Question

If $p\left(x\right)$ is a cubic polynomial with $p\left(1\right)=3,p\left(0\right)=2,p(-1)=4,$ then $\underset{-1}{\overset{1}{\int }}p\left(x\right)dx$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is D $5$

Let $p\left(x\right)=a{x}^3+b{x}^2+cx+d$
$p\left(0\right)=2$
$\Rightarrow d=2$
$p\left(1\right)=3\Rightarrow a+b+c=1...$ (1)
$p(-1)=4\Rightarrow -a+b-c=\mathrm{2....}$ (2)
from (1) and (2)
$2b=3\Rightarrow b=\frac{3}{2}$
$\underset{-1}{\overset{1}{\int }}(a{x}^3+b{x}^2+cx+d)dx=\underset{-1}{\overset{1}{\int }}\left(a{x}^3\right)dx+\underset{-1}{\overset{1}{\int }}\left(b{x}^2\right)dx+\underset{-1}{\overset{1}{\int }}\left(cx\right)dx+\underset{-1}{\overset{1}{\int }}\left(d\right)dx=2\underset{0}{\overset{1}{\int }}(b{x}^2+d)=2[\frac{3}{2}\frac{x^3}{3}{]}_0^1+2[2x{]}_0^1=5$