f(x)=ax^2+bx+c
c=1(∵f(0)=1)
f'(x)=2ax+b
2a+b=4 ( ∵f'(1)=4 )
∫[0to1]f(x)dx=[ax^3/3 + bx^2/2 + x]{0to1}=a/3 + b/2 + 1 = 1
2a+3b=0
b=-2 , a=3