Question

Qual o resultado da integral ∫² ln x/x dx pelo Método dos Trapézios adotamos 5 subdivisões? A= 0,271829; B= 1,359146; C= 0,135915; D= 0,237172

Answer 1

Resposta:

$\boxed{\bold{\displaystyle{d)~0.237172}}}$

Explicação passo-a-passo:

Olá, boa noite.

Para calcularmos esta soma trapezoidal, devemos relembrar alguns conceitos.

Dado um intervalo $[a,~b]$ que deve ter n subdivisões iguais, utilizamos o valor $\Delta x$ para representar o valor de cada uma delas. Por isso, quanto maior for n, melhor é a aproximação.

Sabemos que $\Delta x =\dfrac{b-a}{n}$ e estamos integrando a função no intervalo $[1,~2]$, logo:

$\Delta x=\dfrac{2-1}{5}=\dfrac{1}{5}$

Lembremos que a fórmula que utilizamos para a soma trapezoidal $\math{S}$ é dada por

$S=\displaystyle{\sum_{i=0}^{n-1}\dfrac{(f(x_{i})+f(x_{i+1}))\Delta x}{2}$

Então, substituindo os valores que temos na fórmula, ficaríamos com a seguinte soma:

$S=\dfrac{\left(\dfrac{\ln 1}{1}+\dfrac{\ln\left(1+\dfrac{1}{5}\right)}{\left(1+\dfrac{1}{5}\right)}\right)\cdot\dfrac{1}{5}}{2}$ $+\dfrac{\left(\dfrac{\ln\left(1+\dfrac{1}{5}\right)}{\left(1+\dfrac{1}{5}\right)}+\dfrac{\ln\left(1+\dfrac{2}{5}\right)}{\left(1+\dfrac{2}{5}\right)\right)\cdot\dfrac{1}{5}}}{2}+\dfrac{\left(\dfrac{\ln\left(1+\dfrac{2}{5}\right)}{\left(1+\dfrac{2}{5}\right)}+\dfrac{\ln\left(1+\dfrac{3}{5}\right)}{\left(1+\dfrac{3}{5}\right)\right)\cdot\dfrac{1}{5}}}{2}$$+\dfrac{\left(\dfrac{\ln\left(1+\dfrac{3}{5}\right)}{\left(1+\dfrac{3}{5}\right)}+\dfrac{\ln\left(1+\dfrac{4}{5}\right)}{\left(1+\dfrac{4}{5}\right)\right)\cdot\dfrac{1}{5}}}{2}+\dfrac{\left(\dfrac{\ln\left(1+\dfrac{4}{5}\right)}{\left(1+\dfrac{4}{5}\right)}+\dfrac{\ln\left(1+\dfrac{5}{5}\right)}{\left(1+\dfrac{5}{5}\right)\right)\cdot\dfrac{1}{5}}}{2}$

Some os valores dentro dos parênteses

$S=\dfrac{\left(\ln1+\dfrac{\ln\left(\dfrac{6}{5}\right)}{\left(\dfrac{6}{5}\right)}\right)\cdot\dfrac{1}{5}}{2}$ $+\dfrac{\left(\dfrac{\ln\left(\dfrac{6}{5}\right)}{\left(\dfrac{6}{5}\right)}+\dfrac{\ln\left(\dfrac{7}{5}\right)}{\left(\dfrac{7}{5}\right)\right)\cdot\dfrac{1}{5}}}{2}+\dfrac{\left(\dfrac{\ln\left(\dfrac{7}{5}\right)}{\left(\dfrac{7}{5}\right)}+\dfrac{\ln\left(\dfrac{8}{5}\right)}{\left(\dfrac{8}{5}\right)\right)\cdot\dfrac{1}{5}}}{2}$  $+\dfrac{\left(\dfrac{\ln\left(\dfrac{8}{5}\right)}{\left(\dfrac{8}{5}\right)}+\dfrac{\ln\left(\dfrac{9}{5}\right)}{\left(\dfrac{9}{5}\right)\right)\cdot\dfrac{1}{5}}}{2}+\dfrac{\left(\dfrac{\ln\left(\dfrac{9}{5}\right)}{\left(\dfrac{9}{5}\right)}+\dfrac{\ln2}{2\right)\cdot\dfrac{1}{5}}}{2}$

Simplificando as frações de frações e sabendo que o valor de $\ln1=0$, ficamos com:

$S=\dfrac{\dfrac{5\ln\left(\dfrac{6}{5}\right)}{6}\cdot\dfrac{1}{5}}{2}$ $+\dfrac{\left(\dfrac{5\ln\left(\dfrac{6}{5}\right)}{6}+\dfrac{5\ln\left(\dfrac{7}{5}\right)}{7\right)\cdot\dfrac{1}{5}}}{2}+\dfrac{\left(\dfrac{5\ln\left(\dfrac{7}{5}\right)}{7}+\dfrac{5\ln\left(\dfrac{8}{5}\right)}{8\right)\cdot\dfrac{1}{5}}}{2}$  $+\dfrac{\left(\dfrac{5\ln\left(\dfrac{8}{5}\right)}{8}+\dfrac{5\ln\left(\dfrac{9}{5}\right)}{9\right)\cdot\dfrac{1}{5}}}{2}+\dfrac{\left(\dfrac{5\ln\left(\dfrac{9}{5}\right)}{9}+\dfrac{\ln2}{2\right)\cdot\dfrac{1}{5}}}{2}$

Efetue a propriedade distributiva da multiplicação

$S=\dfrac{\ln\left(\dfrac{6}{5}\right)}{12}+\dfrac{\dfrac{\ln\left(\dfrac{6}{5}\right)}{6}+\dfrac{\ln\left(\dfrac{7}{5}\right)}{7}}{2}+\dfrac{\dfrac{\ln\left(\dfrac{7}{5}\right)}{7}+\dfrac{\ln\left(\dfrac{8}{5}\right)}{8}}{2}+\dfrac{\dfrac{\ln\left(\dfrac{8}{5}\right)}{8}+\dfrac{\ln\left(\dfrac{9}{5}\right)}{9}}{2}+\dfrac{\dfrac{\ln\left(\dfrac{9}{5}\right)}{9}+\dfrac{\ln2}{10}}{2}$

Some as frações e simplifique-as caso possível

$S=\dfrac{\ln\left(\dfrac{6}{5}\right)}{12}+\dfrac{7\ln\left(\dfrac{6}{5}\right)+6\ln\left(\dfrac{7}{5}\right)}{84}+\dfrac{8\ln\left(\dfrac{7}{5}\right)+7\ln\left(\dfrac{8}{5}\right)}{112}+\dfrac{9\ln\left(\dfrac{8}{5}\right)+8\ln\left(\dfrac{9}{5}\right)}{144}+\dfrac{10\ln\left(\dfrac{9}{5}\right)+9\ln2}{180}$

O mínimo múltiplo comum entre os denominadores é 5040, logo

$S=\dfrac{420\ln\left(\dfrac{6}{5}\right)+420\ln\left(\dfrac{6}{5}\right)+360\ln\left(\dfrac{7}{5}\right)+360\ln\left(\dfrac{7}{5}\right)+315\ln\left(\dfrac{8}{5}\right)+315\ln\left(\dfrac{8}{5}\right)}{5040}$$+\dfrac{280\ln\left(\dfrac{9}{5}\right)+280\ln\left(\dfrac{9}{5}\right)+252\ln2}{5040}$

Sabendo que $\ln\left(\dfrac{x}{y}\right)=\ln x-\ln y$, ficamos com

$S=\dfrac{840\ln6 +720\ln7+630\ln 8+560\ln 9+252\ln2-2750\ln5}{5040}$

Com o auxílio de uma calculadora, por fim encontramos a aproximação

$S\approx 0.237172$

Este é o resultado da soma trapezoidal desta função neste intervalo e é a resposta contida na letra d).