Question

O cálculo da integral∫_0^5▒dr/√(5&x) resulta em:

Answer 1

$\displaystyle\int\limits_{0}^{5}{\dfrac{dx}{\sqrt{5^{x}}}}\\ \\ \\ =\int\limits_{0}^{5}{\dfrac{dx}{(5^{x})^{1/2}}}\\ \\ \\ =\int\limits_{0}^{5}{\dfrac{dx}{(5^{x/2})}}\\ \\ \\ =\int\limits_{0}^{5}{5^{-x/2}\,dx}\\ \\ \\ =\int\limits_{0}^{5}{(e^{\mathrm{\ell n}(5)})^{-x/2}\,dx}\\ \\ \\ =\int\limits_{0}^{5}{e^{-\frac{\mathrm{\ell n}(5)}{2}\,x}\,dx}\;\;\;\;\;\;\mathbf{(i)}$


Façamos a seguinte mudança de variável:

$-\dfrac{\mathrm{\ell n}(5)}{2}\,x=u\;\;\Rightarrow\;\;-\dfrac{\mathrm{\ell n}(5)}{2}\,dx=du\;\;\Rightarrow\;\;dx=-\dfrac{2}{\mathrm{\ell n}(5)}\,du$


Mudando os extremos de integração:

$\text{quando }x=0\;\;\Rightarrow\;\;u=0\\ \\ \text{quando }x=5\;\;\Rightarrow\;\;u=-\dfrac{5\,\mathrm{\ell n}(5)}{2}$


Substituindo em $\mathbf{(i)},$ ficamos com

$\displaystyle\int\limits_{0}^{-\frac{5}{2}\,\mathrm{\ell n}(5)}{e^{u}\cdot \left(-\dfrac{2}{\mathrm{\ell n}(5)} \right )\,du}\\ \\ \\ =-\dfrac{2}{\mathrm{\ell n}(5)}\cdot \int\limits_{0}^{-\frac{5}{2}\,\mathrm{\ell n}(5)}{e^{u}\,du}\\ \\ \\ =-\dfrac{2}{\mathrm{\ell n}(5)}\cdot \left.\begin{array}{c}\\ e^{u}\\\\ \end{array}\right|_{0}^{-\frac{5}{2}\,\mathrm{\ell n}(5)}\\ \\ \\ =-\dfrac{2}{\mathrm{\ell n}(5)}\cdot [e^{-\frac{5}{2}\,\mathrm{\ell n}(5)}-e^{0}]\\ \\ \\ =-\dfrac{2}{\mathrm{\ell n}(5)}\cdot (5^{-5/2}-1)\\ \\ \\ =-\dfrac{2}{\mathrm{\ell n}(5)}\cdot \left(\dfrac{1}{5^{5/2}}-1\right)\\ \\ \\ =-\dfrac{2}{\mathrm{\ell n}(5)}\cdot \left(\dfrac{1}{\sqrt{5^{5}}}-1\right)\\ \\ \\ =-\dfrac{2}{\mathrm{\ell n}(5)}\cdot \left(\dfrac{1}{\sqrt{(5^{2})^{2}\cdot 5}}-1\right)$

$=-\dfrac{2}{\mathrm{\ell n}(5)}\cdot \left(\dfrac{1}{5^{2}\cdot \sqrt{5}}-1\right)\\ \\ \\ =-\dfrac{2}{\mathrm{\ell n}(5)}\cdot \left(\dfrac{1}{25\sqrt{5}}-1\right)$

Answer 2

A resposta está em anexo.