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The force applied against a moving object travelling on a linear path is given by #F(x)= x^2+e^x #. How much work would it take to move the object over #x in [2, 5] #?

Answer 1

Adrian Ayala

The work is #=180.02J#

#inte^xdx=e^x+C#

#int(x^n)dx=x^(n+1)/(n+1)+C(n!=-1)#

The work is

#DeltaW=F(x)*Deltax#

#F(x)=x^2+e^x#

#W=int_2^(5)(x^2+e^x)dx#

#=[1/3x^3+e^x]_2^(5)#

#=(1/3*5^3+e^5)-(1/3*2^3+e^2)#

#=125/3-8/3+e^5-e^2#

#=117/3+141.02#

#=180.02J#

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Answer 2

Olivia Askew

To calculate the work done, integrate the force function F(x) over the given interval [2, 5]:

[ \text{Work} = \int_{2}^{5} (x^2 + e^x) ,dx ]

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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.