Calcula el área que delimitan las dos parábolas $$f(x)=x^2+2x-1$$ y $$g(x)=-x^2+2x$$.
Desarrollo:
Buscamos primero los puntos de intersección:
$$$x^2+2x-1=-x^2+2x$$$ $$$2x^2-1=0$$$ $$$x^2=\dfrac{1}{2}$$$ $$$x=\dfrac{1}{\pm\sqrt{2}}$$$
Si además dibujamos las parábolas vemos que $$g(x)$$ está por encima de $$f(x)$$, con lo que el orden será $$g(x)-f(x)$$:
$$\displaystyle \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (-x^2+2x-x^2-2x+1) \ dx = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (-2x^2+1) \ dx =$$
$$=\Big[-2\dfrac{x^3}{3}+x\Big]^{\frac{1}{\sqrt{2}}}_{\frac{-1}{\sqrt{2}}}= -2\cdot\dfrac{(\frac{1}{\sqrt{2}})^3}{3}+\dfrac{1}{\sqrt{2}}-\Big(-2\cdot\dfrac{(\frac{-1}{\sqrt{2}})^3}{3}-\dfrac{1}{\sqrt{2}}\Big)=$$
$$=-2\cdot\dfrac{\frac{1}{2\sqrt{2}}}{3}+\dfrac{1}{\sqrt{2}}-\Big(-2\cdot\dfrac{\frac{-1}{2\sqrt{2}}}{3}-\dfrac{1}{\sqrt{2}}\Big)=$$
$$=\dfrac{-2}{3\cdot2\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\Big(\dfrac{2}{3\cdot2\sqrt{2}}-\dfrac{1}{\sqrt{2}}\Big)=$$
$$=\dfrac{-1}{3\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{2}}=$$
$$=-\dfrac{2}{3\sqrt{2}}+\dfrac{2}{\sqrt{2}}=\dfrac{-2+6}{3\sqrt{2}}=\dfrac{4}{3\sqrt{2}} \ u^2$$
Solución:
El área es de $$\dfrac{4}{3\sqrt{2}} \ u^2$$