Question

Can someone please help me,
How do you find the definite integral of 9-x2 from x equals 0 to 3 using rieman sums? PLEASE, IM SO DESPERATE

Answer 1

Using a left-endpoint sum, you can approximate the definite integral by

\[\begin{align*}\int_0^3 \left(9-x^2\right)\,dx&\approx\sum_{i=1}^nf\left(\frac{3(i-1)}{n}\right)\times\frac{3-0}{n}\\\\
&=\frac{3}{n}\sum_{i=1}^n\left(9-\left(\frac{3i-3}{n}\right)^2\right)\\\\
&=\frac{3}{n}\sum_{i=1}^n\left(9-9\frac{(i-1)^2}{n^2}\right)\\\\
&=\frac{27}{n}\sum_{i=1}^n\left(1-\frac{i^2-2i+1}{n^2}\right)\\\\
&=\frac{27}{n}\sum_{i=1}^n\left(\frac{n^2-i^2+2i-1}{n^2}\right)\\\\
&=\frac{27}{n^3}\sum_{i=1}^n\left(n^2-i^2+2i-1\right)\\\\
&=\frac{27}{n^3}\left(n^3-\frac{n(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2}-n\right)\\\\
&=27-\frac{9n(n+1)(2n+1)}{2n^3}+\frac{27n(n+1)}{n^3}-\frac{27n}{n^3}
\end{align*}\]
What happens as \(n\to\infty\)?