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A force F=a+bx acts on a particle in the...

A force F=a+bx acts on a particle in the x-directioin, where a and b are constants. Find the work done by this force during a displacement form `x=0 to x=d`.

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The correct Answer is:
A, B, D
Given that `F=a+bx`
Where a and b are constants
So, work done by this force during the displacement x=0 to x=d is given by
`W=int_0^d Fdx `
`int_0^d(a+bx)dx`
`=[ax+(bx^2)/2]_0^d`
`=ad+(bd^2)/2`
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