Question

Let $f\left(x\right)$ be a polynomial of degree $2$ satisfying $f\left(0\right)=1,$ f'(0)=-2$and{f}^{\prime \prime }\left(0\right)=6,$ then ${\int }_{-1}^{2}f\left(x\right)dx$ is equal to

Answer 1

Let $f\left(x\right)={ax}^2+bx+c$
$\Rightarrow f^{\prime }\left(x\right)=2ax+b$ and $f^{\prime \prime }\left(x\right)=2a.$
We are given $f\left(0\right)=c=1,f^{\prime }\left(0\right)=b=-2$ and $f^{\prime \prime }\left(0\right)=2a=6\Rightarrow a=3,b=-2$ and $c=1.$
$\therefore f\left(x\right)={3x}^2-2x+1.$
$\therefore {\int }_{-1}^{2}f\left(x\right)dx={\int }_{-1}^2({3x}^2-2x+1)dx={[x^3-x^2+x]}_{-1}^2$
$=(8-4+2)-(-1-1-1)=6+3=9.$