[tex]f(x)=x^3+14x^2+45x,\ g(x)=-x^3-14x^2-45x=-(x^3+14x^2+45x)=-f(x)[/tex]
[tex]x^3+14x^2+45x=-x^3-14x^2-45x\ \ \ \ |+x^3+14x^2+45x\\\\2x^3+28x^2+90x=0\\\\2x(x^2+14x+45)=0\iff2x=0\ \vee\ x^2+14x+45=0\\\\x=0\ \vee\ x^2+9x+5x+45=0\\\\x=0\ \vee\ x(x+9)+5(x+9)=0\\\\\x=0\ \vee\ (x+9)(x+5)=0\\\\x=0\ \vee\ x=-9\ \vee\ x=-5[/tex]
Because g (x) = - f (x), the fields of the areas between the curve f and the x-axis and the curve g and the x-axis are the same, it is enough to count twice the area concerning the curve f.
[tex]2\left(\int\limits_{-9}^{-5}(x^3+14x^2+45x)dx-\int\limits_{-5}^0(x^3+14x^2+45x)dx\right)=(*)\\\\\int(x^3+14x^2+45x)dx=\dfrac{1}{4}x^4+\dfrac{14}{3}x^3+\dfrac{45}{2}x^2\\\\\int\limits_{-9}^{-5}(x^3+14x^2+45x)dx=\left\dfrac{1}{4}x^4+\dfrac{14}{3}x^3+\dfrac{45}{2}x^2\right]_{-9}^{-5}\\\\=\left(\dfrac{1}{4}(-5)^4+\dfrac{14}{3}(-5)^3+\dfrac{45}{2}(-5)^2\right)-\left(\dfrac{1}{4}(-9)^4+\dfrac{14}{3}(-9)^3+\dfrac{45}{2}(-9)^2\right)\\\\=\dfrac{224}{3}[/tex]
[tex]\int\limits_{-5}^{0}(x^3+14x^2+45x)dx=\left\dfrac{1}{4}x^4+\dfrac{14}{3}x^3+\dfrac{45}{2}x^2\right]_{-5}^{0}\\\\=\left(\dfrac{1}{4}0^4+\dfrac{14}{3}0^3+\dfrac{45}{2}0^2\right)-\left(\dfrac{1}{4}(-5)^4+\dfrac{14}{3}(-5)^3+\dfrac{45}{2}(-5)^2\right)\\\\=-\dfrac{1625}{12}[/tex]
[tex]Area=2\cdot\left(\dfrac{224}{3}-\left(-\dfrac{1625}{12}\right)\right)=2\cdot\left(\dfrac{896}{12}+\dfrac{1625}{12}\right)\\\\=2\cdot\dfrac{2521}{12}=\dfrac{2521}{6}=420\dfrac{1}{6}\approx420[/tex]
