Question

Resolve as integrais abaixo

Answer 1

a) $\displaystyle\int\limits_{-1}^{2}{(x^{3}-2x)\,dx}$

$=\left[\dfrac{x^{4}}{4}-x^{2} \right ]_{-1}^{2}\\ \\ \\ =\left[\dfrac{2^{4}}{4}-2^{2} \right ]-\left[\dfrac{(-1)^{4}}{4}-(-1)^{2} \right ]\\ \\ \\ =\left[\dfrac{16}{4}-4 \right ]-\left[\dfrac{1}{4}-1 \right ]\\ \\ \\ =0-\left[\dfrac{1-4}{4} \right ]\\ \\ \\ =0-\left[-\dfrac{3}{4} \right ]\\ \\ \\ =\dfrac{3}{4}$


b) $\displaystyle\int\limits_{0}^{1}{\left(1+\dfrac{1}{2}\,u^{4}-\dfrac{2}{5}\,u^{9}\right)\,du}$

$=\left[u+\dfrac{1}{2}\cdot \dfrac{u^{5}}{5}-\dfrac{2}{5}\cdot \dfrac{u^{10}}{10} \right ]_{0}^{1}\\ \\ \\ =\left[u+\dfrac{u^{5}}{10}-\dfrac{u^{10}}{25} \right ]_{0}^{1}\\ \\ \\ =\left[1+\dfrac{1^{5}}{10}-\dfrac{1^{10}}{25} \right ]-\left[0+\dfrac{0^{5}}{10}-\dfrac{0^{10}}{25} \right ]\\ \\ \\ =\left[1+\dfrac{1}{10}-\dfrac{1}{25} \right ]-0\\ \\ \\ =\dfrac{50+5-2}{50}\\ \\ \\ =\dfrac{53}{50}$


c) $\displaystyle\int\limits_{1}^{8}{\,^{3}\!\!\!\sqrt{x}\,dx}$

$=\displaystyle\int\limits_{1}^{8}{x^{1/3}\,dx}\\ \\ \\ =\left[\dfrac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1} \right ]_{1}^{8}\\ \\ \\ =\left[\dfrac{x^{4/3}}{\frac{4}{3}} \right ]_{1}^{8}\\ \\ \\ =\left[\dfrac{3x^{4/3}}{4} \right ]_{1}^{8}\\ \\ \\ =\left[\dfrac{3(\,^{3}\!\!\!\sqrt{x})^{4}}{4} \right ]_{1}^{8}\\ \\ \\ =\left[\dfrac{3(\,^{3}\!\!\!\sqrt{8})^{4}}{4} \right ]-\left[\dfrac{3(\,^{3}\!\!\!\sqrt{1})^{4}}{4} \right ]\\ \\ \\ =\left[\dfrac{3\cdot 2^{4}}{4} \right ]-\left[\dfrac{3\cdot 1^{4}}{4} \right ]\\ \\ \\ =12-\dfrac{3}{4}\\ \\ \\ =\dfrac{45}{4}$


d) $\displaystyle\int\limits_{1}^{2}{\dfrac{v^{3}+3v^{6}}{v^{4}}\,dv}$

$=\displaystyle\int\limits_{1}^{2}{\left(\dfrac{v^{3}}{v^{4}}+\dfrac{3v^{6}}{v^{4}} \right )\,dv}\\ \\ \\ =\displaystyle\int\limits_{1}^{2}{\left(\dfrac{1}{v}+3v^{2} \right )\,dv}\\ \\ \\ =\left[\mathrm{\ell n\,}|v|+v^{3}\right]_{1}^{2}\\ \\ =\left[\mathrm{\ell n\,}|2|+2^{3}\right]-\left[\mathrm{\ell n\,}|1|+1^{3}\right]\\ \\ =\left[\mathrm{\ell n\,}(2)+8\right]-\left[0+1\right]\\ \\ =\mathrm{\ell n\,}(2)+8-1\\ \\ =\mathrm{\ell n\,}(2)+7$