Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is D $5$

Let $p\left(x\right)=a{x}^3+b{x}^2+cx+d$

$p\left(0\right)=2$

$d=2$

$p\left(1\right)=2$

$a+b+c+2=3$

$a+b+c=1....\left(1\right)$

$p(-1)=4$

$-a+b-c+2=4$

$-a+b-c=2....\left(2\right)$

from $\left(1\right)$ and $\left(2\right)$

$2b=3\Rightarrow b=\frac{3}{2}$

$P\left(x\right)=a{x}^3+\frac{3}{2}{x}^2+cx+2$

${\int }_{-1}^{1}p\left(x\right)dx={\int }_{-1}^1(a{x}^2+\frac{3}{2}{x}^2+cx+2)dx=5$