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The force in Newtons acting on a particle located x meters from the origin is given by the function F(x)=12x2+7. What is the work done in Joules when moving the particle from x=2 to x=5?

Sagot :

loneguy

[tex]W = \displaystyle \int^{5}_{2} F(x) ~ dx\\\\\\= \displaystyle \int^{5}_{2} \left(12x^2 +7\right) ~dx\\\\\\= 12\displaystyle \int^{5}_{2} x^2 ~dx + \displaystyle \int^{5}_{2} 7~ dx\\\\\\=12 \left[\dfrac{x^3}3 \right]^{5}_{2} + 7\left[x\right]^{5}_{2}\\\\\\=4(5^3-2^3) + 7(5-2) = 468 + 21 = 489~ J[/tex]

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