(1+y^3)^(1/2) −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 −3 -2.5 −2 -1.5 −1 -0.5 0 0.5 1 1.5 2 2.5 3 x y
Exercício∫ x 2 0 1 + y 3 2 d y \int_{x^2}^0\sqrt[2]{1+y^3}dy ∫ x 2 0 2 1 + y 3 d y
Solução explicada passo a passo
1
Aplicamos a regra: ∫ 1 + θ 3 d x \int\sqrt{1+\theta ^3}dx ∫ 1 + θ 3 d x = 2 5 θ 3 + 1 ( θ 4 + − 1 6 ( 3 ) 3 4 − 1 6 ( θ + 1 ) θ 2 − θ + 1 F ( arcsin ( − ( − 1 ) 5 6 ( θ + 1 ) 3 4 ) ∣ − 1 3 ) + θ ) + C =\frac{2}{5\sqrt{\theta ^3+1}}\left(\theta ^4+\sqrt[6]{-1}\sqrt[4]{\left(3\right)^{3}}\sqrt{\sqrt[6]{-1}\left(\theta +1\right)}\sqrt{\theta ^2-\theta +1}F\left(\arcsin\left(\frac{\sqrt{-\sqrt[6]{\left(-1\right)^{5}}\left(\theta +1\right)}}{\sqrt[4]{3}}\right)\Big\vert \sqrt[3]{-1}\right)+\theta \right)+C = 5 θ 3 + 1 2 ( θ 4 + 6 − 1 4 ( 3 ) 3 6 − 1 ( θ + 1 ) θ 2 − θ + 1 F ( arcsin ( 4 3 − 6 ( − 1 ) 5 ( θ + 1 ) ) ∣ ∣ 3 − 1 ) + θ ) + C 1 + x 3 = 1 + y 3 1+x^3=1+y^3 1 + x 3 = 1 + y 3 1 / 2 = 1 2 1/2=\frac{1}{2} 1/2 = 2 1 d x = d y dx=dy d x = d y x = y x=y x = y x 3 = y 3 x^3=y^3 x 3 = y 3
[ 2 5 y 3 + 1 ( y 4 + − 1 6 ( 3 ) 3 4 − 1 6 ( y + 1 ) y 2 − y + 1 F ( arcsin ( − ( − 1 ) 5 6 ( y + 1 ) 3 4 ) ∣ − 1 3 ) + y ) ] x 2 0 \left[\frac{2}{5\sqrt{y^3+1}}\left(y^4+\sqrt[6]{-1}\sqrt[4]{\left(3\right)^{3}}\sqrt{\sqrt[6]{-1}\left(y+1\right)}\sqrt{y^2-y+1}F\left(\arcsin\left(\frac{\sqrt{-\sqrt[6]{\left(-1\right)^{5}}\left(y+1\right)}}{\sqrt[4]{3}}\right)\Big\vert \sqrt[3]{-1}\right)+y\right)\right]_{x^2}^{0} ⎣ ⎡ 5 y 3 + 1 2 ⎝ ⎛ y 4 + 6 − 1 4 ( 3 ) 3 6 − 1 ( y + 1 ) y 2 − y + 1 F ⎝ ⎛ arcsin ⎝ ⎛ 4 3 − 6 ( − 1 ) 5 ( y + 1 ) ⎠ ⎞ ∣ ∣ 3 − 1 ⎠ ⎞ + y ⎠ ⎞ ⎦ ⎤ x 2 0
Resposta final para o problema [ 2 5 y 3 + 1 ( y 4 + − 1 6 ( 3 ) 3 4 − 1 6 ( y + 1 ) y 2 − y + 1 F ( arcsin ( − ( − 1 ) 5 6 ( y + 1 ) 3 4 ) ∣ − 1 3 ) + y ) ] x 2 0 \left[\frac{2}{5\sqrt{y^3+1}}\left(y^4+\sqrt[6]{-1}\sqrt[4]{\left(3\right)^{3}}\sqrt{\sqrt[6]{-1}\left(y+1\right)}\sqrt{y^2-y+1}F\left(\arcsin\left(\frac{\sqrt{-\sqrt[6]{\left(-1\right)^{5}}\left(y+1\right)}}{\sqrt[4]{3}}\right)\Big\vert \sqrt[3]{-1}\right)+y\right)\right]_{x^2}^{0} ⎣ ⎡ 5 y 3 + 1 2 ⎝ ⎛ y 4 + 6 − 1 4 ( 3 ) 3 6 − 1 ( y + 1 ) y 2 − y + 1 F ⎝ ⎛ arcsin ⎝ ⎛ 4 3 − 6 ( − 1 ) 5 ( y + 1 ) ⎠ ⎞ ∣ ∣ 3 − 1 ⎠ ⎞ + y ⎠ ⎞ ⎦ ⎤ x 2 0