<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] So the area is $$\begin{align}A&=\int_{1}^2\big[(4x-x^2-1)-(x+1)\big]dx\\
&=\int_{1}^2(-x^2+3x-2)dx\\
&=\Big[\frac{-x^3}{3}+\frac{3x^2}{2}-2x\Big]_{1}^{2}\\
&=\left(-\frac{8}{3}+6-4\right)-\left(-\frac{1}{3}+\frac{3}{2}-2\right)\\
&=-\frac{8}{3}+2+\frac{1}{3}-\frac{3}{2}+2\\
&=-\frac{7}{3}+4-\frac{3}{2}\\
&=-\frac{7}{3}+\frac{5}{2}\\
&=\frac{1}{6}\\
\end{align}$$
Edit: I got down voted. Well, there is nothing I can do. My only point is to show to the OP that his integral $$\int_{1}^2(3x-x^2)dx$$ is wrong. Maybe he thought that $1$ cancels out. So, I showed how he should have the correct computations.
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