Question

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

Answer 1

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,

$y=x^3-3x^2+2x$

It can be written as,

$f(x)=x^3-3x^2+2x$

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

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

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

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

$x(x-2)(x-1)=0$

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,

$A=\int_{0}^{1}f(x)dx+\int_{1}^{2}[-f(x)]dx$

$A=\int_{0}^{1}f(x)dx-\int_{1}^{2}f(x)dx$

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.

$A=2\int_{0}^{1}f(x)dx$

$A=2\int_{0}^{1}[x^3-3x^2+2x]dx$

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".