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# 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 + 3x^{2}

**Solution:**

Given: Function f = 7 - 2x + 3x^{2} 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(x^{2}/2) + 3(x^{3}/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(1^{2}/2) + 3(1^{3}/3)] - [7(0) - 2(0^{2}/2) + 3(0^{3}/3)].

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

w = 7

## 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 + 3x^{2}

**Summary:**

Work does by the force f = 7 - 2x + 3x^{2} as the particle moves along the x-axis from x = 0 to x = 1 is 7.

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