Answer to Question #345168 in Calculus for Jane

Question

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

Answer 1

$$x^3-2x^2-3x=2x^3-3x^2-9x$$

$$x^3-x^2-6x=0$$

$$x(x^2-x-6)=0$$

$$x(x+2)(x-3)=0$$

$$x_1=-2, x_2=0, x_3=3$$

$$A_1=\displaystyle\int_{-2}^{0}(2x^3-3x^2-9x-(x^3-2x^2-3x))dx$$ $$=\displaystyle\int_{-2}^{0}(x^3-x^2-6x)dx$$

$$=[\dfrac{x^4}{4}-\dfrac{x^3}{3}-3x^2]\begin{matrix} 0 \\ -2 \end{matrix}$$

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

$$=\dfrac{16}{3}({units}^2)$$ $$A_2=\displaystyle\int_{0}^{3}(x^3-2x^2-3x-(2x^3-3x^2-9x))dx$$

$$=\displaystyle\int_{0}^{3}(-x^3+x^2+6x)dx$$

$$=[-\dfrac{x^4}{4}+\dfrac{x^3}{3}+3x^2]\begin{matrix} 3 \\ 0 \end{matrix}$$

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

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