Write the two curves

The parabolas are: $y_1=x^2-5x$ $y_2=7x-x^2$

Find the points of intersection

For intersection: $x^2-5x=7x-x^2$

Bring all terms to one side: $2x^2-12x=0$ $2x(x-6)=0$

Hence, $x=0 \quad \text{or} \quad x=6$

Determine which curve is above the other

Take a point between $0$ and $6$, say $x=3$.

$y_1(3)=3^2-5(3)=9-15=-6$ $y_2(3)=7(3)-3^2=21-9=12$

So on $[0,6]$, the upper curve is: $y=7x-x^2$ and the lower curve is: $y=x^2-5x$

Set up the area integral

Area enclosed is: $A=\int_0^6 \Big[(7x-x^2)-(x^2-5x)\Big]dx$

Simplify the integrand: $A=\int_0^6 (12x-2x^2)\,dx$

Evaluate the integral

$A=\left[6x^2-\frac{2x^3}{3}\right]_0^6$

At $x=6$: $6(6^2)-\frac{2(6^3)}{3}=6(36)-\frac{2(216)}{3}=216-144=72$

At $x=0$: $0$

Therefore, $A=72$

Compare with stored answer

My derived answer is $72$, but the stored correct answer is $198$.

Hence, the stored answer does not match the correct computation.