<s>[INST] Find centroid of region of two curves Find the coordinates (to three decimal places) of the centroid of $y=2^x$, and $y=x^2$.
EDIT: $(0\le x\le2)$
I understand this with a triangle, not with curves.
 [/INST] The computation of the centroid in $R^2$, of a region bounded by two continuous functions, goes, by definition, as follows. (Note that, over $[0,2]$,  $x^2 \le 2^x$.) 
First, one has to calculate the area, $\mathscr a$, of the region $$A=\{x,y\ ;\ 0\le x\le 2, \ x^2\le y\le 2^x\}.$$ 
  $ \color{white}{bbbbbbbbbbb}$ 
$$\text{The region with the centroid to be calculated below.}$$
$$\color{white}{nnn}$$
We have for the area:
$$\mathscr a= \iint_Adydx=\int_0^2\ \left[\int_{x^2}^{2^x}dy\right]\ dx=\int_0^2 2^xdx-\int_0^2x^2dx.$$
Then, for the coordinates of the centroid:
$$\overline x=\frac{1}{a}\int_0^2 x(2^x-x^2)\ dx=\frac{1}{\mathscr a}\left(\int_0^2x2^xdx-\int_0^2x^3dx\right),$$ 
$$\overline y=\frac{1}{2\mathscr a}\int_0^2 (2^x+x^2)(2^x-x^2)\ dx=\frac{1}{2\mathscr a}\left(\int_0^22^{2x}dx-\int_0^2x^4dx\right).$$ 
Now, we calculate the following integrals 
$$\int_0^2 2^xdx=\frac{3}{\ln2},$$
$$\int_0^2x^2dx=\frac{8}{3},$$
$$\int_0^2 x2^x\ dx=\frac{8}{\ln2}-\frac{3}{(\ln2)^2},$$
$$\int_0^2x^3\ dx=4.$$
$$\int_0^22^{2x}\ dx=\frac{15}{2\ln2},$$
$$\int_0^2x^4\ dx=\frac{32}{5}.$$
Having done all this, one can easily put together the pieces:
$$a=\frac{3}{ln2}-\frac{8}{3}=1.6614,$$
$$\overline x=\frac{1}{\mathscr a}\left(\frac{8}{\ln2}-\frac{3}{(\ln2)^2}-4\right)=0.7809,$$
$$\overline y=\frac{1}{2\mathscr a}\left(\frac{15}{2\ln2}-\frac{32}{5}\right)=1.3303.$$
</s>