If A(x) = begin{vmatrix} x+1 & 2x+1 & 3x+1 2 | vedclass

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(B) Given $A(x) = \begin{vmatrix} x+1 & 2x+1 & 3x+1 \ 2x+1 & 3x+1 & x+1 \ 3x+1 & x+1 & 2x+1 \end{vmatrix}$.Applying $C_2 ightarrow C_2 - C_1$ and

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$-\frac{15}{2}$

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(B) Given $A(x) = \begin{vmatrix} x+1 & 2x+1 & 3x+1 \ 2x+1 & 3x+1 & x+1 \ 3x+1 & x+1 & 2x+1 \end{vmatrix}$.Applying $C_2 ightarrow C_2 - C_1$ and $C_3 ightarrow C_3 - C_2$:$A(x) = \begin{vmatrix} x+1 & x & x \ 2x+1 & x & -2x \ 3x+1 & -2x & x \end{vmatrix}$.Taking $x$ common from $C_2$ and $C_3$:$A(x) = x^2 \begin{vmatrix} x+1 & 1 & 1 \ 2x+1 & 1 & -2 \ 3x+1 & -2 & 1 \end{vmatrix}$.Applying $C_2 ightarrow C_2 - C_3$:$A(x) = x^2 \begin{vmatrix} x+1 & 0 & 1 \ 2x+1 & 3 & -2 \ 3x+1 & -3 & 1 \end{vmatrix}$.Applying $R_2 ightarrow R_2 + R_3$:$A(x) = x^2 \begin{vmatrix} x+1 & 0 & 1 \ 5x+2 & 0 & -1 \ 3x+1 & -3 & 1 \end{vmatrix}$.Expanding along $C_2$:$A(x) = x^2 \cdot (-(-3)) \cdot \begin{vmatrix} x+1 & 1 \ 5x+2 & -1 \end{vmatrix} = 3x^2 \{(-x-1) - (5x+2)\} = 3x^2(-6x-3) = -18x^3 - 9x^2$.Now, $\int_0^1 A(x) dx = \int_0^1 (-18x^3 - 9x^2) dx = \left[ -\frac{18x^4}{4} - \frac{9x^3}{3} ight]_0^1 = \left[ -\frac{9}{2}x^4 - 3x^3 ight]_0^1 = -\frac{9}{2} - 3 = -\frac{15}{2}$.

If $A(x) = \begin{vmatrix} x+1 & 2x+1 & 3x+1 \\ 2x+1 & 3x+1 & x+1 \\ 3x+1 & x+1 & 2x+1 \end{vmatrix}$, then $\int_0^1 A(x) dx$ is equal to

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