Question

qual o valor dessa integral?

Answer 1

$\int\limits_{-1}^{1} \int\limits_{0}^{x} \int\limits_{x}^{3x} \left(y*z*cos(x^5)\right) \;dzdydx\\\\ = \int\limits_{-1}^{1} \left[\; \int\limits_{0}^{x} y\;dy * \int\limits_{x}^{3x} z\;dz \right] cos(x^5)dx\\\\ \int\limits_{-1}^{1} \left[\; \left \frac{y^2}{2} \right |_{0}^{x} \; *\; \left \frac{z^2}{2} \right |_{x}^{3x} \right] cos(x^5)dx \\\\ \int\limits_{-1}^{1} \left [ \left( \frac{x^2}{2} - \frac{0^2}{2} \right) *\left( \frac{(3x)^2}{2} - \frac{x^2}{2} \right) \right ] \; cos(x^5) dx$

$\int\limits_{-1}^{1} \left [ \left( \frac{x^2}{2} \right) *\left( \frac{8x^2}{2} \right) \right ] \; cos(x^5) dx\\\\ \int\limits_{-1}^{1} \left [2x^4 \right ] \; cos(x^5) dx \\\\ = \int\limits_{-1}^{1}2 x^4cos(x^5)\;dx$


substituição
u = x^5
du = 5x^4 dx
du/(5x^4)  = dx

$2 \int\limits_{-1}^{1}\; x^4 cos(u)\; \frac{du}{5x^4} \\\\ 2\int\limits_{-1}^{1}\; cos(u)\; \frac{du}{5} \\\\2 * \frac{1}{5}\int\limits_{-1}^{1}\; cos(u)\; du \\\\ \frac{2}{5}\int\limits_{-1}^{1} cos(u)\; du = \frac{2}{5}*\;\left ( sen(u) \right )|^{1}_{-1} = \frac{2(sen(1)-sen(-1))}{5}= \frac{4sen(1)}{5}$