whats the function off such that f'(x)=x^2 and f(-1)=0

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how do I get the answer to the above question?

Jun 29th, 2015

Thank you for the opportunity to help you with your question!

take integration of 

integrate[f'(x)]=integrate[x^2]

by integrating it we get
f(x) = (x^3)/3 + c
f(x) = (x^3)/3+ c
by putting 

f(-1)=0

we get
0 = [(-1)^3]/3 + c
c=1
so 
f(x)= (x^3)/3 + 1


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Jun 29th, 2015

why does c=1?

Jun 29th, 2015

because of this

 f(-1)=0

 initial value which is given in question 

 

Jun 29th, 2015

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