Resolva as integrais definidas e indefinidas abaixo:
\int_{1 / 2}^{2} \frac{\sqrt{x^{7}}-\sqrt[3]{x}}{6 \sqrt[4]{x}} d x

Passo 1

O enunciado pede a integral definida
\int_{1/2}^{2}\frac{\sqrt{x^{7}}-\sqrt[3]{x}}{6\sqrt[4]{x}}\,dx,
que envolve manipular potências de x e usar a fórmula de antiderivada de um monômio:
\int x^{n}\,dx = \frac{x^{n+1}}{n+1},\quad n\neq -1.

Passo 2

Reescrevemos cada raiz como potência de x: \sqrt{x^{7}} = x^{7/2}, \sqrt[3]{x} = x^{1/3}, \sqrt[4]{x} = x^{1/4}.
Assim o integrando fica
\frac{x^{7/2}-x^{1/3}}{6\,x^{1/4}} = \frac{1}{6}\Bigl(x^{7/2-1/4}-x^{1/3-1/4}\Bigr) = \frac{1}{6}\bigl(x^{13/4}-x^{1/12}\bigr).

Passo 3

Encontramos uma primitiva F(x) de \frac{1}{6}(x^{13/4}-x^{1/12}).
Para x^{13/4} temos
\int x^{13/4}\,dx = \frac{x^{13/4+1}}{13/4+1} = \frac{x^{17/4}}{17/4} = \frac{4}{17}\,x^{17/4},
e para x^{1/12}
\int x^{1/12}\,dx = \frac{x^{1/12+1}}{1/12+1} = \frac{x^{13/12}}{13/12} = \frac{12}{13}\,x^{13/12}.
Multiplicando pelo fator 1/6,
F(x)=\frac{1}{6}\Bigl(\frac{4}{17}x^{17/4}-\frac{12}{13}x^{13/12}\Bigr) =\frac{2}{51}\,x^{17/4}-\frac{2}{13}\,x^{13/12}.

Passo 4

Pelo Teorema Fundamental do Cálculo,
\int_{1/2}^{2}\frac{\sqrt{x^{7}}-\sqrt[3]{x}}{6\sqrt[4]{x}}\,dx =F(2)-F\bigl(\tfrac12\bigr).
Calculamos
F(2)=\frac{2}{51}\,2^{17/4}-\frac{2}{13}\,2^{13/12}, \quad F\bigl(\tfrac12\bigr) =\frac{2}{51}\,\bigl(\tfrac12\bigr)^{17/4} -\frac{2}{13}\,\bigl(\tfrac12\bigr)^{13/12}.

Passo 5

Juntando as duas parcelas,
\int_{1/2}^{2}\frac{\sqrt{x^{7}}-\sqrt[3]{x}}{6\sqrt[4]{x}}\,dx =\frac{2}{51}\Bigl(2^{17/4}-2^{-17/4}\Bigr) -\frac{2}{13}\Bigl(2^{13/12}-2^{-13/12}\Bigr).

Resposta

\frac{2}{51}\bigl(2^{17/4}-2^{-17/4}\bigr) -\frac{2}{13}\bigl(2^{13/12}-2^{-13/12}\bigr).

Exercícios relacionados