Given force as a function of position

$F(x) = 3x^2 + 2x - 5$

The body moves from:

$x=2\text{ m to }x=4\text{ m}$

Formula for work done by a variable force

Work done is the integral of force with respect to displacement:

$W = \int_{2}^{4} F(x)\,dx = \int_{2}^{4} (3x^2+2x-5)\,dx$

Integrate the function

$\int (3x^2+2x-5)\,dx = x^3 + x^2 - 5x$

Apply the limits

$W = \left[x^3 + x^2 - 5x\right]_{2}^{4}$

At $x=4$:

$4^3 + 4^2 - 5(4) = 64 + 16 - 20 = 60$

At $x=2$:

$2^3 + 2^2 - 5(2) = 8 + 4 - 10 = 2$

Therefore,

$W = 60 - 2 = 58\text{ J}$

Final answer

$\boxed{58}$

Comparison with stored correct answer

Stored correct answer = $58$

My derived answer also = $58$, so they agree.