Answer to Question #345114 in Calculus for Jane

Question

Find the area, take the elements of the area parallel to the x-axis. y= 2x³-3x³-9x; y=x²-2x²-3x.

Answer 1

$$x^3-2x^2-3x=2x^3-3x^2-9x$$ $$x^3-x^2-6x=0$$ $$x(x^2-x-6)=0$$ $$x(x+2)(x-3)=0$$ $$x_1=-2, x_2=0, x_3=3$$ $$A_1=\displaystyle\int_{-2}^{0}(2x^3-3x^2-9x-(x^3-2x^2-3x))dx$$ $$=\displaystyle\int_{-2}^{0}(x^3-x^2-6x)dx$$ $$=[\dfrac{x^4}{4}-\dfrac{x^3}{3}-3x^2]\begin{matrix} 0 \\ -2 \end{matrix}$$

$$=0-(\dfrac{(-2)^4}{4}-\dfrac{(-2)^3}{3}-3(-2)^2)$$

$$=\dfrac{16}{3}({units}^2)$$ $$A_2=\displaystyle\int_{0}^{3}(x^3-2x^2-3x-(2x^3-3x^2-9x))dx$$ $$=\displaystyle\int_{0}^{3}(-x^3+x^2+6x)dx$$ $$=[-\dfrac{x^4}{4}+\dfrac{x^3}{3}+3x^2]\begin{matrix} 3 \\ 0 \end{matrix}$$

$$=-\dfrac{(3)^4}{4}+\dfrac{(3)^3}{3}+3(3)^2-0$$

$$=\dfrac{63}{4}({units}^2)$$