Question

resposta completa na intregal definida ∫01 (x2 +√¯ x) dx

Answer 1

$\int\limits_{0}^{1}(x^{2}+\sqrt{x})dx=\int(x^{2}+\sqrt{x})dx]_{0}^{1}$

Ou seja, podemos calcular essa integral definida primeiramente achando a integral indefinida, e depois aplicá-la nos limites de integração dados.

Achando a integral indefinida (sem adicionar a constante):

$\int(x^{2}+\sqrt{x})dx=\int(x^{2}+x^{1/2})dx\\\\\int(x^{2}+\sqrt{x})dx=\int x^{2}dx+\int x^{1/2}dx\\\\\int(x^{2}+\sqrt{x})dx=\frac{x^{3}}{3}+\frac{x^{(1/2)+1}}{(1/2)+1}\\\\\int(x^{2}+\sqrt{x})dx=\frac{x^{3}}{3}+\frac{x^{3/2}}{(3/2)}\\\\\int(x^{2}+\sqrt{x})dx=\frac{x^{3}}{3}+\frac{2x^{3/2}}{3}$
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$\int\limits_{0}^{1}(x^{2}+\sqrt{x})dx=\left\dfrac{x^{3}}{3}+\dfrac{2x^{3/2}}{3}\right]_{0}^{1}\\\\\\\int\limits_{0}^{1}(x^{2}+\sqrt{x})dx=\left(\dfrac{1^{3}}{3}+\dfrac{2(1)^{3/2}}{3}\right)-\left(\dfrac{0^{3}}{3}+\dfrac{2(0)^{3/2}}{3}\right)\\\\\\\int\limits_{0}^{1}(x^{2}+\sqrt{x})dx=\dfrac{1}{3}+\dfrac{2}{3}-0\\\\\\\int\limits_{0}^{1}(x^{2}+\sqrt{x})dx=\dfrac{3}{3}\\\\\\\boxed{\boxed{\int\limits_{0}^{1}(x^{2}+\sqrt{x})dx=1}}$