Seja f:[0, \pi] \rightarrow \mathbb{R} a função definida por f(x)=\sin x e \mathscr{P} a partição de [0, \pi ] dada por \mathscr{P}=\left\{0, \frac{\pi}{n}, \frac{2 \pi}{n}, \ldots, \frac{(n-1) \pi}{n}, \frac{n \pi}{n}\right\}.
  • Se \xi_{j}=\frac{j \pi}{n}, j=1, \ldots, n mostre que a soma de Riemann de f com respeito à partição \mathscr{P} será dada por
\Sigma\left(f, \mathscr{P}^{\star}\right)=\sum_{j=1}^{n}\left(\sin \xi_{j}\right) \frac{\pi}{n}=\frac{\pi}{n} \frac{\cos \left(\frac{\pi}{2 n}\right)}{\sin \left(\frac{\pi}{2 n}\right)}
  • Calcule \lim _{n \rightarrow \infty} \Sigma\left(f, \mathscr{P}^{\star}\right).
  • Calcule \int_{0}^{\pi} f(x) d x.
  • Calcule \int_{0}^{\pi}|f(x)| d x.

Passo 1

O enunciado define f:[0,\pi]\to\mathbb R por f(x)=\sin x e a partição uniforme \mathscr P=\{0,\tfrac{\pi}{n},\tfrac{2\pi}{n},\dots,\pi\}.
Queremos calcular a soma de Riemann com pontos \xi_j=\tfrac{j\pi}{n}, depois o seu limite quando n\to\infty, em seguida \displaystyle\int_0^\pi\sin x\,dx e \displaystyle\int_0^\pi|\sin x|\,dx.
Para isso vamos usar a identidade para soma finita de senos e o cálculo de integrais elementares.

Passo 2

Defina \displaystyle\alpha=\frac{\pi}{n}. Então
\Sigma(f,\mathscr P^\star) =\sum_{j=1}^n\sin\bigl(j\alpha\bigr)\,\alpha =\alpha\sum_{j=1}^n\sin\bigl(j\alpha\bigr).
A identidade para soma de senos é
\sum_{j=1}^n\sin(j\alpha) =\frac{\sin\!\bigl(\tfrac{n\alpha}{2}\bigr)\, \sin\!\bigl(\tfrac{(n+1)\alpha}{2}\bigr)} {\sin\!\bigl(\tfrac{\alpha}{2}\bigr)}.
Aqui n\alpha/2=\pi/2, logo \sin(n\alpha/2)=1, e (n+1)\alpha/2=\pi/2+\alpha/2 implica \sin\bigl((n+1)\alpha/2\bigr)=\cos(\alpha/2).
Substituindo em \Sigma obtemos
\Sigma(f,\mathscr P^\star) =\alpha\;\frac{1\cdot\cos(\tfrac{\alpha}{2})} {\sin(\tfrac{\alpha}{2})} =\frac{\pi}{n}\,\frac{\cos\!\bigl(\tfrac{\pi}{2n}\bigr)} {\sin\!\bigl(\tfrac{\pi}{2n}\bigr)}.

Passo 3

Para n\to\infty temos \tfrac{\pi}{2n}\to0, então \sin\!\bigl(\tfrac{\pi}{2n}\bigr)\sim\tfrac{\pi}{2n} e \cos\!\bigl(\tfrac{\pi}{2n}\bigr)\to1. Logo
\lim_{n\to\infty}\Sigma(f,\mathscr P^\star) =\lim_{n\to\infty}\frac{\pi}{n}\, \frac{\cos(\tfrac{\pi}{2n})}{\sin(\tfrac{\pi}{2n})} =\lim_{n\to\infty}\frac{\pi}{n}\, \frac{1}{\tfrac{\pi}{2n}} =2.

Passo 4

Calculemos
\int_0^\pi\sin x\,dx =\bigl[-\cos x\bigr]_0^\pi =(-\cos\pi)-(-\cos0) =(-(-1))-(-1) =1+1 =2.

Passo 5

Em [0,\pi] tem-se \sin x\ge0, logo |\sin x|=\sin x. Assim
\int_0^\pi|\sin x|\,dx =\int_0^\pi\sin x\,dx =2.

Resposta

\displaystyle\Sigma(f,\mathscr P^\star) =\frac{\pi}{n}\,\frac{\cos\!\bigl(\tfrac{\pi}{2n}\bigr)} {\sin\!\bigl(\tfrac{\pi}{2n}\bigr)} \displaystyle\lim_{n\to\infty}\Sigma(f,\mathscr P^\star)=2 \displaystyle\int_0^\pi\sin x\,dx=2 \displaystyle\int_0^\pi|\sin x|\,dx=2

Exercícios relacionados