Question

Dois exercícios de Integral em anexo abaixo:

Answer 1

Explicação passo-a-passo:

Integral definida :

Para fazer o cálculo d'uma integral definida vamos obedecer a seguinte designação :

${\displaystyle \int^{B}_{A}\mathsf{f(x) , dx}} ~=~\mathsf{F(x)\Bigg|^{B}_{A}=F(B)-F(A)}$

$\mathsf{I}~=~ {\displaystyle \int^{90\°}_{0}\mathsf{-\sin(x) , dx}} </p><p>$

$\mathsf{I~=~\cos(x) \Bigg|^{90°}_{0°}}$

$\mathsf{I~=~\cos(90°)-\cos(0°) }$

$\mathsf{I~=~0-1~=~-1 }$

e)

${\displaystyle \int^{2}_{1} \sqrt{x} dx }$

$\mathsf{I}~=~ {\displaystyle \int^{2}_{1} \mathsf{x^{\frac{1}{2}}} dx }$

$\mathsf{I~=~\dfrac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \Bigg|^{2}_{1} }$

$\mathsf{I~=~\dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} \Bigg|^{2}_{1} }$

$\mathsf{I~=~x^{\frac{3}{2}}.\dfrac{2}{3} \Bigg|^{2}_{1} }$

$\mathsf{I~=~\dfrac{2\sqrt{x^3}}{3} \Bigg|^{2}_{1} }$

$\mathsf{I~=~\dfrac{2\sqrt{2^3}}{3}-\dfrac{2\sqrt{1^3}}{3} }$

$\mathsf{I~=~\dfrac{4\sqrt{2}}{3}-\dfrac{2}{3} }$

$\mathsf{I~=~\dfrac{4\sqrt{2}-2}{3} }$

Espero ter ajudado bastante!)

Answer 2

• A ↔

$\sf \displaystyle {\int _0^{90^{\circ \:}}-sin \left(x\right)dx}\\\\\\{Remova\:a\:constante}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\\\\\=-\int _0^{90^{\circ \:}}\sin \left(x\right)dx\\\\\\=-\left[-cos \left(x\right)\right]^{90^{\circ \:}}_0\\\\\\\to \boxed{\sf =-1}$

• B ↔

$\sf \displaystyle \int _1^2\sqrt{x}dx\\\\\\=\int _1^2x^{\frac{1}{2}}dx\\\\\\{Aplicar\:a\:regra\:da\:pot\hat{e}ncia}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\\\=\left[\dfrac{x^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}\right]^2_1\\\\\\=\left[\frac{2}{3}x^{\frac{3}{2}}\right]^2_1\\\\\\\to \boxed{\sf =\frac{4\sqrt{2}}{3}-\frac{2}{3} \approx1,21895...}$