Q.Variation of a force in a certain region is given by $F = 6x^2 - 4x - 8$ (F in newton, x in metre). It displaces an object from x = 1 m to x = 2 m in this region. Calculate the amount of work done.

[!TLDR] $W=\int_1^2(6x^2-4x-8),dx=[2x^3-2x^2-8x]_1^2=(16-8-16)-(2-2-8 …

$W=\int_{1}^{2}F\,dx=\int_1^2(6x^2-4x-8)\,dx=\left[2x^3-2x^2-8x\right]_1^2$. At $x=2$: $2(8)-2(4)-8(2)=16-8-16=-8$. At $x=1$: $2(1)-2(1)-8(1)=2-2-8=-8$. So $W=(-8)-(-8)=0$ J. The work done is exactly zero -- this happens because the force is negative (net opposing) over most/all of this interval, and the specific polynomial integrates to give a net cancellation between $x=1$ and $x=2$; it does NOT mean the force itself was ever zero throughout the motion, only that it …

Directly integrate the given polynomial force with respect to displacement between the stated l …

Arithmetic slips evaluating the antiderivative at the two limits, or forgetting to subtract the lower-limit value from the upper-limit va …