Question

Se alguém puder responder por favor, assunto = DERIVADA
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Answer 1

$\mathrm{Seja\ uma\ curva\ dada\ por}\ y=x\sqrt{1-x}.$

$\textrm{Pede-se para calcular a}\ \mathrm{\acute{a}rea\ da\ regi\tilde{a}o\ delimitada\ por}\ y,\\ \mathrm{o\ eixo}\ x\ \mathrm{e\ as\ retas}\ x=0\ \text{e}\ x=1.$

$\mathrm{Nesse\ caso,\ calcularemos\ a\ integral\ de finida}\ S=\int_{x_1}^{x_2}{y\ \mathrm{d}x}\\ \mathrm{na\ qual\ os\ limites\ de\ integra\c{c}\tilde{a}o\ s\tilde{a}o}\ x_1=0\ \text{e}\ x_2=1.$

$\boxed{S=\int\limits_{x_1}^{x_2}{y\ \mathrm{d}x}}\Longrightarrow S=\int\limits_0^1{x\sqrt{1-x}\ \mathrm{d}x}$

$\mathrm{Substitui\c{c}\tilde{a}o\ de\ vari\acute{a}veis}\Longrightarrow u=1-x\Longrightarrow x=1-u$

$\Longrightarrow \dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{\mathrm{d}}{\mathrm{d}x}(1-x)=-1\Longrightarrow \mathrm{d}x=-\mathrm{d}u$

$\mathrm{Al\acute{e}m\ disso,\ os\ novos\ limites\ de\ integra\c{c}\tilde{a}o\ ser\tilde{a}o}\text{:}$

$u_1=1-x_1=1-0\Longrightarrow u_1=1$

$u_2=1-x_2=1-1\Longrightarrow u_2=0$

$\textrm{Desse modo:}$

$S=\int\limits_1^0{(1-u)\sqrt{u}\ (-\mathrm{d}u})=\int\limits_1^0{\left(u^{\frac{3}{2}}-u^{\frac{1}{2}}\right) \mathrm{d}u}$

$\Longrightarrow S=\left(\int{u^{\frac{3}{2}}\mathrm{d}u}-\int{u^{\frac{1}{2}}\mathrm{d}u}\right)\bigg|_1^0=\left(\dfrac{2}{5}u^{\frac{5}{2}}-\dfrac{2}{3}u^{\frac{3}{2}}\right)\bigg|_1^0$

$\Longrightarrow S=0-\left(\dfrac{2}{5}-\dfrac{2}{3}\right)=-\bigg(\dfrac{6-10}{15}\bigg)\ \therefore\ \boxed{S=\dfrac{4}{15}\ \mathrm{u.a.}}$