<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] There is a mistake in the line below:


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

Actually you should be integrating the difference of the $2$ curves within that limit, i.e. $(4x-x^2-1)-(x+1)=3x-x^2-2$, since each of them represents the area under the curve and bounded by the x-axis.
The magnitude of the enclosed area will, accordingly, be
$$\begin{align} & \int_1^2 \{(4x-x^2-1)-(x+1)\} \mathrm{dx} \\ & 
=\int_1^2 (3x-x^2-2) \mathrm{dx} \\ & 
=\left[\frac{3x^2}{2}-\frac{x^3}{3}-2x\right]_1^2 \\ & 
=\frac{3(2^2-1)}{2}-\frac{2^3-1}{3}-2(2-1) \\ & 
=\frac{9}{2}-\frac{7}{3}-2 \\ & 
=\frac{27-14-12}{6} \\ & 
=\color{blue}{\frac{1}{6}}\end{align}$$
</s>