Question

determine a área entre a curva, ajuda o mais rápido possível s2

Answer 1

$\text{O gr\'afico da fun{\c c}\~ao tg(x) de }-\dfrac{\pi}{4} \text{\ a 0 est\'a abaixo do eixo x (y=0), logo a integral ser\'a: }\\ \\ -\int\limits^{0}_{-\frac{\pi}{4}} {tg(x)} \, dx \\ \\ \text{J\'a de 0 at\'e\ } \dfrac{\pi}{3} \text{a fun{\c c}\~ao tg(x) est\'a acima do eixo x (y=0), logo:} \\ \\ \int\limits^{\frac{\pi}{3}}_0 {tg(x)} \, dx \\ \\ \text{Logo a \'area total vai ser: }\\ \\ -\int\limits^0_{-\frac{\pi}{4}} {tg(x)} \, dx + \int\limits^{\frac{\pi}{3}}_0 {tg(x)} \, dx$

$\text{Agora temos que determinar a primitiva da tg(x)} \\ \\ \int{tg(x)} \, dx = \int{\dfrac{sen(x)}{cos(x)}} \, dx \\ \\ \text{Substitui{\c c}\~ao u du} \\ u = cos(x) \\ du = -sen(x)dx\\ \\ -\int{\dfrac{1}{u}} \, du = -ln(u)+c = ln(u^{-1})+c = ln(\dfrac{1}{u})+c = ln(\dfrac{1}{cos(x)})+c = \int{tg(x)} \, dx \\ \\ -\int\limits^0_{-\frac{\pi}{4}} {tg(x)} \, dx = ln(\dfrac{1}{cos(x)})|_{-\frac{\pi}{4}}^0 = -ln(\dfrac{1}{cos(0)}) + ln(\dfrac{1}{cos(-\dfrac{\pi}{4})}) = -ln(\sqrt{2}) + ln(2)$

$\int\limits^{\frac{\pi}{3}}_0 {tg(x)} \, dx = ln(\dfrac{1}{cos(x)})|_0^{\frac{\pi}{3}} =ln(\dfrac{1}{cos(\frac{\pi}{3})}) - ln(\dfrac{1}{cos(0)}) = ln(2) \\ \\ \'Area \ total \\ \\ -ln(\sqrt{2}) + ln(2) + ln(2) \cong 1.04\ (u.a)$