<s>[INST] Finding the area between a line and a curve 
The two equations are $x+1$ and $4x-x^2-1$.
The answer is $\frac{1}{6}$, but I've done it 4 different times and gotten -$\frac{15}{2}$ each time.
My working:


*

*$x+1$ = $4x-x^2-1$

*$x^2-3x+2 = 0$

*$(x-1)(x-2)$ means $x=1$ or $x=2$

*$\int_1^2$ $3x-x^2$

*$[\frac{3x^2}{2}-\frac{x^3}{3}]_1^2$

*$\frac{3(1)^2}{2}-\frac{(1)^3}{3}$ = $\frac{3}{2}-\frac{1}{3}$

*$\frac{3(2)^2}{2}-\frac{(2)^3}{3}$ = $\frac{12}{2}-\frac{8}{3}$

*($\frac{3}{2}-\frac{1}{3}$)-($\frac{12}{2}-\frac{8}{3}$) = $-\frac{9}{2}-\frac{9}{3}$

*-$\frac{15}{2}$

 [/INST] Hint: You need to be doing top curve minus bottom curve i.e;
$$(4x-x^2-1) - (x+1) = 3x-x^2-2$$
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