Question

Calcule a área entre as curvas y=2x^2-2 e y=x+1

Answer 1

$\boxed{\begin{array}{l}\underline{\rm pontos~de~intersecc_{\!\!,}\tilde ao\!:}\\\sf 2x^2-2=x+1\\\sf 2x^2-x-2-1=0\\\sf 2x^2-x-3=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-1)^2-4\cdot2\cdot(-3)\\\sf\Delta=1+24\\\sf\Delta=25\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf x=\dfrac{-(-1)\pm\sqrt{25}}{2\cdot2}\\\sf x=\dfrac{1\pm5}{4}\begin{cases}\sf x_1=\dfrac{1+5}{4}=\dfrac{6\div2}{4\div2}=\dfrac{3}{2}\\\sf x_2=\dfrac{1-5}{4}=-\dfrac{4}{4}=-1\end{cases}\end{array}}$

$\boxed{\begin{array}{l}\sf se~x=\dfrac{3}{2}:\\\sf y=\dfrac{3}{2}+1=\dfrac{5}{2}\\\sf se~x=-1:\\\sf y=-1+1=0\\\sf A\bigg(\dfrac{3}{2},\dfrac{5}{2}\bigg)~~B(-1,0)\end{array}}$

$\boxed{\begin{array}{l}\sf a~\acute area~da~regi\tilde ao~\acute e~dada~por\\\displaystyle\sf A=\int_{-1}^{\frac{3}{2}}(x+1-[2x^2-2])~dx\\\displaystyle\sf A=\int_{-1}^{\frac{3}{2}}(x+1-2x^2+2)~dx\\\displaystyle\sf A=\int_{-1}^{\frac{3}{2}}(-2x^2+x+3)~dx\\\\\sf A=\bigg[-\dfrac{2}{3}x^3+\dfrac{1}{2}x^2+3x\bigg]_{-1}^{\frac{3}{2}}\end{array}}$

$\small\boxed{\begin{array}{l}\sf A=-\dfrac{2}{3}\cdot\bigg(\dfrac{3}{2}\bigg)^3+\dfrac{1}{2}\cdot\bigg(\dfrac{3}{2}\bigg)^2+3\cdot\dfrac{3}{2}-\bigg[-\dfrac{2}{3}\cdot(-1)^3+\dfrac{1}{2}\cdot(-1)^2+3\cdot(-1)\bigg]\\\sf A=-\dfrac{2}{3}\cdot\dfrac{27}{8}+\dfrac{1}{2}\cdot\dfrac{9}{4}+\dfrac{9}{2}-\dfrac{2}{3}-\dfrac{1}{2}+3\\\\\sf A=-\dfrac{54}{24}+\dfrac{9}{8}+\dfrac{9}{2}-\dfrac{2}{3}-\dfrac{1}{2}+3\\\\\sf A=\dfrac{-54+27+108-16-12+72}{24}\\\\\sf A=\dfrac{125}{24}~u\cdot a\end{array}}$