Question

If $A\left(x\right)=\left|\begin{array}{ccc}x+1& 2x+1& 3x+1\\ 2x+1& 3x+1& x+1\\ 3x+1& x+1& 2x+1\end{array}\right|$
then ${\int }_0^{1}A\left(x\right)dx=$

• A. $-15$
• B. $\frac{-15}{2}$
• C. $-30$
• D. $-5$

Answer 1

Answer: (B)

$\frac{-15}{2}$

$A\left(x\right)=\left|\begin{array}{ccc}x+1& 2x+1& 3x+1\\ 2x+1& 3x+1& x+1\\ 3x+1& x+1& 2x+1\end{array}\right|=(x+1)\left(\right(3x+1\left)\right(2x+1)-(x+1\left)\right(x+1\left)\right)-(2x+1)\left(\right(2x+1\left)\right(2x+1)-(x+1\left)\right(3x+1\left)\right)+(3x+1)\left(\right(x+1\left)\right(2x+1)-(3x+1\left)\right(3x+1\left)\right)A\left(x\right)=-18x^3-9x^2{\int }_0^{1}A\left(x\right)dx=\frac{-18}{4}{x}^4-3x^3=\frac{-18}{4}-3=\frac{-30}{4}=\frac{-15}{2}$