Question

2∫3 0∫x (x/2 +5xy2/2+4y) dy dx

Answer 1

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Calcular a integral iterada:

$\mathsf{\displaystyle\int_2^3 \int_0^x\left(\frac{x}{2}+\frac{5xy^2}{2}+4y\right)dy\,dx}$


Na ordem dada, primeiro integramos na variável  y,  e por último na variável  x:

$\mathsf{=\displaystyle\int_2^3 \left[\int_0^x\left(\frac{x}{2}+\frac{5xy^2}{2}+4y\right)dy \right ]dx}\\\\\\ \mathsf{=\displaystyle\int_2^3 \left[\int_0^x\frac{x}{2}\,dy+\int_0^x \frac{5xy^2}{2}\,dy+\int_0^x 4y\,dy \right ]dx}\\\\\\ \mathsf{=\displaystyle\int_2^3 \left[\frac{x}{2}\int_0^x dy+\frac{5x}{2}\int_0^x y^2\,dy+4\int_0^x y\,dy \right ]dx}\\\\\\ \mathsf{=\displaystyle\int_2^3 \left[\frac{x}{2}\cdot y\Big|_0^x+\frac{5x}{2}\cdot \frac{y^{2+1}}{2+1}\bigg|_0^x+4\cdot \frac{y^{1+1}}{1+1}\bigg|_0^x \right ]dx}\\\\\\ \mathsf{=\displaystyle\int_2^3 \left[\frac{x}{2}\cdot y\Big|_0^x+\frac{5x}{2}\cdot \frac{y^3}{3}\bigg|_0^x+4\cdot \frac{y^2}{2}\bigg|_0^x \right ]dx}$

$\mathsf{=\displaystyle\int_2^3 \left[\frac{x}{2}\cdot (x-0)+\frac{5x}{2}\cdot\left(\frac{x^3}{3}-\frac{0^3}{3}\right)+4\cdot\left(\frac{x^2}{2}-\frac{0^2}{2}\right)\right]dx}\\\\\\ \mathsf{=\displaystyle\int_2^3 \left[\frac{x}{2}\cdot x+\frac{5x}{2}\cdot\frac{x^3}{3}+4\cdot\frac{x^2}{2}\right]dx}\\\\\\ \mathsf{=\displaystyle\int_2^3 \left[\frac{x^2}{2}+\frac{5x^4}{6}+2x^2\right]dx}\\\\\\ \mathsf{=\displaystyle\int_2^3 \left[\left(\frac{1}{2}+2\right)\!x^2+\frac{5x^4}{6}\right]dx}\\\\\\ \mathsf{=\displaystyle\int_2^3 \left[\frac{5x^2}{2}+\frac{5x^4}{6}\right]dx}$

$\mathsf{=\displaystyle\int_2^3 \frac{5x^2}{2}\,dx+\int_2^3\frac{5x^4}{6}\,dx}\\\\\\ \mathsf{=\displaystyle\frac{5}{2}\int_2^3 x^2\,dx+\frac{5}{6}\int_2^3 x^4\,dx}\\\\\\ \mathsf{=\dfrac{5}{2}\cdot\dfrac{x^{2+1}}{2+1}\bigg|_2^3+\dfrac{5}{6}\cdot \dfrac{x^{4+1}}{4+1}\bigg|_2^3}\\\\\\ \mathsf{=\dfrac{5}{2}\cdot\dfrac{x^3}{3}\bigg|_2^3+\dfrac{5}{6}\cdot \dfrac{x^5}{5}\bigg|_2^3}$

$\mathsf{=\dfrac{5}{2}\cdot\left(\dfrac{3^3}{3}-\dfrac{2^3}{3}\right )+\dfrac{5}{6}\cdot \left(\dfrac{3^5}{5}-\dfrac{2^5}{5}\right)}\\\\\\ \mathsf{=\dfrac{5}{2}\cdot\left(\dfrac{27}{3}-\dfrac{8}{3}\right )+\dfrac{5}{6}\cdot \left(\dfrac{243}{5}-\dfrac{32}{5}\right)}\\\\\\ \mathsf{=\dfrac{5}{2}\cdot\dfrac{19}{3}+\dfrac{\diagup\!\!\!\! 5}{6}\cdot \dfrac{211}{\diagup\!\!\!\! 5}}\\\\\\ \mathsf{=\dfrac{95}{6}+\dfrac{211}{6}}\\\\\\ \mathsf{=\dfrac{306}{6}}\\\\\\ \mathsf{=51}\quad\longleftarrow\quad\textsf{esta \'e a resposta.}$


Bons estudos! :-)