Question

Encontre a área da região limitada pelas curvas y= X ao quadrado + 1 y = x+ 1
1/9 u.a
1 u.a
1/6 u.a
2/9 u.a
2 u.a

Answer 1

$\mathrm{f(x)=x^2+1\ \ \|\ \ g(x)=x+1}\\\\ \textbf{Abscissas de intersec\c{c}\~ao:}\\\\ \mathrm{f(x)=g(x)\ \to\ x^2+1=x+1\ \to\ x^2-x=0}\\ \mathrm{x(x-1)=0\ \ \|\ \ x=0\ \ \|\ \ x-1=0\ \to\ x=1}$

$\textbf{\'Area da regi\~ao limitada pelas curvas:}\\\\ \mathrm{\int\limits_a^bf(x)-g(x)\ dx=\int\limits_0^1x^2+1-(x+1)\ dx=}\\\\\\ \mathrm{=\int\limits_0^1x^2+1-x-1\ dx=\int\limits_0^1x^2-x\ dx=}\\\\\\ \mathrm{=\bigg(\int x^2\ dx-\int x\ dx\bigg)\bigg|_0^1=\bigg(\dfrac{x^3}{3}-\dfrac{x^2}{2}\bigg)\bigg|_0^1=}\\\\\\ \mathrm{=\bigg(\dfrac{1^3}{3}-\dfrac{1^2}{2}\bigg)-\bigg(\dfrac{0^3}{3}-\dfrac{0^2}{2}\bigg)=\dfrac{2-3}{6}=-\dfrac{1}{6}}$

$\mathrm{S=\bigg|-\dfrac{1}{6}\ \bigg|\ \to\ \boxed{\mathbf{S=\dfrac{1}{6}\ u.a.}}}$