Find area under $y=-x^3$ from x=-3 to x=0 using the Riemann sum

Question

How can I find the area under $y=-x^3$ from $x=-3$ to $x=0$ using the Riemann sum?

let $h = 3/n$

$x_i =-3 + i/n$

\begin{align} A &= R_n \\ &= \lim_{n\to \infty} {\sum_{i=1}^{n} h (-(x_i)^3)} \\ &= \lim_{n\to \infty} h{\sum_{i=1}^{n} -(x_i)^3 } \\ &= \lim_{n\to \infty} h{\sum_{i=1}^{n} -(-3 + i/n)^3 } \end{align}

Hopefully somebody can provide some guidance here.

Answer 1

Hint: There's a formula: $\sum_{i=1}^n i^3=(\dfrac {n(n+1)}2)^2$.

And a formula: $\sum_{i=1}^n i^2=\dfrac {n(n+1)(2n+1)}6$.

And there's the Gauß sum: $\sum_{i=1}^ni=\dfrac {n(n+1)}2$.

Finally, use $(-3+\dfrac in)^3=(-3)^3+3(-3)^2(\dfrac in)+3(-3)(\dfrac in)^2+(\dfrac in)^3$.

Answer 2

Because $$\int\limits_0^3x^3dx=\lim_{n\rightarrow+\infty}\sum_{k=1}^n\left(\frac{3k}{n}\right)^3\frac{3}{n}=\lim_{n\rightarrow+\infty}\frac{81n^2(n+1)^2}{4n^4}=\frac{81}{4}.$$