Which of the following definite integrals could be used to calculate the total area bounded by the graph of y = x^3 – 3x^2 + 2x and the x-axis? one half times the integral from 0 to 2 of the quantity x cubed minus 3 times x squared plus 2 times x, dx 2 times the integral from 0 to 1 of the quantity x cubed minus 3 times x squared plus 2 times x, dx the integral from 0 to 1 of the quantity squared of x cubed minus 3 times x squared plus 2 times x, dx the integral from 0 to 2 of the quantity x cubed minus 3 times x squared plus 2 times x, dx

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,

It can be written as,

Find the zeros of the equation. Equation the function equal to 0.

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,

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.

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".

answer: the equation matches the regression

step-by-step explanation:

1: -2 3/4, -2 1/2, -14, 0, 3, 8 1/3

2: -4, -3, -2 1/2, -2 1/4, 3 1/4,

step-by-step explanation: