Calcule as integrais usando funções Gamma e Beta
\int_{0}^{1} \sqrt{x}(1-x)^{3} d x

Passo 1

O enunciado pede para calcular
\int_{0}^{1} \sqrt{x}\,(1-x)^{3}\,dx
usando as funções Beta e Gamma. Vamos usar que
B(p,q)=\int_{0}^{1}x^{p-1}(1-x)^{q-1}\,dx
e que
B(p,q)=\frac{\Gamma(p)\,\Gamma(q)}{\Gamma(p+q)}.

Passo 2

Reescrevemos o integrando da forma x^{p-1}(1-x)^{q-1}. Como \sqrt{x}=x^{1/2}, temos
x^{1/2}=x^{(3/2)-1}
e (1-x)^{3}=(1-x)^{4-1}. Logo podemos identificar
p=\tfrac{3}{2}, \quad q=4,
e assim
\int_{0}^{1}\sqrt{x}\,(1-x)^{3}\,dx = B\bigl(\tfrac{3}{2},4\bigr).

Passo 3

Pela relação com Gamma,
B\bigl(\tfrac{3}{2},4\bigr) =\frac{\Gamma\bigl(\tfrac{3}{2}\bigr)\,\Gamma(4)}{\Gamma\bigl(\tfrac{3}{2}+4\bigr)} =\frac{\Gamma\bigl(\tfrac{3}{2}\bigr)\,\Gamma(4)}{\Gamma\bigl(\tfrac{11}{2}\bigr)}.

Passo 4

Calculamos cada Gamma:
\Gamma\bigl(\tfrac{3}{2}\bigr)=\frac{\sqrt{\pi}}{2}, \quad \Gamma(4)=3!=6,
\Gamma\bigl(\tfrac{11}{2}\bigr) =\bigl(\tfrac{9}{2}\bigr)\bigl(\tfrac{7}{2}\bigr)\bigl(\tfrac{5}{2}\bigr)\bigl(\tfrac{3}{2}\bigr)\bigl(\tfrac{1}{2}\bigr)\,\Gamma\bigl(\tfrac{1}{2}\bigr) =\frac{9\cdot7\cdot5\cdot3\cdot1}{2^{5}}\sqrt{\pi} =\frac{945}{32}\sqrt{\pi}.

Passo 5

Substituindo:
\int_{0}^{1}\sqrt{x}\,(1-x)^{3}\,dx =\frac{\frac{\sqrt{\pi}}{2}\cdot6}{\frac{945}{32}\sqrt{\pi}} =\frac{6}{2}\cdot\frac{32}{945} =\frac{96}{945} =\frac{32}{315}.

Resposta

\displaystyle \frac{32}{315}

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