Given f(x)=A.x^2+B, integration of f(x)=7
differentiate on both sides
f`(x)=2*x*A, But f`(1)=2
Therefore f`(1)=1*2*A
2=2A
Hence,A=1
Integrate the function:f(x)=A.x^2+B with limits 0 to 3
Therefore[( A.x^3)/3+Bx] 0 to 3=7
[(x^3)/3+Bx] 0 to 3=7
(27/3+B.3)-(0)=7
9+3B=7
3B=-2
B=-2/3
Therefore A=1 and B=-2/3