Question

If $f\left(x\right)$ be a quadratic polynomial such that $f\left(0\right)=2,f^{\prime }\left(0\right)=-3$ and $f^{\prime \prime }\left(0\right)=4$
then ${\int }_{-1}^{1}f\left(x\right)dx$ is equal to

• A. $-3$
• B. $16/3$
• C. $0$
• D. none of these

Answer 1

Answer: (B)

$16/3$

$f\left(x\right)={ax}^2+bx+cf\left(0\right)=2\Rightarrow c=2f^{\prime }\left(x\right)=2ax+b\therefore f^{\prime }\left(0\right)=-3\Rightarrow b=-3f^{\prime \prime }\left(x\right)=2a\therefore f^{\prime \prime }\left(0\right)=4\Rightarrow a=2f\left(x\right)={2x}^2-3x+2{\int }_{-1}^{1}f\left(x\right)dx={\int }_{-1}^1({2x}^2+3x+2)dx$

$={[\frac{{2x}^3}{3}-\frac{{3x}^2}{2}+2x]}_{-1}^1$

$=\frac{2}{3}-\frac{3}{2}+2+\frac{2}{3}+\frac{3}{2}+2=\frac{16}{3}$