Question

PLZ HELP!!! Use limits to evaluate the integral.

Answer 1

Split up the interval [0, 2] into n equally spaced subintervals:

`\left[0,\frac2n\right],\left[\frac2n,\frac4n\right],\left[\frac4n,\frac6n\right],\ldots,\left[\frac{2(n-1)}n,2\right]`

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

`r_i=\frac{2i}n`

where
`1\le i\le n`. Each interval has length
`\Delta x_i=\frac{2-0}n=\frac2n`.

At these sampling points, the function takes on values of

`f(r_i)=7{r_i}^3=7\left(\frac{2i}n\right)^3=(56i^3)/(n^3)`

We approximate the integral with the Riemann sum:

`\displaystyle\sum_(i=1)^nf(r_i)\Delta x_i=\frac{112}n\sum_(i=1)^ni^3`

Recall that

`\displaystyle\sum_(i=1)^ni^3=\frac{n^2(n+1)^2}4`

so that the sum reduces to

`\displaystyle\sum_(i=1)^nf(r_i)\Delta x_i=(28n^2(n+1)^2)/(n^4)`

Take the limit as n approaches infinity, and the Riemann sum converges to the value of the integral:

`\displaystyle\int_0^27x^3\,\mathrm dx=\lim_(n\to\infty)(28n^2(n+1)^2)/(n^4)=\boxed{28}`

Just to check:

`\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28`