Answer: The correct option is second, i.e. ,"2 times the integral from 0 to 1 of the quantity x cubed minus 3 times x squared plus 2 times x, dx".
Explanation:
The given equation is,
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It can be written as,
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Find the zeros of the equation. Equation the function equal to 0.
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
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
So, the three zeros are 0, 1 and 2.
The graph of the equation is shown below.
From the given graph it is noticed that the enclosed by the curve and x- axis is lies between 0 to 2, but the area from 0 to 1 lies above the x-axis and area from 1 to 2 lies below the x-axis. So the function will be negative from 1 to 2.
The area enclosed by curve and x-axis is,
![A=\int_{0}^{1}f(x)dx+\int_{1}^{2}[-f(x)]dx](https://tex.z-dn.net/?f=A%3D%5Cint_%7B0%7D%5E%7B1%7Df%28x%29dx%2B%5Cint_%7B1%7D%5E%7B2%7D%5B-f%28x%29%5Ddx)
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From the graph it is noticed that the area from 0 to 1 is symmetric or same as area from 1 to 2. So the total area is the twice of area from 0 to 1.

![A=2\int_{0}^{1}[x^3-3x^2+2x]dx](https://tex.z-dn.net/?f=A%3D2%5Cint_%7B0%7D%5E%7B1%7D%5Bx%5E3-3x%5E2%2B2x%5Ddx)
Therefore, The correct option is "2 times the integral from 0 to 1 of the quantity x cubed minus 3 times x squared plus 2 times x, dx".