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WILL BECOME A FAN AND GIVE A MEDAL If f'(x)= 3x^2 and f(-1)=2, then integral from 0 to 2 of f(x) dx= (A) 8/3 (B) 4 (C) 7 (D) 10 (E) 28 Could you please show your work as well as select an answer? I'd really appreciate it!
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f'(x)=3x^2 f(x)=integral 3x^2dx=3x^3 /3 +c=x^3+c f(-1)=(-1)^3+c 2=-1+c c=3 f(x)=x^3 +3 again f'(x)=3x^2 f(x)=integral 0 to 2 3x^2 dx=[3x^3 /3 ] ,0 to2 =[x^3],0 to2 =[(2)^3-0^3]=8
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