<s>[INST] Find the area between the functions I have a homework where I have to find the area between the functions $y=x-2$ and $y=\sqrt{7-2x}$. 
Computing the integral I have found $A=\frac {x^2}2-2x-\frac13 (7-2x)^{3/2}$; The upper limit is $\frac72$ and the downlimit is $2$. 
After make the computing I have found $A= -0.607051$. 
I think there is something wrong with the answer. 
 [/INST] First, you need to draw a picture:

The two curves cross, so you will need to divide this into two separate integrals. They cross when $x-2=\sqrt{7-2x}$, so that $(x-2)^2 = x^2-4x+4 = 7-2x$. Solving gives $x=3$ (and $x=-1$, which you don't care about here). The integrals of the two functions are
$$\int (x-2)\,dx = \frac{x^2}{2} - 2x + C,\qquad
  \int \sqrt{7-2x}\,dx = -\frac{1}{3}(7-2x)^{3/2} + C$$
(note that the integral in your post above is incorrect). Therefore the enclosed area is
$$
\int_2^3 (\sqrt{7-2x}-(x-2))\,dx + \int_3^{7/2} ((x-2)-\sqrt{7-2x})\,dx,$$
which evaluates to $-\frac{13}{24} + \sqrt{3}\approx 1.19038$.
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