Question

Show all work and reasoning

which of the following limits is equal to
`\int\limits^5_2 {x^2} \, dx` ?

A.
`\lim_(n \to \infty) \sum^n_(k=1)(2+ (k)/(n))^2 (1)/(n)`

B.
`\lim_(n \to \infty) \sum^n_(k=1)(2+ (k)/(n))^2 (3)/(n)`

C.
`\lim_(n \to \infty) \sum^n_(k=1)(2+ (3k)/(n))^2 (1)/(n)`

D.
`\lim_(n \to \infty) \sum^n_(k=1)(2+ (3k)/(n))^2 (3)/(n)`

Answer 1

Split up the interval [2, 5] into equally spaced subintervals, then consider the value of at the right endpoint of each subinterval.

The length of the interval is , so the length of each subinterval would be . This means the first rectangle's height would be taken to be when , so that the height is , and its base would have length . So the area under over the first subinterval is .

Continuing in this fashion, the area under over the th subinterval is approximated by , and so the Riemann approximation to the definite integral is

and its value is given exactly by taking . So the answer is D (and the value of the integral is exactly 39).

Answer 2

Split up the interval [2, 5] into
`n` equally spaced subintervals, then consider the value of
`f(x)` at the right endpoint of each subinterval.

The length of the interval is
`5-2=3`, so the length of each subinterval would be
`\frac3n`. This means the first rectangle's height would be taken to be
`x^2` when
`x=2+\frac3n`, so that the height is
`\left(2+\frac3n\right)^2`, and its base would have length
`\frac{3k}n`. So the area under
`x^2` over the first subinterval is
`\left(2+\frac3n\right)^2\frac3n`.

Continuing in this fashion, the area under
`x^2` over the
`k`th subinterval is approximated by
`\left(2+\frac{3k}n\right)^2\frac{3k}n`, and so the Riemann approximation to the definite integral is


`\displaystyle\sum_(k=1)^n\left(2+\frac{3k}n\right)^2\frac{3k}n`

and its value is given exactly by taking
`n\to\infty`. So the answer is D (and the value of the integral is exactly 39).