Question

calcule a integral dada pela funcao.

ee98c62b918bb7aaa94f008d0fcc6cb0.docx

Answer 1

Boa tarde!

Solução!

$Comprimento\overline{AB}=\displaystyle \int_{2}^{4}(log_{2}x)dx\\\\\\ Vou~~ fazer~~ algumas~~ etapas ~~separadas.\\\\\\ log_{2}x\\\\\\ y= log_{2}x\\\\\\ 2^{y}=x\\\\\ yln2=lnx\\\\\\ ln2 \frac{dy}{dx}= \frac{1}{x}\\\\\\\ \boxed{\frac{dy}{dx }=\frac{1}{xln2}}$

Vamos fazer a integração por partes!

$\boxed{\displaystyle \int (u~dv)=uv-\displaystyle \int vdu}\\\\\\ u=log_{2}x~~~~~~~dv=1dx\\\\\\ du= \frac{1}{xln2}~~~~~~~v=x\\\\\\\\\ \displaystyle \int_{2}^{4}(log_{2}x)dx=x.log_{2}x-\displaystyle \int1. \frac{1}{xln2}dx\\\\\\\ \displaystyle \int_{2}^{4}(log_{2}x)dx=x.log_{2}x- \frac{1}{ln2} \displaystyle \int1.dx\\\\\\\ \displaystyle \int_{2}^{4}(log_{2}x)dx=x.log_{2}x- \frac{x}{ln2} +c$


$Comprimento\overline{AB}=(x.log_{2}x- \frac{x}{ln2})-(x.log_{2}x- \frac{x}{ln2})\bigg|_{2}^{4} \\\\\\\\\\ Comprimento\overline{AB}=(4.log_{2}4- \frac{4}{ln2})-(2.log_{2}2- \frac{2}{ln2}) \\\\\\\\\\ Comprimento\overline{AB}=(4.2- 5,77)-(2.1- 2,88) \\\\\\\\\\ Comprimento\overline{AB}=(2,23)-(-0,88) \\\\\\\\\\ Comprimento\overline{AB}=(2,23+0,88) \\\\\\\\\\ Comprimento\overline{AB}=(3,11) \\\\\\\\\\ \boxed{RespostaComprimento\overline{AB}=3,11~u.c}$

Não é nenhuma das alternativa!

Boa tarde!
Bons estudos!