I can't evaluate $\int_{-3}^3 \sqrt{9-x^2} dx$ using the limit definition I need some help here's what I did:
Formula to use:
$$\int_a^bf(x)dx= \lim_{n\rightarrow \infty} \sum_{k=1}^n f(c_k)\Delta x_k$$
Get the variables:
$$\Delta x_k : \frac{b-a}{n} = \frac{3-(-3)}{n} = \frac{6}{n}$$
$$c_k : a + k(\Delta x_k) = \left(-3+\frac{6k}{n}\right) = \left(\frac{-3n+6k}{n}\right)$$
Plug variables into formula and simplify observing the properties of sigma:
$$\begin{align}
\int_{-3}^3 \sqrt{9-x^2} dx &= \lim_{n\rightarrow \infty} \sum_{k=1}^n \left(\sqrt{9-\left(\frac{-3n+6k}{n}\right)^2}\right)\left(\frac{6}{n}\right)\\
&=\lim_{n\rightarrow \infty} \left(\frac{6}{n}\right)\sum_{k=1}^n \left(\sqrt{9-\left(\left(\frac{-3n+6k}{n}\right)\left(\frac{-3n+6k}{n}\right)\right)}\right)\\
&=\lim_{n\rightarrow \infty} \left(\frac{6}{n}\right)\sum_{k=1}^n \left(\sqrt{9-\left(\frac{9n^2-36nk+36k}{n^2}\right)}\right)\\
&=\lim_{n\rightarrow \infty} \left(\frac{6}{n}\right)\sum_{k=1}^n \left(\sqrt{9-\left(\frac{9n^2}{n^2}-\frac{36nk}{n^2}+\frac{36k}{n^2}\right)}\right)\\
&=\lim_{n\rightarrow \infty} \left(\frac{6}{n}\right)\sum_{k=1}^n \left(\sqrt{9-\left(9-\frac{36k}{n}+\frac{36k}{n^2}\right)}\right)\\
&=\lim_{n\rightarrow \infty} \left(\frac{6}{n}\right)\sum_{k=1}^n \left(\sqrt{\left(\frac{36k}{n}-\frac{36k}{n^2}\right)}\right)\\
&=\lim_{n\rightarrow \infty} \left(\frac{6}{n}\right)\sum_{k=1}^n \left(\sqrt{\frac{36nk-36k}{n^2}}\right)\\
\end{align}$$
That's as far as I got. According to the textbook, the answer to $\int_{-3}^3 \sqrt{9-x^2} dx$ is ${}^{9\pi}/_2$ square units. Where did I go wrong?
Best Answer
Let's simplify the integral \begin{align*} \int_{-3}^3 \sqrt{9-x^2} dx &= \int_{-3}^3 \sqrt{9(1-\frac{x^2}{9})} dx = 3 \int_{-3}^3 \sqrt{1-\left(\frac{x}{3}\right)^2} dx = 9\int_{-1}^1 \sqrt{1-x^2} dx \end{align*} We have \begin{align*} \frac{2}{n} \sum_{k=0}^{n-1} \sqrt{1-\left(-1 + k \times \frac{2}{n}\right)^2} &= \frac{2}{n} \sum_{k=0}^{n-1} \sqrt{\frac{4k}{n} - \frac{4k^2}{n^2}}\\ &=\frac{4}{n} \sum_{k=0}^{n-1} \sqrt{\frac{k}{n} - \frac{k^2}{n^2}}\\ &=\frac{4}{n^2} \sum_{k=0}^{n-1} \sqrt{k(n-k)}\\ \end{align*} Well, it seems to me that it is not possible to further simplify the expression. But we can still determine the value of the integral with a geometric interpretation: $\displaystyle\int_{-1}^1 \sqrt{1-x^2} dx$ corresponds to half of the area of a circle of radius $1$. Therefore $$ \int_{-3}^3 \sqrt{9-x^2} dx =\frac{9\pi}{2} $$ Hope it helps.