How to convert into a definite integral
Could you show me how to convert the following into a definite integral:
$$\lim\limits_{n \to \infty} \sum_{k=1}^{3n} \frac{1}{n}\cos\left(\frac{k\pi}{n}\right)\sin\left(\frac{2k\pi}{n}\right)$$
Thank you!
$\endgroup$2 Answers
$\begingroup$It's literally
$$\int_0^3\cos(\pi x)\sin(2\pi x)dx$$
The substitution $u=\pi x$ re-expresses it as
$${1\over\pi}\int_0^{3\pi}\cos(u)\sin(2u)du$$
The basic idea is to think of the $1\over n$ as $\Delta x$, which in the limit becomes a $dx$, and $k/n$ as $x$, which in the limit varies from $0$ when $k=1$ to $3$ when $k=3n$.
$\endgroup$ 3 $\begingroup$Recall the Riemann sum $$\int_{x=a}^b f(x) \, dx = \lim_{n \to \infty} \sum_{k=1}^n f\left(a + \frac{b-a}{n} k\right) \frac{b-a}{n}.$$ Write your limit as $$S = \lim_{m \to \infty} \sum_{k=1}^{3m} \frac{1}{m} \cos \frac{k\pi}{m} \sin \frac{2k\pi}{m}.$$ Then with the choice $n = 3m$, $a = 0$, $b = 1$, $f(x) = \cos 3\pi x \sin 6\pi x,$ we get $$S = 3 \int_{x=0}^1 \cos 3\pi x \sin 6\pi x \, dx = \frac{1}{\pi} \int_{x=0}^{3\pi} \cos x \sin 2x \, dx.$$
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