Question

Calcule a integral abaixo: $\int\limits^3_1$$\int\limits^x_x$ xy² dydx

Answer 1

Integrais duplas devem ser resolvidas "de dentro para fora":

$\int_{1}^{3}\int_{x^{2}}^{x}xy^{2}dydx = \left [ x . \frac{y^{3}}{3} \right ]_{x^{2}}^{x} = x . \frac{x^{3}}{3} - x . \frac{(x^{2})^{3}}{3} = \frac{x^{4}}{3} - \frac{x^{7}}{3} = \frac{1}{3}(x^{4}-x^{7})$

$\int_{1}^{3}\frac{1}{3}(x^{4}-x^{7})dx = \frac{1}{3}\int_{1}^{3}(x^{4}-x^{7})dx =\frac{1}{3}\left [ \frac{x^{5}}{5}-\frac{x^{8}}{8} \right ]_{1}^{3}$

$= \frac{1}{3}\left [ \left ( \frac{3^{5}}{5}-\frac{3^{8}}{8} \right )-\left ( \frac{1^{5}}{5}-\frac{1^{8}}{8} \right ) \right ]\\ \\= \frac{1}{3}\left [ \left ( \frac{243}{5}-\frac{6561}{8} \right )-\left ( \frac{1}{5}-\frac{1}{8} \right ) \right ]\\ \\ = \frac{1}{3}\left [ -\frac{30861}{40}- \frac{3}{40} \right ]\\ \\ = \frac{1}{3}\left [ -\frac{30864}{40} \right ]\\ \\ = -\frac{10288}{40} = -\frac{1286}{5}$