)dx`

Using the formula

a=1

`(b-a)/(n)=(b-1)/(n)`

`(b-1)/(n)=(1)/(n)`

`b-1=1`

`b=1+1=2`

`\lim_(n\rightarrow\infty)\sum_(k=1)^(n)(1+(k)/(n))^2* (1)/(n)=\int_(1)^(2)x^2 dx`

Option C is true.