Question

Calcular as integrais: $\int\limits^4_1 {\frac{2x^{3}-x^{2}\sqrt{x}+4}{3x^{2} } } \, dx$

Answer 1

$\displaystyle\mathsf{\int\limits_{1}^{4}\dfrac{2x^3-x^2\sqrt{x}+4}{3x^2}dx}\\\displaystyle\int\limits_{1}^{4}\mathsf{\dfrac{2}{3}x-\dfrac{1}{3}x^{\frac{1}{2}}+\dfrac{4}{3x^2}dx}$

$\mathsf{\left(\dfrac{1}{3}x^2-\dfrac{2}{9}x^{\frac{3}{2}}-\dfrac{4}{3x}\right)\Bigg|_{1}^{4}}$

$\mathsf{\dfrac{1}{3}.4^2-\dfrac{2}{9}.4^{\frac{3}{2}}-\dfrac{4}{3.4}-\left[\dfrac{1}{3}.1^2-\dfrac{2}{9}.1^{\frac{3}{2}}-\dfrac{4}{3.1}\right]}\\\mathsf{\dfrac{16}{3}-\dfrac{16}{9}-\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{2}{9}+\dfrac{4}{3}}$

$\mathsf{\dfrac{48-16-3-3+2+12}{9}=\dfrac{42}{9}=\dfrac{14}{3}}$

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$\displaystyle\mathsf{Life=\int\limits_{happy}^{death}\dfrac{happiness}{time}dtime}$