Answer
$\dfrac{7}{3}$
Work Step by Step
Here, we have: $4x=x \sqrt {25-x^2} \implies x^2(9-x^2)=0$
Since, the intersection points lie on the intervals or points $(0,0)$, $(-3,-12)$ and $(3, 12)$ and the area is in the first quadrant. So, we have the required intervals $[0, 3]$ because $f(x) \geq g(x)$
The area between the curves can be computed as:
$A=\int_m^n [f(x)-g(x)] \ dx\\=|\int_0^3 [x \sqrt {25-x^2}-4x] \ dx|\\=|-\dfrac{1}{3} (25-x^2)^{3/2}-2x^2|_0^3\\=\dfrac{7}{3}$
