How much work does this force do as the particle moves along the x-axis from x = 0 to x = 1? If F = 7 - 2x + 3x²

Solution:

Given: Function f = 7 - 2x + 3x² and the limits a = 0 and b = 1

Work Done by a force is given by w = \(\int_{x=a}^{b}f(x)dx\)

w = \(\int_{x=0}^{1}(7 - 2x + 3x^{2}). dx\)

w = \(\int_{x=0}^{1}7dx - 2\int_{x=0}^{1}xdx+3\int_{x=0}^{1}x^{2}dx\)

\(\int xdx=\frac{x^{2}}{2}\)

\(\int x^{2}dx =\frac{x^{3}}{3}\)

\(\int adx =ax\)

\(\int 7dx =7x\)

w = [7(x) - 2(x²/2) + 3(x³/3)] with the limits from 0 to 1

W is the difference between the values of the integral of a given function f(x) for an upper value b and a lower value a of the independent variable x.

Here upper limit is x = 1 and lower limit x = 0

w =[7(1) - 2(1²/2) + 3(1³/3)] - [7(0) - 2(0²/2) + 3(0³/3)].

w = [7 - 1 + 1] - [ 0 ]

w = 7