Question

Find the area between the curves: y=x3−14x2+45x and y=−x3+14x2−45x

Answer 1

`f(x)=x^3+14x^2+45x,\ g(x)=-x^3-14x^2-45x=-(x^3+14x^2+45x)=-f(x)`

`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`

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.

`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=(1)/(4)x^4+(14)/(3)x^3+(45)/(2)x^2\\\\\int\limits_(-9)^(-5)(x^3+14x^2+45x)dx=\left(1)/(4)x^4+(14)/(3)x^3+(45)/(2)x^2\right]_(-9)^(-5)\\\\=\left((1)/(4)(-5)^4+(14)/(3)(-5)^3+(45)/(2)(-5)^2\right)-\left((1)/(4)(-9)^4+(14)/(3)(-9)^3+(45)/(2)(-9)^2\right)\\\\=(224)/(3)`

`\int\limits_(-5)^(0)(x^3+14x^2+45x)dx=\left(1)/(4)x^4+(14)/(3)x^3+(45)/(2)x^2\right]_(-5)^(0)\\\\=\left((1)/(4)0^4+(14)/(3)0^3+(45)/(2)0^2\right)-\left((1)/(4)(-5)^4+(14)/(3)(-5)^3+(45)/(2)(-5)^2\right)\\\\=-(1625)/(12)`

`Area=2\cdot\left((224)/(3)-\left(-(1625)/(12)\right)\right)=2\cdot\left((896)/(12)+(1625)/(12)\right)\\\\=2\cdot(2521)/(12)=(2521)/(6)=420(1)/(6)\approx420`