Area between 2 curves and $X$-axis I have trouble doing this. I have to find the area of the region bounded by the parabola ($4X-X^2$), $y=X$ and the $X$-axis. I found the interception points but don't understand what the common area is between all of them. The answer is $\frac{37}{6}$
Thanks guys. 
 A: One easily computes that the intersection points of $f(X):=-X^2+4X$ and $X$-Axis are $(0,0)$ and $(4,0)$. In addition, the intersection point of $f(X)$ and $g(X):=X$ is $(3,0)$.
The integral that you want to compute is therefore: $$\int_{0}^{3}g(X) + \int_{3}^{4}f(X) = \Big[\frac{1}{2}X^2\Big]_0^3+\Big[\frac{1}{3}X^3+2X^2\Big]_3^4=\frac{37}{6}$$
A: It is helpful to visualize the intersection between the two curves and $x$-axis by a graphing program such as Wolfram Alpha or Desmos.

Labeling $f(x)=4x-x^2, g(x)=x, h(x)=0$, we see that $g(x) < f(x)$ when $0< x < 3$. So, for $0\le x \le 3$ the area of the region bounded by the curves and the $x$-axis is the triangle between the points $(0,0), (3,3),$ and $(3,0)$. The area of the triangle is 
$$\frac{1}{2}{(\text{base})}{(\text{height})}=\frac{1}{2}(3)(3)=\frac{9}{2}$$
Next, when $3<x\le 4$ we have that $f(x) < g(x)$. Therefore, the area of the region bounded by the curves and the $x$-axis is the integral between $f(x)$ and $h(x)$
$$\int_3^4 \big(f(x)-h(x)\big)dx=\int_3^4 \big(4x-x^2\big)dx=\left(2x^2-\frac{x^3}{3}\right)\bigg|_3^4=\frac{5}{3}$$
the total area is the sum of these two regions
$$\frac{9}{2}+\frac{5}{3}=\frac{27}{6}+\frac{10}{6}=\frac{37}{6}$$
