Evaluate integral 3x(X-2) from 0 to 2 and comment on your answer.
?? This is just a simple application of the power rule
∫3x(x-2) dx = 3∫x^2 - 2x dx
Surely that isn't too hard?
Perhaps the "comment on your answer" refers to the fact that
the answer turned out negative, and Raj did not realize that the region
lies below the x-axis.
Always make a sketch to understand what the question deals with.
To evaluate the integral ∫3x(X-2) dx from 0 to 2, we can use the properties of integrals and linearity.
Step 1: Expand the expression inside the integral:
∫3x(X-2) dx = ∫3x^2 - 6x dx
Step 2: Apply the power rule of integration:
∫3x^2 - 6x dx = x^3 - 3x^2 + C
Step 3: Evaluate the integral from 0 to 2:
Evaluate the antiderivative at the upper limit and subtract the value at the lower limit:
[2^3 - 3(2)^2] - [0^3 - 3(0)^2]
= [8 - 12] - [0 - 0]
= -4
Therefore, the value of the integral ∫3x(X-2) dx from 0 to 2 is -4.
Comment:
The negative value of the integral suggests that the total area under the curve is below the x-axis, indicating that the net area is negative. This means that the function f(x) = 3x(X-2) dips below the x-axis in the evaluated interval.