Question #345168

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

1
Expert's answer
2022-05-31T08:52:47-0400
x32x23x=2x33x29xx^3-2x^2-3x=2x^3-3x^2-9x

x3x26x=0x^3-x^2-6x=0

x(x2x6)=0x(x^2-x-6)=0

x(x+2)(x3)=0x(x+2)(x-3)=0

x1=2,x2=0,x3=3x_1=-2, x_2=0, x_3=3

A1=20(2x33x29x(x32x23x))dxA_1=\displaystyle\int_{-2}^{0}(2x^3-3x^2-9x-(x^3-2x^2-3x))dx=20(x3x26x)dx=\displaystyle\int_{-2}^{0}(x^3-x^2-6x)dx


=[x44x333x2]02=[\dfrac{x^4}{4}-\dfrac{x^3}{3}-3x^2]\begin{matrix} 0 \\ -2 \end{matrix}




=0((2)44(2)333(2)2)=0-(\dfrac{(-2)^4}{4}-\dfrac{(-2)^3}{3}-3(-2)^2)




=163(units2)=\dfrac{16}{3}({units}^2)A2=03(x32x23x(2x33x29x))dxA_2=\displaystyle\int_{0}^{3}(x^3-2x^2-3x-(2x^3-3x^2-9x))dx


=03(x3+x2+6x)dx=\displaystyle\int_{0}^{3}(-x^3+x^2+6x)dx


=[x44+x33+3x2]30=[-\dfrac{x^4}{4}+\dfrac{x^3}{3}+3x^2]\begin{matrix} 3 \\ 0 \end{matrix}




=(3)44+(3)33+3(3)20=-\dfrac{(3)^4}{4}+\dfrac{(3)^3}{3}+3(3)^2-0




=634(units2)=\dfrac{63}{4}({units}^2)




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