Assume that a force of (3x^2 - x + 1)N is pulling an object at each point of the x-axis. find the work done in moving the object from x = 1 to x = 4
this is my answer
ans = 125/2
pls do you think am right?
I don't get that, near, but not that.
The answer is the integral of F dx from x=1 to 4.
That would equal the value of the indefinite integral x^3 -x^2/2 + x at x=4 MINUS the value at x+1.
[64 -8 + 4] - [1 - 1/2 + 1]
= 60 - (3/2) = 117/2
Check both our calculations; I am getting sloppy and have posted several wrong answers recently.
To find the work done in moving the object from x = 1 to x = 4, you need to calculate the definite integral of the force function over the given interval.
The work done by a force is given by the formula:
Work = ∫[x=a to x=b] F(x) dx
In this case, the force function is given by F(x) = 3x^2 - x + 1, and you are finding the work done from x = 1 to x = 4.
So, the integral becomes:
Work = ∫[x=1 to x=4] (3x^2 - x + 1) dx
To evaluate this integral, you need to find the antiderivative of the force function, which will give you the work function. Then you can plug in the upper and lower limits of integration to find the work done.
The antiderivative of 3x^2 - x + 1 is x^3 - (1/2)x^2 + x.
Now you can evaluate the work function at the upper and lower limits of integration:
Work = [x^3 - (1/2)x^2 + x] evaluated from x = 1 to x = 4
Plugging in the values, you get:
Work = [4^3 - (1/2)4^2 + 4] - [1^3 - (1/2)1^2 + 1]
= [64 - 8 + 4] - [1 - (1/2) + 1]
= 60 - (2/2) - 1
= 59 - 1
= 58
Therefore, the correct answer is 58, not 125/2.