Question

If $\Delta \left(x\right)=\left|\begin{array}{ccc}1+x+2x^2& x+3& 1\\ x+2x^2& x& 3\\ 3x+6x^2& 3x+11& 9\end{array}\right|$ then ${\int }_0^1\Delta \left(x\right)dx$ is

• A. $\frac{176}{5}$
• B. $-\frac{176}{3}$
• C. $\frac{186}{3}$
• D. $-\frac{192}{3}$

Answer 1

The correct option is B $-\frac{176}{3}$

$R_2\to R_2-R_1,R_3\to R_3-3R_1$
$\Delta \left(x\right)=\left|\begin{array}{ccc}1+x+2x^2& x+3& 1\\ x+2x^2& -3& 2\\ -3& 2& 6\end{array}\right|$
$R_1\to R_1-\frac{R_2}{2}$
$\Delta \left(x\right)=\left|\begin{array}{ccc}1+x+2x^2+\frac{1}{2}& x+3+\frac{3}{2}& 0\\ -1& -3& 2\\ -3& 2& 6\end{array}\right|$
$R_2\to R_2-\frac{R_3}{3}$
$\Delta \left(x\right)=\left|\begin{array}{ccc}\frac{3}{2}+x+2x^2& x+\frac{9}{2}& 0\\ 0& -3-\frac{2}{3}& 0\\ -3& 2& 6\end{array}\right|$
$\Delta \left(x\right)=\left|\begin{array}{ccc}\frac{3}{2}+x+2x^2& x+\frac{9}{2}& 0\\ 0& -\frac{11}{3}& 0\\ -3& 2& 6\end{array}\right|$
$\Delta \left(x\right)=6(-\frac{11}{3}(\frac{3}{2}+x+2x^2))$
${\int }_0^1\Delta \left(x\right)={\left[\frac{-66}{3}[\frac{3}{2}x+\frac{x^2}{2}+\frac{2x^3}{3}]\right]}_0^1$
$=\frac{-66}{3}(\frac{3}{2}+\frac{1}{2}+\frac{2}{3})=-22\left[\frac{9+3+4}{6}\right]=\frac{-176}{3}$