Calcule o valor exato das integrais definidas:
\int_{-1}^{0}(\sqrt{3-2 x}) x^{2} d x

Passo 1

O enunciado pede para calcular a integral definida
\int_{-1}^{0}\sqrt{3-2x}\,x^{2}\,dx.
O conteúdo principal envolve a técnica de substituição para simplificar a raiz e depois integrar potências de u.

Passo 2

Fazemos a substituição
u = 3 - 2x,
de modo que
x = \frac{3 - u}{2}, \quad dx = -\frac{1}{2}\,du.
Quando x=-1, u=3-2(-1)=5; quando x=0, u=3-2\cdot0=3.

Passo 3

Em termos de u, temos
x^{2} = \Bigl(\tfrac{3-u}{2}\Bigr)^{2}, \quad \sqrt{3-2x} = \sqrt{u}, \quad dx = -\tfrac12\,du.
Portanto
\int_{-1}^{0}\sqrt{3-2x}\,x^{2}\,dx =-\tfrac18\int_{5}^{3}(3-u)^{2}\,u^{\tfrac12}\,du =\tfrac18\int_{3}^{5}(3-u)^{2}\,u^{\tfrac12}\,du.

Passo 4

Expandimos (3-u)^{2}=u^{2}-6u+9, de modo que o integrando fica
u^{\tfrac52}-6u^{\tfrac32}+9u^{\tfrac12}.
Logo
\tfrac18\int_{3}^{5}(3-u)^{2}u^{\tfrac12}\,du =\tfrac18\int_{3}^{5}\bigl(u^{\tfrac52}-6u^{\tfrac32}+9u^{\tfrac12}\bigr)\,du.

Passo 5

Calculamos a primitiva de cada termo:
\int u^{\tfrac52}du = \tfrac{2}{7}u^{\tfrac72}, \quad \int u^{\tfrac32}du = \tfrac{2}{5}u^{\tfrac52}, \quad \int u^{\tfrac12}du = \tfrac{2}{3}u^{\tfrac32}.
Assim
\tfrac18\Bigl[\tfrac{2}{7}u^{\tfrac72} -6\cdot\tfrac{2}{5}u^{\tfrac52} +9\cdot\tfrac{2}{3}u^{\tfrac32}\Bigr]_{3}^{5} =\Bigl[\tfrac{1}{28}u^{\tfrac72} -\tfrac{3}{10}u^{\tfrac52} +\tfrac{3}{4}u^{\tfrac32}\Bigr]_{3}^{5}.

Passo 6

Avaliaremos em u=5 e u=3:
\Bigl(\tfrac{1}{28}\,5^{\tfrac72} -\tfrac{3}{10}\,5^{\tfrac52} +\tfrac{3}{4}\,5^{\tfrac32}\Bigr) - \Bigl(\tfrac{1}{28}\,3^{\tfrac72} -\tfrac{3}{10}\,3^{\tfrac52} +\tfrac{3}{4}\,3^{\tfrac32}\Bigr) =\frac{25\sqrt{5}-18\sqrt{3}}{35}.

Resposta

\displaystyle \frac{25\sqrt{5}-18\sqrt{3}}{35}

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