finding area of a curve

two curves have equations y=x^2-4x+7 and y=6x-x^2-1.

Find the area between the two curves

$\begin{array}{|rcll|} \hline f\left(x\right)=6x-x^2-1 \\ g\left(x\right)=x^2-4x+7 \\ \hline \end{array}$

intersections:

$\begin{array}{|rcll|} \hline g\left(x\right) &=& f\left(x\right) \\ x^2-4x+7 &=& 6x-x^2-1 \\ \dots \\ 2x^2-10x+8 &=& 0 \quad & | \quad : 2 \\ x^2-5x+4 &=& 0 \\ (x-1)(x-4) &=& 0 \\ x_1 = 1 &\text{and}& x_2 = 4\\ \hline \end{array}$

area between the two curves:

$\begin{array}{|rcll|} \hline && \int \limits_{1}^{4}(~ f(x) - g(x) ~)\ dx \\ &=& \int \limits_{1}^{4}(~ 6x-x^2-1 - ( x^2-4x+7 ) ~)\ dx \\ &=& \int \limits_{1}^{4}(~ 6x-x^2-1 - x^2+4x-7 ~)\ dx \\ && \dots \\ &=& \int \limits_{1}^{4}(~ 10x-2x^2-8 ~)\ dx \\ &=& [~ 10\frac{x^2}{2}-2\frac{x^3}{3}-8x ~]_1^4 \\ &=& [~ 5x^2 -\frac{2}{3}x^3-8x ~]_1^4 \\ &=& [~ 5\cdot (4)^2 -\frac{2}{3}(4)^3-8\cdot 4 ~]-[~ 5(1)^2 -\frac{2}{3}(1)^3-8\cdot 1 ~] \\ &=& 5\cdot 16 -\frac{2}{3}\cdot 64-32 -5 +\frac{2}{3}+8 \\ &=& 51 -\frac{2}{3}\cdot 64 +\frac{2}{3} \\ &=& 51-\frac{2}{3}(64 - 1) \\ &=& 51-\frac{2}{3}(63) \\ &=& 51-2\cdot 21\\ &=& 9 \\ \hline \end{array}$

The area between the two curves is 9