Calculate the area enclosed between the following two functions:
$$f(x)=-x^2+4x$$ and $$g(x)=x-4$$.
Development:
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 $$f(x)-g(x)$$:
$$\displaystyle \int_{-1}^4 (-x^2+4x-x+4) \ dx = \int_{-1}^4 (-x^2+3x+4) \ dx =$$
$$=\Big[-\dfrac{x^3}{3}+3\dfrac{x^2}{2}+4x\Big]^4_{-1}= -\dfrac{4^3}{3}+3\cdot\dfrac{4^2}{2}+4\cdot4-\Big(-\dfrac{(-1)^3}{3}+3\cdot\dfrac{(-1)^2}{2}+4(-1)\Big)=$$
$$=-\dfrac{64}{3}+24+16-\Big(+\dfrac{1}{3}+\dfrac{3}{2}-4\Big)=\dfrac{125}{6} \ u^2$$
Solution:
The area is $$\dfrac{125}{6} \ u^2$$