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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