( \mathbf{3 , 0} ptos) Calcule as integrais usando funções Gamma e Beta
\int_{0}^{1} 3 x^{2} \sqrt{1-x} d x

Passo 1

O enunciado pede para calcular
\int_{0}^{1}3\,x^{2}\sqrt{1-x}\,dx
usando as funções Beta e Gamma. Para isso precisamos lembrar que
B(a,b)=\int_{0}^{1}t^{a-1}(1-t)^{b-1}dt \quad\text{e}\quad B(a,b)=\frac{\Gamma(a)\,\Gamma(b)}{\Gamma(a+b)}.

Passo 2

Reescrevemos a integral de forma a identificar um Beta. Note que
\sqrt{1-x}=(1-x)^{1/2},
logo
\int_{0}^{1}3\,x^{2}\,(1-x)^{1/2}dx =3\int_{0}^{1}x^{3-1}(1-x)^{\tfrac32-1}dx =3\,B\bigl(3,\tfrac32\bigr).

Passo 3

Usamos a relação entre Beta e Gamma:
B\bigl(3,\tfrac32\bigr) =\frac{\Gamma(3)\,\Gamma(\tfrac32)}{\Gamma\bigl(3+\tfrac32\bigr)} =\frac{\Gamma(3)\,\Gamma(\tfrac32)}{\Gamma(\tfrac92)}.

Passo 4

Calculamos cada Gamma:
\Gamma(3)=2!, \quad \Gamma\bigl(\tfrac12\bigr)=\sqrt{\pi}, \quad \Gamma\bigl(\tfrac32\bigr)=\tfrac12\,\Gamma\bigl(\tfrac12\bigr)=\tfrac12\sqrt{\pi},
e
\Gamma\bigl(\tfrac92\bigr) =\frac{7}{2}\,\Gamma\bigl(\tfrac72\bigr) =\frac{7}{2}\cdot\frac{5}{2}\,\Gamma\bigl(\tfrac52\bigr) =\frac{7}{2}\cdot\frac{5}{2}\cdot\frac{3}{2}\,\Gamma\bigl(\tfrac32\bigr) =\frac{7\cdot5\cdot3}{2^{3}}\cdot\frac12\sqrt{\pi} =\frac{105}{16}\sqrt{\pi}.

Passo 5

Substituindo em B(3,\tfrac32):
B\bigl(3,\tfrac32\bigr) =\frac{2\cdot\bigl(\tfrac12\sqrt{\pi}\bigr)}{\tfrac{105}{16}\sqrt{\pi}} =\frac{\sqrt{\pi}}{\tfrac{105}{16}\sqrt{\pi}} =\frac{16}{105}.
Então a integral original é
3\,B\bigl(3,\tfrac32\bigr) =3\cdot\frac{16}{105} =\frac{48}{105} =\frac{16}{35}.

Resposta

\displaystyle \frac{16}{35}

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