Question

Qual a integral definida de $\sqrt[3]{x}$ , sendo a=1 e b=8

Answer 1

Resposta:

∛x=x^(1/3)

∫ x^(1/3)  dx  = x^(1/3+1) /  (1/3)+1  =x^(4/3)  / (4/3) = (3/4)* x^(4/3)

                          8

=[  (3/4)* x^(4/3)]

                           1

=(3/4)* 8^(4/3)] - (3/4) *1^(4/3)

=(3/4)* (2³)^(4/3)] - (3/4) *1^(4/3)

=(3/4)* (2)^(4)] - (3/4) *1^(4/3)

=(3/4) *16 -3/4 * 1

=12 -3/4 = (48-3)/4 =45/4  unid. área

Answer 2

$\int\limits^8_1\sqrt[3]{x}dx$

$\int\limits^8_1{x}^{\frac{1}{3}}dx$

$\frac{3}{4}{x}^{\frac{4}{3}}$ com x variando de 1 a 8.Substituindo os limites de integração temos

$\frac{3}{4}.{8}^{\frac{4}{3}}-\frac{3}{4}{1}^{\frac{3}{4}}$

$\frac{3}{4}.{{2}^{\cancel{3}}}^{\frac{4}{\cancel{3}}} -\frac{3}{4}.{1}^{\frac{4}{3}}$

$\frac{48}{4}-\frac{3}{4}$

$\boxed{\boxed{\int\limits^8_1\sqrt[3]{x}dx=\frac{45}{4}}}$