Question

Integral definida de ∫ (√x-x²) dx em [4, 1]

Answer 1

Olá ! 

Note que √x = x^(1/2) 

Primeiro integro e depois defino : 

$\int\limits^4_1 { (\sqrt{x} -x^{2})} \, dx \\\\\\ \int\limits^4_1 {(x^{ \frac{1}{2} }+x^{2})} \, dx \\\\\\Resolvendo...\\\\\\ \frac{x^{ \frac{1}{2} +1}}{ \frac{1}{2} +1} + \frac{x^{2+1}}{2+1} \\\\\\ \frac{x^{ \frac{1}{2} + \frac{2}{2} }}{ \frac{1}{2} + \frac{2}{2} } + \frac{x^{3}}{3} \\\\\\ \frac{x^{ \frac{3}{2} }}{ \frac{3}{2} } + \frac{x^{3}}{3}$
$\frac{x^{ \frac{3}{2} }}{1} . \frac{2}{3} + \frac{x^{3}}{3} \\\\\\ \frac{2x^{ \frac{3}{2} }}{3} + \frac{x^{3}}{3} \\\\\\ \frac{2x^{ \frac{3}{2} }+x^{3}}{3} =\boxed{ \frac{2 \sqrt[2]{x^{3}}+x^{3} }{3} }\\\\\\Agora\ limito...\\\\\\| \frac{2 \sqrt[2]{4^{3}}+4^{3} }{3} |-| \frac{2 \sqrt[2]{1^{3}}+1^{3} }{3} |\\\\\\| \frac{2 \sqrt{64} +64}{3} |-| \frac{2 \sqrt{1}+1 }{3} |$

$| \frac{2.8+64}{3} |-| \frac{1+1}{3} |\\\\\\| \frac{16+64}{3} |-| \frac{2}{3} |\\\\\\ \frac{80}{3} - \frac{2}{3} \\\\\\ \frac{80-2}{3} \\\\\\ \frac{78}{3} =\boxed{\boxed{\boxed{26}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ok$