Question

cálculo 2 integral derivada

Answer 1

$\mathbf{a)~~}\displaystyle\int\limits_{1}^{3}{\left(1+\dfrac{1}{x}+\dfrac{1}{x^{2}} \right )dx}\\ \\ \\ =\int\limits_{1}^{3}{1\,dx}+\int\limits_{1}^{3}{\dfrac{1}{x}\,dx}+\int\limits_{1}^{3}{\dfrac{1}{x^{2}}\,dx}\\ \\ \\ =\int\limits_{1}^{3}{1\,dx}+\int\limits_{1}^{3}{\dfrac{1}{x}\,dx}+\int\limits_{1}^{3}{x^{-2}\,dx}\\ \\ \\ =\left.\left(x+\mathrm{\ell n}|x|+\dfrac{x^{-2+1}}{-2+1} \right )\right|_{1}^{3}$


Como $x$ é sempre positivo no intervalo de integração, podemos dispensar o módulo no logaritmo. E ficamos com

$=\displaystyle\left.\left(x+\mathrm{\ell n\,}x+ \dfrac{x^{-1}}{-1} \right )\right|_{1}^{3}\\ \\ \\ =\left.\left(x+\mathrm{\ell n\,}x-\dfrac{1}{x} \right )\right|_{1}^{3}\\ \\\\ =\left(3+\mathrm{\ell n\,}3-\dfrac{1}{3} \right )-\left(1+\mathrm{\ell n\,}1- \dfrac{1}{1} \right )\\ \\ \\ =\mathrm{\ell n\,}3+\dfrac{9-1}{3}-1+0+1\\ \\ \\ =\mathrm{\ell n\,}3+\dfrac{8}{3}$


$\mathbf{b)~~}\displaystyle\int\limits_{-1}^{1}{3t^{4}\,dt}\\ \\ \\ =3\cdot \left.\dfrac{t^{4+1}}{4+1}\right|_{-1}^{1}\\ \\ \\ =3\cdot \left.\dfrac{t^{5}}{5}\right|_{-1}^{1}\\ \\ \\ =3\cdot \left(\dfrac{1^{5}}{5}-\dfrac{(-1)^{5}}{5} \right )\\ \\ \\ =3\cdot \left(\dfrac{1}{5}-\dfrac{(-1)}{5} \right )\\ \\ \\ =3\cdot \left(\dfrac{1}{5}+\dfrac{1}{5} \right )\\ \\ \\ =3\cdot\dfrac{2}{5} \\ \\ \\ =\dfrac{6}{5}$