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 $x=2$

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

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

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

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

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

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

 A: You got the two $x$ points right, but you make a mistake in the next step.  The correct version should be:
$$\int_1^2[\text{top function}-\text{bottom function}]\ dx$$
Here, the top function is $4x-x^2-1$ and the bottom function is $x+1$, hence
$$\begin{align}\text{Area}&=\int_1^2[(4x-x^2-1)-(x+1)]\ dx\\&=\int_1^2[4x-x^2-1-x-1]\ dx\\&=\int_1^2[3x-x^2-2]\ dx\end{align}$$
Can you take it from here?
A: 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.
A: Hint: You need to be doing top curve minus bottom curve i.e;
$$(4x-x^2-1) - (x+1) = 3x-x^2-2$$
A: 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
$$\begin{align} & \int_1^2 \{(4x-x^2-1)-(x+1)\} \mathrm{dx} \\ & 
=\int_1^2 (3x-x^2-2) \mathrm{dx} \\ & 
=\left[\frac{3x^2}{2}-\frac{x^3}{3}-2x\right]_1^2 \\ & 
=\frac{3(2^2-1)}{2}-\frac{2^3-1}{3}-2(2-1) \\ & 
=\frac{9}{2}-\frac{7}{3}-2 \\ & 
=\frac{27-14-12}{6} \\ & 
=\color{blue}{\frac{1}{6}}\end{align}$$
A: Step 5. onwards is not ok
$$\int_a^bf(x)  dx = F(b)-F(a) $$
$$\int_a^bf(x)  dx = F(b-a) $$
The first result is correct but second one is in general wrong.
EDIT1:
The constant term in $ (x^2-3 x +2) $ is missing for integration. Factorization is not beneficial,
