Respostas

2014-06-09T01:17:01-03:00
 \boxed{\int\limits^4_{-4} {( \sqrt{25-x^2})^2-(3)^2 } \, dx }

simplificando antes de integrar

\sqrt{25-x^2})^2-(3)^2\\\\\not\sqrt{25-x^2})^\not2-9\\\\25-x^2-9\\\\16-x^2

agora temos
 \int\limits^4_{-4} {16-x^2} \, dx =|16x- \frac{x^3}{3} |

fazendo o calculo
(16(4)- \frac{4^3}{3})-(16(-4)- \frac{(-4)^3}{3}) = \frac{256}{3}

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mas se for

 \int\limits^4_{-4} {( \sqrt{25}-x^2)^2-3^2 } \, dx

simplificando primeiro se resolve esse produto notavel

\boxed{(a-b)^2=a^2-2*a*b +b^2}

 (\sqrt{25}-x^2)^2-3^2 \\\\(5-x^2)^2 -9\\\\(5^2-2*5*x^2+(x^2)^2)-9\\\\25-10x^2+x^4-9\\\\16-10x^2+x^4

 \int\limits^4_{-4} {16-10x^2+x^4} \, dx =|16x- \frac{10x^3}{3} + \frac{x^5}{5} |

fazendo o calculo

(16(4)- \frac{10(4)^3}{3} + \frac{(4)^5}{5} )-(16(-4)- \frac{10(-4)^3}{3} + \frac{(-4)^5}{5} )= \frac{1664}{15}

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