Areas delimited by two functions

Let's first look for the points of intersection:

$$x^2+2x-1=-x^2+2x$$ $$2x^2-1=0$$ $$x^2=\dfrac{1}{2}$$ $$x=\dfrac{1}{\pm\sqrt{2}}$$

If we draw the parables we see that $g(x)$ is above $f(x)$, so the order has to be $g(x)-f(x)$:

$$\displaystyle \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (-x^2+2x-x^2-2x+1) \ dx = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (-2x^2+1) \ dx =$$

$$=\Big[-2\dfrac{x^3}{3}+x\Big]^{\frac{1}{\sqrt{2}}}_{\frac{-1}{\sqrt{2}}}= -2\cdot\dfrac{(\frac{1}{\sqrt{2}})^3}{3}+\dfrac{1}{\sqrt{2}}-\Big(-2\cdot\dfrac{(\frac{-1}{\sqrt{2}})^3}{3}-\dfrac{1}{\sqrt{2}}\Big)=$$

$$=-2\cdot\dfrac{\frac{1}{2\sqrt{2}}}{3}+\dfrac{1}{\sqrt{2}}-\Big(-2\cdot\dfrac{\frac{-1}{2\sqrt{2}}}{3}-\dfrac{1}{\sqrt{2}}\Big)=$$

$$=\dfrac{-2}{3\cdot2\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\Big(\dfrac{2}{3\cdot2\sqrt{2}}-\dfrac{1}{\sqrt{2}}\Big)=$$

$$=\dfrac{-1}{3\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{2}}=$$

$$=-\dfrac{2}{3\sqrt{2}}+\dfrac{2}{\sqrt{2}}=\dfrac{-2+6}{3\sqrt{2}}=\dfrac{4}{3\sqrt{2}} \ u^2$$

Let's first look for the points of intersection:

$$-x^2+4x=x-4$$ $$-x^2+3x+4=0$$ $$x=\dfrac{-3\pm\sqrt{9-4(-1)(4)}}{2(-1)}=\dfrac{-3\pm\sqrt{25}}{-2}=\dfrac{-3\pm5}{-2}$$ $$x=-1 \ \text{and} \ x=4$$

If we draw the functions we see that $f(x)$ is above $g(x)$, so the order has to be $f(x)-g(x)$:

$$\displaystyle \int_{-1}^4 (-x^2+4x-x+4) \ dx = \int_{-1}^4 (-x^2+3x+4) \ dx =$$

$$=\Big[-\dfrac{x^3}{3}+3\dfrac{x^2}{2}+4x\Big]^4_{-1}= -\dfrac{4^3}{3}+3\cdot\dfrac{4^2}{2}+4\cdot4-\Big(-\dfrac{(-1)^3}{3}+3\cdot\dfrac{(-1)^2}{2}+4(-1)\Big)=$$

$$=-\dfrac{64}{3}+24+16-\Big(+\dfrac{1}{3}+\dfrac{3}{2}-4\Big)=\dfrac{125}{6} \ u^2$$