Question

responder com a resolução

Answer 1

Achando a integral indefinida (sem a constante de integração):

$\int(3x-\frac{x^{3}}{4})dx=\int3xdx-\int\frac{x^{3}}{4}dx\\\\\int(3x-\frac{x^{3}}{4})dx=3\int x^{1}dx-\frac{1}{4}\int x^{3}dx\\\\\int(3x-\frac{x^{3}}{4})dx=3\cdot\frac{x^{1+1}}{1+1}-\frac{1}{4}\cdot\frac{x^{3+1}}{3+1}\\\\\boxed{\int(3x-\frac{x^{3}}{4})dx=\dfrac{3x^{2}}{2}-\frac{x^{4}}{16}}$
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$\int\limits_{0}^{4}(3x-\frac{x^{3}}{4})dx=\left\dfrac{3x^{2}}{2}-\dfrac{x^{4}}{16}\right]_{0}^{4}\\\\\\\int\limits_{0}^{4}(3x-\frac{x^{3}}{4})dx=\left(\dfrac{3(4)^{2}}{2}-\dfrac{4^{4}}{16}\right)-\left(\dfrac{3(0)^{2}}{2}-\dfrac{0^{4}}{16}\right)\\\\\\\int\limits_{0}^{4}(3x-\frac{x^{3}}{4})dx=\dfrac{3(16)}{2}-\dfrac{256}{16}-0\\\\\\\int\limits_{0}^{4}(3x-\frac{x^{3}}{4})dx=24-16\\\\\\\boxed{\boxed{\int\limits_{0}^{4}(3x-\frac{x^{3}}{4})dx=8}}$