Question

A área sombreada no gráfico abaixo é limitada superiormente pela função:

Answer 1

$A=-\int\limits^4_1 {\frac{12lnx}{x^2} } \, dx =-12\int\limits^4_1 {\frac{lnx}{x^2} } \, dx$

u = lnx        dv = 1/x^2

du = 1/x          v = -1/x

$-12(-\frac{lnx}{x}- \int\limits^4_1 {-\frac{1}{x^2} } \, dx \\-12(-\frac{lnx}{x}+ \int\limits^1_4 {x^{-2} }) \, dx \\-12(-\frac{lnx}{x} -\frac{1}{x} )$

Variando de 1 a 4

A = [-12(-ln4/4 - 1/4)] - [-12(-ln1/1 - 1/1)]

A= (3ln4 + 3) - (12)  = 3ln4 - 9

A ~= |-4,84| ~= 4,84m^2

Answer 2

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$\tt vamos~avaliar~a~integral~\displaystyle\sf\int\dfrac{12\cdot \ell n(x)}{x^2}~dx\\\tt em~seguida~substituir~os~limites~de~\rm{integrac_{\!\!,}\tilde ao}.$

$\displaystyle\sf\int\dfrac{12\cdot \ell n(x)}{x^2}~dx=12\cdot\int\dfrac{\ell n(x)}{x^2}~dx\\\rm fac_{\!\!,}a~u=\ell n(x)\implies du=\dfrac{1}{x}dx\\\rm dv=\dfrac{dx}{x^2}\implies v=-\dfrac{1}{x}\\\displaystyle\sf12\int\dfrac{\ell n(x)}{x^2}=12\left[\ell n(x)\cdot\left(-\dfrac{1}{x}\right)-\int-\dfrac{1}{x}\cdot\dfrac{1}{x}~dx\right]\\\displaystyle\sf12\left[-\dfrac{\cdot\ell n(x)}{x}+\int\dfrac{1}{x^2}~dx\right]\\\displaystyle\sf12\left[-\dfrac{\ell n(x)}{x}-\dfrac{1}{x}\right]+k$

$\displaystyle\sf\int_1^4\dfrac{12\cdot\ell n(x)}{x^2}~dx=12\cdot\left[-\dfrac{\ell n(x)}{x}-\dfrac{1}{x}\right]_1^4\\\sf 12\cdot\left[-\dfrac{\ell n(4)}{4}-\dfrac{1}{4}-\left(-\dfrac{\ell n(1)}{1}-\dfrac{1}{1}\right)\right]\\\sf 12\left[-\dfrac{\ell n(4)}{4}-\dfrac{1}{4}+\dfrac{\ell n(1)}{1}+\dfrac{1}{1}\right]\\\sf 12\left[\dfrac{-\ell n(4)-1+\underbrace{4\ell n(1)}_{0}+4}{4}\right]\\\sf\diagup\!\!\!\!\!12^3\cdot\left[\dfrac{-\ell n(4)+3}{\diagup\!\!\!4}\right]\\\sf 3\cdot(3-\ell n(4))=3\cdot1,61$

$\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int_1^4\dfrac{12\ell n(x)}{x^2}~dx=4,83~m^2\checkmark}}}}$