Question

Integral definida de raiz cubica de t a=-1 b=1

Answer 1

$\displaystyle\int_{-1}^{1}\,^{3}\!\!\!\sqrt{t}\,dt\\\\\\ =\int_{-1}^{1}t^{1/3}\,dt~~~~~~\mathbf{(i)}$


Regra para encontrar primitiva de potência:

$\displaystyle\int{t^n\,dt}=\dfrac{t^{n+1}}{n+1}\,,~~~~n\ne -1.$


Portanto, voltando a $\mathbf{(i)}$ temos

$=\left.\left(\dfrac{t^{(1/3)+1}}{\frac{1}{3}+1} \right )\right|_{-1}^{1}\\\\\\ =\left.\left(\dfrac{t^{4/3}}{\frac{4}{3}} \right )\right|_{-1}^{1}\\\\\\ =\left.\left(\dfrac{3}{4}\,t^{4/3}\right)\right|_{-1}^{1}\\\\\\ =\dfrac{3}{4}\cdot 1^{4/3}-\dfrac{3}{4}\cdot (-1)^{4/3}\\\\\\ =\dfrac{3}{4}\cdot \,^{3}\!\!\!\sqrt{1^4}-\dfrac{3}{4}\cdot \,^{3}\!\!\!\sqrt{(-1)^4}\\\\\\ =\dfrac{3}{4}\cdot \,^{3}\!\!\!\sqrt{1}-\dfrac{3}{4}\cdot \,^{3}\!\!\!\sqrt{1}\\\\\\ =\dfrac{3}{4}-\dfrac{3}{4}\\\\\\ =0$

como era de se esperar.

(Função ímpar integrada sobre um intervalo simétrico em relação à origem dá zero)