What is the area under $9-x^2$ between $0$ and $1$? I want to find the area under $9-x^2$ between $0$ and $1$. I found $1/3$ but wolfram says that it isn't correct and gives me $26/3$.
I evaluated my area by using the limit of $n$ going to infinity of the sum of $k^2/n^3$ from $k=1$ to $n$.
 A: $$\int_0^1 (9-x^2)\ dx = \left(9x - \frac{x^3}3{}\right)_0^1$$
$$ = 9 - \frac{1}{3} = \frac{26}{3}$$
A: Instead of a simple $\lim_{n\to\infty}\sum_{k=0}^{n-1}\frac{k^2}{n^3}$,
$$\begin{align*}\int_0^1(9-x^2)\ dx 
&= \lim_{n\to\infty}\sum_{k=1}^{n}\frac1n\cdot\left[9-\left(\frac{k}{n}\right)^2\right]\\
&= \lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac9n-\frac{k^2}{n^3}\right)\\
&= \lim_{n\to\infty}\left(\frac9n\sum_{k=1}^{n}1-\frac1{n^3}\sum_{k=1}^{n}k^2\right)\\
&= \lim_{n\to\infty}\left[\frac9n\cdot n-\frac1{n^3}\cdot\frac{n(n+1)(2n+1)}6\right]\\
&= \lim_{n\to\infty}\left[9-\frac{1\left(1+\frac1n\right)\left(2+\frac1n\right)}6\right]\\
&= 9-\frac13\\
&= \frac{26}3
\end{align*}$$
A: By "under the curve" I assume you mean bound by the curve and the x-axis. The requested area is $\int_{0}^{1} 9-x^2\, dx$. We can easily calculate this integral using the fact that $\int x^n\, dx=\frac{x^{n+1}}{n+1}+C$, and we get $9x-\frac{x^3}{3}+C$. Evaluating this from the limits $0,1$ we get $\frac{26}{3}-0=\frac{26}{3}$
