What is #f(x) = int (x-4)^3 dx# if #f(3)=-1 #?

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Answer 1 · 2016-03-17T02:51:26

#f(x)=((x-4)^4-5)/4#

To integrate, use substitution, then use the rule

#intu^ndu=u^(n+1)/(n+1)+C#

If we set #u=x-4#, then we have #du=dx# and the simplified integral:

#f(x)=intu^3du#

Applying the rule, this becomes

#f(x)=u^4/4+C#

#f(x)=(x-4)^4/4+C#

Now, we can determine #C# since we know that #f(3)=-1#:

#-1=(3-4)^4/4+C#

#=-1=(-1)^4/4+C#

#=-1=1/4+C#

#-5/4=C#

So,

#f(x)=(x-4)^4/4-5/4=((x-4)^4-5)/4#