A one-dimensional conservative force is given by the function F (x) = (2.00 N/m) x + (1.00 N/m3) x3. What is the change in potential energy of the object as it moves from x = 1.00 m to x = 2.00 m? Enter your answer in Joules.

Given that

F (x) = (2.00 N/m) x + (1.00 N/m3) x3.

Which can be written as

F (x) = 2x + x³ N/m

Potential energy from x=1 to x=2

P. E is given as

P. E=∫F (x) dx

P. E=∫2x+x³ dx. From x=1 to x=2

P. E = 2x²/2 + x⁴/4 from x=1 to x=2

P. E = x² + x⁴/4 from x=1 to x=2

P. E = (2²+2⁴/4) - (1² + 1⁴/4)

P. E = (4+4) - (1+1/4)

P. E, = 8-5/4

P. E=27/4

P. E=6.75J

Since it is moving from lower position to upper position, then P. E should be negative

P. E = - 6.75J