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The force applied against a moving object travelling on a linear path is given by #F(x)=3x^2+x #. How much work would it take to move the object over #x in [1,3 ] #?

Answer 1

here,work will be done against the force to move the object,

so,work done #dW =F dx# (as,angle between #F# and #dx# is zero,as moving along a linear pathway)
so, #dW =(3x^2 +x)dx#
or, #dW =3x^2 dx +x dx#
so, #int _0^W = 3 int_1^3 x^2dx + int_1^3 x dx#
so, #W=[x^3]_1^3 + 1/2 [x^2]_1^3=30J#
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Answer 2

#W = 30# J

Simply integrate using the bounds provided: #W = \int_1^3 F(x)dx#
#= \int_1^3 (3x^2+x) dx#
#=3\int_1^3x^2 dx + \int_1^3xdx#
#= (x^3+1/2x^2)_(x=1)^(x=3)#
#=27+9/2-(1+1/2)#
#=30# J
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Answer 3

To find the work done, integrate the force function F(x) over the interval [1,3] with respect to x:

∫[1 to 3] (3x^2 + x) dx

= [x^3 + (1/2)x^2] evaluated from 1 to 3

= [(3^3 + (1/2)(3^2)) - (1^3 + (1/2)(1^2))]

= [(27 + 9/2) - (1 + 1/2)]

= [27 + 9/2 - 1 - 1/2]

= 27 + 9/2 - 1 - 1/2

= 27 + 4.5 - 1 - 0.5

= 30 - 2

= 28 units of work.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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